1.6: Skeletal System
Learning Objectives
- Examine the anatomy of the skeletal system
- Determine the main functions of the skeletal system
- Differentiate the medical terms of the skeletal system
- Discover common diseases, disorders, and procedures related to the skeletal system
- Recognize common medical specialties associated with the skeletal system
Skeletal System Word Parts
Click on prefixes, combining forms, and suffixes to reveal a list of word parts to memorize for the Skeletal System.
Introduction to the Skeletal System
The skeletal system forms the framework of the body. It is the body system composed of bones, cartilage, and ligaments. Each bone serves a particular function and varies in size, shape, and strength. Bones are weight-bearing structures in your body and can therefore change in thickness as you gain or lose weight. The skeletal system performs the following critical functions for the human body:
- supports the body
- facilitates movement
- protects internal organs
- produces blood cells
- stores and releases minerals and fat
Watch this video:
Media 6.1. The Skeletal System: Crash Course A&P #19 [Online video]. Copyright 2015 by CrashCourse .
Practice Medical Terms Related to the Skeletal System
Anatomy (Structures) of the Skeletal System
The skeletal system includes all of the bones, cartilages, and ligaments of the body that support and give shape to the body and body structures. The skeleton consists of the bones of the body. For adults, there are 206 bones in the skeleton. Younger individuals have higher numbers of bones because some bones fuse together during childhood and adolescence to form an adult bone. The primary functions of the skeleton are to provide a rigid, internal structure that can support the weight of the body against the force of gravity, and to provide a structure upon which muscles can act to produce movements of the body.
In addition to providing for support and movements of the body, the skeleton has protective and storage functions. It protects the internal organs, including the brain, spinal cord, heart, lungs, and pelvic organs. The bones of the skeleton serve as the primary storage site for important minerals such as calcium and phosphate. The bone marrow found within bones stores fat and houses the blood-cell-producing tissue of the body.
The skeleton is subdivided into two major divisions: the axial and appendicular .
The Axial Skeleton
The axial skeleton forms the vertical, central axis of the body and includes all bones of the head, neck, chest, and back (see Figure 6.1 ). It serves to protect the brain, spinal cord, heart, and lungs. It also serves as the attachment site for muscles that move the head, neck, and back, and for muscles that act across the shoulder and hip joints to move their corresponding limbs.
The axial skeleton of the adult consists of 80 bones, including the skull , the vertebral column , and the thoracic cage . The skull is formed by 22 bones. Also associated with the head are an additional seven bones, including the hyoid bone and the ear ossicles (three small bones found in each middle ear). The vertebral column consists of 24 bones, each called a vertebra , plus the sacrum and coccyx . The thoracic cage includes the 12 pairs of ribs and the sternum , the flattened bone of the anterior chest.
The cranium or skull supports the face and protects the brain. It is subdivided into the bones of the skull and the bones of the face.
Bones of the Skull
- Frontal – forms the forehead
- Parietal – the upper lateral sides of the cranium
- Occipital – the posterior skull and base of the cranial cavity
- Temporal – the lower lateral sides of the cranium
- Sphenoid -the ‘keystone’ bone that forms part of the base of the skull and eye sockets
- Ethmoid – forms part of the nose and orbit and base of the cranium
- Auditory ossicles – the small bones of the middle ear
- External auditory meatus – the external opening of the ear and temporal bone
Bones of the Face
- Zygomatic – the cheekbone
- Maxillary – the upper jaw and hard palate
- Palatine – the lateral walls of the nose
- Lacrimal – the walls of the orbit
- Inferior conchae – the lower lateral wall of the nasal cavity
- Vomer – the bone that separates the left and right nasal cavity
- Mandible – the lower jaw bone (the only movable bone of the skull)
- Hyoid – the bone located between the mandible and larynx, not connected to other bones
Did you know?
The axial skeleton has 80 bones and includes bones of the skull (and face), vertebral column, and thoracic cage.
Bones of the Vertebral Column
The vertebral column is also known as the spinal column or spine (see Figure 6.2 ). It consists of a sequence of vertebrae (singular = vertebra), each of which is separated and united by an intervertebral disc . Together, the vertebrae and intervertebral discs form the vertebral column. It is a flexible column that supports the head, neck, and body and allows for their movements. It also protects the spinal cord, which passes down the back through openings in the vertebrae.
Types of Vertebrae
- Cervical – C1 to C7, the first 7 vertebrae in the neck region
- Thoracic – T1 to T12, the next 12 vertebrae that form the outward curvature of the spine
- Lumbar – L1 to L5, the next 5 vertebrae that form the inner curvature of the spine
- Sacrum – the triangular-shaped bone at the base of the spine
- Coccyx – the tailbone
Bones of the Thoracic Cavity
The thoracic cage (rib cage) forms the thorax (chest) portion of the body. It consists of the 12 pairs of ribs with their costal cartilages and the sternum (see Figure 6.3 ). The ribs are anchored posteriorly to the 12 thoracic vertebrae (T1–T12). The thoracic cage protects the heart and lungs.
Ribs
There are 12 sets of ribs and can be divided as such:
- 7 true ribs – they are attached to the front of the sternum
- 3 false ribs – they are attached to the cartilage that joins the sternum
- 2 floating ribs – they are not attached to the front of the sternum
Sternum
The sternum , also known as the breast bone, is divided into 3 parts:
- manubrium – the upper portion of the breast bone
- body – the middle portion of the breast bone
- xiphoid process – the lower portion of the breast bone and is made up of cartilage
Concept Check
- What is the medical term for the upper jaw bone and the lower jaw bone?
- What medical term is used for the bones of the inner ear?
- How many bones make up the cervical region of the vertebral column?
The Appendicular Skeleton
The appendicular skeleton includes all bones of the upper and lower limbs, plus the bones that attach each limb to the axial skeleton. There are 126 bones in the appendicular skeleton of an adult.
Bones of the Pectoral Girdle
- Scapula – the shoulder blades
- Clavicle – the collarbone, which connects the sternum to the scapula
- Acromion – the extension that forms the bony point of the shoulder
Bones of the Upper Limbs
The bones of the upper limbs include the bones of the arms, wrists, and hands.
Bones of the Arm
- Humerus – the bone in the upper arm
- Radius – the bone that runs thumb-side of the forearm
- Ulna – the bone that runs on the side of the little finger of the forearm ( Figure 6.4 )
Bones of the Wrist and Hand
- Carpals – the wrist bones
- Metacarpals – the bones in the palm
- Phalanges – the finger and toe bones
Each phalanx has three bones: the distal, medial, and proximal. The exception is the thumb and big toe which has two bones: the distal and proximal ( Figure 6.5 ). There are 30 bones in each upper limb. Can you count them on your limb?
Did you know?
The appendicular skeleton has 126 bones. It is divided into the bones of the u pper limbs and l ower limbs that attach each limb to the skeleton.
Bones of the Pelvic Region
The bones of the pelvic region protect the reproductive, urinary, and excretory organs.
- Pelvic girdle – the hip or coxal bone; it is formed by the fusion of three bones during adolescence
- Illium – the largest part of the hip bone
- Ischium – the lower portion of pelvic girdle
- Pubis – the anterior portion of pelvic girdle
- Pelvis – consists of four bones: the left and right hip bones as well as the sacrum and coccyx
- Acetabulum – the large socket in the pelvic bones that holds the head of the femur
The shape of the pelvic girdle is different for males than females. In the male, it is a funnel shape. In the female, it is shaped like a basin to accommodate the fetus during pregnancy.
Bones of the Lower Limbs
The bones of the lower limb include bones of the leg and the feet.
Bones of the Leg
- Femur – the thigh bone and is also referred to as the upper leg bone; it is the longest and strongest bone in the human body
- Patella – the kneecap
- Tibia – the shin bone; it is a medial bone and the main weight-bearing bone of the lower leg
- Fibula – the smaller of the lower leg bones (see Figure 6.6 )
Bones of the Ankles and Feet
- Tarsals – the ankle bones (7 total)
- Malleolus – the bony protrusions of the ankle bones
- Talus – the superior ankle bones
- Calcaneus – the heel bones
- Metatarsals – the foot bones
- Phalanges – the bones of the toes (see Figure 6.7 )
Did you know?
The femur is the longest and strongest bone of the body and accounts for approximately one-quarter of a person’s total height.
Concept Check
Answer the following questions:
- Is the humerus the same as the funny bone?
- What is the medical term for the kneecap?
Anatomy Labeling Activity
Physiology (Function) of the Skeletal System
The bones of the skeletal system are comprised of an inner spongy tissue referred to as bone marrow. There are two types of bone marrow: red and yellow. The red bone marrow produces the red blood cells, and it does so by a process called hematopoiesis . The yellow bone marrow contains adipose tissues which can be a source of energy. The bones of the skeletal system also store minerals such as calcium and phosphate. These minerals are important for the physiological processes in the body and are released into the bloodstream when levels are low in the body.
Joints
Watch this video:
Media 6.2. Joints: Crash Course A&P #20 [Online video]. Copyright 2015 by CrashCourse .
Most bones connect to at least one other bone in the body. The area where bones meet bones or where bones meet cartilage are called articulations . Joints can be classified based on their ability to move. At movable joints, the articulating surfaces of the adjacent bones can move smoothly against each other. However, other joints may be connected by connective tissue or cartilage. These joints are designed for stability and provide for little or no movement. Importantly, joint stability and movement are related to each other. This means that stable joints allow for little or no mobility between the adjacent bones. Conversely, joints that provide the most movement between bones are the least stable.
Based on the function of joints , there are 3 types of joints:
-
Synarthrosis
joints allow no movement.
- For example, joints of the skull
-
Amphiarthrosis
joints allow some movement.
- For example, joints of the pubic symphysis
-
Diarthrosis
joints allow for free movement.
- For example, joints of the knee
Structures associated with joints are:
- Cartilage – the elastic connective tissue that is found at the ends of bones, nose tip, et cetera
- Synovial membrane – the lining or covering of synovial joints
- Synovial fluid – the lubricating fluid found between synovial joints
- Ligaments – the tough, elastic connective tissue that connects bone to bone
- Tendons – the fibrous connective tissue that attaches muscle to bone
- Bursa – the closed, fluid-filled sacs that work as a cushion
- Meniscus – C-shaped cartilage that acts as shock absorbers between bones
Did you know?
The left and right hip bones are connected by an amphiarthrosis joint.
Body Movements
Synovial joints are movable joints and provide most of the body movements. Body movement occurs when the bones, joints, and muscles work together.
Flexion and Extension
Flexion and extension are movements that take place within the sagittal plane and involve anterior or posterior movements of the body or limbs. For the vertebral column, flexion (anterior flexion) is an anterior (forward) bending of the neck or body, while extension involves a posterior-directed motion, such as straightening from a flexed position or bending backward. Lateral flexion is the bending of the neck or body toward the right or left side. These movements of the vertebral column involve both the joints as well as the associated intervertebral disc.
In the limbs, flexion decreases the angle between the bones (bending of the joint), while extension increases the angle and straightens the joint (see Figures 6.8(a-d) ). You will discover in the muscular system chapter that the associated muscles to these movements are flexor and extensor.
Abduction and Adduction
Abduction and adduction motions occur within the coronal plane and involve medial-lateral motions of the limbs, fingers, toes, or thumb. For example, abduction is raising the arm at the shoulder joint, moving it laterally away from the body, while adduction brings the arm down to the side of the body (see Figure 6.8(e) ). In the muscular system chapter, you will discover that the associated muscles to these movements are the abductor and adductor.
Circumduction
Circumduction is the movement of a body region in a circular manner, in which one end of the body region being moved stays relatively stationary while the other end describes a circle. It involves the sequential combination of flexion, adduction, extension, and abduction at a joint (see Figure 6.8(e) ).
Rotation
Rotation can occur within the vertebral column, at a pivot joint , or at a ball-and-socket joint . Rotation of the neck or body is the twisting movement produced by the summation of the small rotational movements available between adjacent vertebrae. At a pivot joint, one bone rotates in relation to another bone.
Rotation can also occur at the ball-and-socket joints of the shoulder and hip. Here, the humerus and femur rotate around their long axis, which moves the anterior surface of the arm or thigh either toward or away from the midline of the body (see Figure 6.8(f) ).
Supination and Pronation
Supination and pronation are movements of the forearm. In the anatomical position, the upper limb is held next to the body with the palm facing forward. This is the supinated position of the forearm. In this position, the radius and ulna are parallel to each other. When the palm faces backward, the forearm is in the pronated position , and the radius and ulna form an X-shape.
Pronation is the movement that allows the palm to face backward while in supination the palm faces forward. It helps to remember that supination is the motion you use when scooping up soup with a spoon (see Figure 6.9(g) ).
Dorsiflexion and Plantar Flexion
Dorsiflexion and plantar flexion are movements at the ankle joint, which is a hinge joint. Lifting the front of the foot, so that the top of the foot moves (upward) toward the anterior leg is dorsiflexion, while lifting the heel of the foot from the ground or pointing the toes downward is plantar flexion. These are the only movements available at the ankle joint (see Figure 6.9(h) ).
Inversion and Eversion
Inversion and eversion are complex movements that involve the multiple plane joints among the tarsal bones of the posterior foot (intertarsal joints) and thus are not motions that take place at the ankle joint. Inversion is the turning of the foot to angle the bottom of the foot toward the midline, while eversion turns the bottom of the foot away from the midline. The foot has a greater range of inversion than eversion motion. These are important motions that help to stabilize the foot when walking or running on an uneven surface and aid in the quick side-to-side changes in direction used during active sports such as basketball, racquetball, or soccer (see Figure 6.9(i) ).
Protraction and Retraction
Protraction and retraction are anterior-posterior movements of the scapula or mandible. Protraction of the scapula occurs when the shoulder is moved forward, as when pushing against something or throwing a ball. Retraction is the opposite motion, with the scapula being pulled posteriorly and medially, toward the vertebral column. For the mandible, protraction occurs when the lower jaw is pushed forward, to stick out the chin, while retraction pulls the lower jaw backward (see Figure 6.9(j) ).
Depression and Elevation
Depression and elevation are downward and upward movements of the scapula or mandible. The upward movement of the scapula and shoulder is elevation, while a downward movement is depression. These movements are used to shrug your shoulders. Similarly, elevation of the mandible is the upward movement of the lower jaw used to close the mouth or bite on something, and depression is the downward movement that produces the opening of the mouth (see Figure 6.9(k) ).
Concept Check
- Discuss the joints involved and movements required for you to cross your arms together in front of your chest.
- Differentiate between pronation and supination .
Practice Skeletal System Movement Terms
Medical Terms in Context
Diseases and Disorders of the Skeletal System
Osteoporosis
The National Institute of Health’s Osteoporosis and Related Bone Diseases National Resource Center describes osteoporosis as bone loss that causes bones to become weak and thin over time. This weakness can lead to fractures from simple movements and occur often in the wrist, shoulder, spine, and hip (National Institute of Arthritis and Musculoskeletal and Skin Diseases, n.d.-b). To learn more, please visit the National Institute of Health’s web page on osteoporosis .
Arthritis
Arthritis often presents as edema , arthralgia , and ankylosis (National Institute of Arthritis and Musculoskeletal and Skin Diseases, n.d.-a). Common types of arthritis are osteoarthritis (OA), rheumatoid arthritis (RA), Gout and lupus. To learn more about arthritis visit this web page from the National Institute of Arthritis and Musculoskeletal and Skin Diseases .
Osteoarthritis
Osteoarthritis is the most common form of arthritis and according to the Centers for Disease Control and Prevention (CDC), affects over 32.5 million adults in the United States. The breakdown of cartilage and bone occurs over time when joints are exposed to heavy workloads either through occupation, obesity, and/or prior injury to a joint. Common signs and symptoms are pain, stiffness, and aching that worsens over time. While there is no cure, symptoms can be managed through exercise, medications, and in severe cases, joint replacements (Centers for Disease Control and Prevention, n.d.-a).
Rheumatoid Arthritis
The CDC describes rheumatoid arthritis (RA) as an autoimmune and inflammatory disease. Autoimmune diseases are disorders in which the immune system overreacts and begins to attack itself. In the case of RA, inflammation of the joint tissues of the hands, wrists, and knees is painful and debilitating. Treatments may include immunosuppressive drugs and anti-inflammatory drugs (Betts et al., 2013). RA can also affect other tissues throughout the body and cause problems in organs such as the lungs, heart, and eyes. RA can affect children; in this case, it is referred to as juvenile rheumatoid arthritis (Centers for Disease Control and Prevention, n.d.-b).
Gout
Gout is an inflammatory arthritis caused by the buildup of uric acid crystals in a joint. Gout has periods of flares and remission and is commonly treated through lifestyle changes and medication. While any joint can be affected, it is common in the lower extremities and most often in the big toe (Centers for Disease Control and Prevention, n.d.-c). To learn more about the causes and treatments please visit the Arthritis Foundation’s web page about gout .
Myasthenia Gravis
The National Institute of Neurological Disorders and Strokes describes myasthenia gravis as a “ chronic autoimmune neuromuscular disease that causes weakness in the skeletal muscles” (Office of Communications and Public Liaison, 2020). To learn more, read the National Institute of Neurological Disorders and Stroke’s myasthenia gravis fact sheet.
Fibromyalgia
Fibromyalgia is a challenging disease to diagnose since symptoms manifest differently and are similar to other diseases. Signs and symptoms may include widespread pain, chronic fatigue, gastrointestinal problems, and headaches. It is not known what causes fibromyalgia. A doctor may need to order tests to rule out other conditions before making a diagnosis of fibromyalgia (National Institute of Arthritis and Musculoskeletal and Skin Diseases, n.d.-c). To learn more about the diagnosis and treatment for fibromyalgia, please read this handout from the National Institute of Arthritis and Musculoskeletal and Skin Diseases (pdf).
Osteomyelitis
Osteomyelitis is a bone infection caused when staphylococcus bacteria travel through the bloodstream from an infection in one part of the body to the bone. Staphylococcus bacteria are found on the skin, and they can transfer to the bone through a wound and/or surgical contamination. The risk increases as people age or if their immune system is compromised (Momodu & Savaliya, 2021). To learn more, visit the Mayo Clinic’s web page on osteomyelitis.
Disorders of the Curvature of the Spine
Developmental anomalies, pathological changes, or obesity can enhance the normal vertebral column curves, resulting in the development of abnormal or excessive curvatures (see Figure 6.10 ). Disorders associated with the curvature of the spine include:
- Kyphosis: Also referred to as humpback, it is an excessive posterior curvature of the thoracic region. This can develop when osteoporosis causes weakening and erosion of the anterior portions of the upper thoracic vertebrae, resulting in their gradual collapse (see Figure 6.11 ).
- Lordosis: Also referred to as swayback, it is an excessive anterior curvature of the lumbar region and is most commonly associated with obesity or late pregnancy. The accumulation of body weight in the abdominal region results in an anterior shift in the line of gravity that carries the weight of the body. This causes an anterior tilt of the pelvis and a pronounced enhancement of the lumbar curve.
- Scoliosis: An abnormal, lateral curvature, accompanied by twisting of the vertebral column. Scoliosis is the most common vertebral abnormality among girls. The cause is usually unknown, but it may result from weakness of the back muscles, defects such as differential growth rates in the right and left sides of the vertebral column, or differences in the length of the lower limbs. When present, scoliosis tends to get worse during adolescent growth spurts. Although most individuals do not require treatment, a back brace may be recommended for growing children. In extreme cases, surgery may be required.
Fractures
A fracture is a broken bone. It will heal whether or not a physician resets it in its anatomical position. If the bone is not reset correctly, the healing process will keep the bone in its deformed position. Crepitation or crepitus is the creaking or popping sound that is heard when fractured bones move against each other. Fractures are classified by their complexity, location, and other features (see Figure 6.12 ). Some fractures may be described using more than one term because they may have the features of more than one type (e.g., an open transverse fracture).
Types of fractures include:
- Closed or simple – bones are broken but do not protrude the skin
- Open or compound – bones are broken and pierce through the skin
- Transverse – bone is broken straight across
- Spiral – the bone has twisted apart
- Comminuted – bones are broken and crushed into pieces
- Greenstick – bones are partially broken; occurs mainly in children
- Oblique – bones are broken at an angle
- Coles – bones are broken at the wrist or distal radius
- Stress – a small crack in the bone
Bone Cancer
There are three types of primary bone cancers: osteosarcoma, Ewing Sarcomas, and chondrosarcoma. These are considered primary cancers because they originate in the bones. Osteosarcoma and Ewing Sarcomas primarily affect children, teenagers, and young adults. Chondrosarcoma primarily affects older adults (National Cancer Institute, n.d.-a). To learn more, visit the American Cancer Society’s web page on bone cancer.
Medical Specialties and Procedures Related to the Skeletal System
Orthopedic Surgeon
Orthopedic surgeons are medical doctors who have specialized training in the prevention, diagnosis, treatment, and surgery of disorders and diseases related to the musculoskeletal systems (Bureau of Labor Statistics, 2021a). For more details, please visit the American College of Surgeons’ page on Orthopedic Surgery .
Rheumatologist
Rheumatologists are medical doctors who specialize in the diagnosis and treatment of disorders of the joints, muscles, and bones. They diagnose and treat diseases such as arthritis, musculoskeletal disorders, osteoporosis, plus autoimmune diseases like ankylosing spondylitis, a chronic spinal inflammatory disease, and rheumatoid arthritis (Fowler et al., 2013). For more details, please follow the link to the American College of Rheumatology’s page on rheumatology .
Doctor of Chiropractic (DC)/Chiropractor
Chiropractors are required to have a Doctor of Chiropractic (D.C.) degree, which is a 4-year postgraduate professional degree, and a state license. Chiropractors focus on spinal adjustments, nutrition, and preventing injury without the use of pharmaceuticals or surgical procedures (Bureau of Labor Statistics, 2021b). To learn more, visit the Bureau of Labor Statistics’ website .
Physical Therapist
A physical therapist is a licensed professional who develops individualized treatment plans for their clients. These plans can include exercises, hands-on therapy, and equipment, such as canes or wheelchairs. Although current licensure laws require that those entering the field have a doctor of physical therapy degree, physical therapists who began working before those requirements went into effect may have a bachelor’s or master’s degree (Bureau of Labor Statistics, 2021c). To learn more, please visit the American Physical Therapy Association website .
Diagnostic Procedures
Common diagnostic procedures related specifically to the skeletal system include x-rays, bone mineral density testing, and arthroscopy.
- X-rays are common diagnostic tests used to confirm or rule out fractures and broken bones. The radiation dose is low so it is considered a safe diagnostic test (MedlinePlus, 2021).
- Dual x-ray absorptiometry (BMD) , also called a bone mineral density test, is a test to determine osteoporosis by measuring the amount of bone mineral in a particular amount of bone (National Cancer Institute, n.d.-b).
- Arthroscopy is a procedure that involves a small incision and the insertion into the joint of an arthroscope, a pencil-thin instrument that allows for visualization of the joint interior. Small surgical instruments are also inserted via additional incisions. These tools allow a surgeon to remove or repair a torn meniscus or to reconstruct a ruptured cruciate ligament.
Practice Terms Related to the Skeletal System
Skeletal System Vocabulary
Abduction : Moving the limb or hand laterally away from the body, or spreading the fingers or toes.
Adduction : Movement that brings the limb or hand toward or across the midline of the body, or brings the fingers or toes together.
Amphiarthrosis : A slightly mobile joint.
Ankylosis: Fixation and immobility of a joint.
Appendicular skeleton: All bones of the upper and lower limbs, plus the girdle bones that attach each limb to the axial skeleton.
Arthralgia: Joint pain.
Arthritis: Chronic inflammation of the synovial joints.
Arthrocentesis : Surgical puncture to aspirate fluid from a joint.
Arthrodesis : Surgical fixation of a joint.
Arthrography : Process of recording a joint.
Arthroplasty : Joint replacement surgery.
Arthroscopy : Process of viewing a joint using an endoscope.
Articulations : Where two bone surfaces meet.
Autoimmune diseases/disorders : Disorders in which the immune system overreacts and begins to attack itself.
Axial skeleton: The central, vertical axis of the body, including the skull, vertebral column, and thoracic cage.
Bradykinesia : Condition of slow movement.
Bursitis : Inflammation of a bursa near a joint.
Chondromalacia : Degeneration of cartilage.
Chronic: A condition that lasts a long time with periods of remission and exacerbation.
Craniotomy : An operation in which a piece of the skull is removed.
Diarthrosis : Freely mobile joints.
Diskectomy : Excision of the intervertebral disk.
Discitis : Inflammation of the intervertebral disk.
Dyskinesia : Abnormal involuntary movements of the extremities, trunk, or jaw.
Edema: Swelling due to excessive liquid in the tissues.
Eversion : Foot movement in which the bottom of the foot is turned laterally, away from the midline.
Extension : Movement in the sagittal plane that increases the angle of a joint (straightens the joint).
Flexion : Movement in the sagittal plane that decreases the angle of a joint (bends the joint).
Hematopoiesis : The production of blood cells.
Hyperkinesia : Excessive movement of muscles of the body as a whole.
Hypertrophy : The enlargement of muscles.
Inversion : Foot movement in which the bottom of the foot is turned toward the midline.
Kyphosis : An excessive posterior curvature of the thoracic region; also called humpback.
Lordosis : Excessive anterior curvature of the lumbar vertebral column region; also called swayback.
Lumbar : Pertaining to the lumbar region of the spine (L1 to L5).
Lumbosacral : Pertaining to the region of the back that includes the lumbar vertebrae, sacrum, and nearby structures.
Muscular dystrophy : A general term for the group of inherited myopathies that are characterized by wasting and weakness of the skeletal muscle.
Osteitis : Inflammation of bone.
Osteoarthritis: The most common type of arthritis; associated with aging and “wear and tear” of the articular cartilage.
Osteoblast : The cell responsible for forming new bone.
Osteochondritis : Inflammation of bone and cartilage.
Osteocyte : Bone cell.
Osteomalacia : A softening of adult bones due to Vitamin D deficiency.
Osteomyelitis : Inflammation of bone and bone marrow.
Osteonecrosis : Abnormal condition of bone death (lack of blood supply).
Osteopenia : Abnormally low bone mass or bone mineral density.
Osteopetrosis : Abnormal condition of porous bones.
Osteoporosis: A disease characterized by a decrease in bone mass that occurs when the rate of bone resorption exceeds the rate of bone formation.
Osteosarcoma : Malignant tumor of bone.
Pelvic : Pertaining to the pelvis.
Pronation : Forearm motion that moves the palm of the hand from the palm forward to the palm backward position.
Rotation : Movement of a bone around a central axis or around its long axis.
Sarcopenia : Age-related muscle atrophy.
Scoliosis : Lateral curvature of the spine.
Spondyloarthritis : Inflammation of the joints of the spine.
Spondylosis : A degenerative spinal disease that can involve any part of the vertebra, intervertebral disk, and surrounding soft tissue.
Supination : Forearm motion that moves the palm of the hand from the palm backward to the palm forward position.
Synarthrosis : An immobile or nearly immobile joint.
Synovectomy : Excision of the synovial membrane.
Synovial sarcoma : Malignant tumor of the synovial membrane.
Tendinitis : Inflammation of the tendon.
Tenosynovitis : Inflammation of the synovial membrane of a tendon.
Vertebroplasty : A procedure used to repair a bone in the spine that has a break caused by cancer, osteoporosis, or trauma.
Test Yourself
References
Bureau of Labor Statistics. (2021a). Physicians and surgeons. In Occupational outlook handbook . U.S. Department of Labor. https://www.bls.gov/ooh/healthcare/physicians-and-surgeons.htm
Bureau of Labor Statistics. (2021b). Chiropractors. In Occupational outlook handbook . U.S. Department of Labor. https://www.bls.gov/ooh/healthcare/chiropractors.htm
Bureau of Labor Statistics. (2021c). Physical therapists. In Occupational outlook handbook . U.S. Department of Labor. https://www.bls.gov/ooh/healthcare/physical-therapists.htm
Centers for Disease Control and Prevention. (n.d.-a). Osteoarthritis . https://www.cdc.gov/arthritis/basics/osteoarthritis.htm
Centers for Disease Control and Prevention. (n.d.-b). Rheumatoid arthritis. CDC Arthritis Program. https://www.cdc.gov/arthritis/basics/rheumatoid-arthritis.html
Centers for Disease Control and Prevention. (n.d.-c). Gout . https://www.cdc.gov/arthritis/basics/gout.html
Fowler, S., Roush, R., & Wise, J. (2013). Concepts of biology . OpenStax. Access for free at https://openstax.org/books/concepts-biology/pages/1-introduction
MedlinePlus. (2021). X-rays . U.S. National Library of Medicine. https://medlineplus.gov/xrays.html
Momodu, I. I., & Savaliya, V. (2021). Osteomyelitis. In StatPearls [Internet] . https://www.ncbi.nlm.nih.gov/books/NBK532250/
National Cancer Institute. (n.d.-a). Primary bone cancer [Fact sheet]. National Institutes of Health, U.S. Department of Health and Human Services. https://www.cancer.gov/types/bone/bone-fact-sheet
National Cancer Institute. (n.d.-b). Dual x-ray absorptiometry . National Institutes of Health, U.S. Department of Health and Human Services. https://www.cancer.gov/publications/dictionaries/cancer-terms/def/dual-x-ray-absorptiometry
National Institute of Arthritis and Musculoskeletal and Skin Diseases. (n.d.-a). Arthritis . National Institutes of Health, U.S. Department of Health and Human Services. https://www.niams.nih.gov/health-topics/arthritis
National Institute of Arthritis and Musculoskeletal and Skin Diseases. (n.d.-b). Osteoporosis overview . National Institutes of Health, U.S. Department of Health and Human Services. https://www.bones.nih.gov/health-info/bone/osteoporosis/overview
National Institutes of Arthritis and Musculoskeletal and Skin Diseases. (n.d-c). Health topics: Fibromyalgia [PDF]. National Institutes of Health, U.S. Department of Health and Human Services. https://www.niams.nih.gov/print/view/pdf/advanced_reading_pdf_/advanced?view_args%5B0%5D=1957
Office of Communications and Public Liaison. (2020). Myasthenia gravis fact sheet. National Institute of Neurological Disorders and Stroke, U.S. Department of Health and Human Services. https://www.ninds.nih.gov/Disorders/Patient-Caregiver-Education/Fact-Sheets/Myasthenia-Gravis-Fact-Sheet
Image Descriptions
Figure 6.1 image description: This diagram shows the human skeleton and identifies the major bones. The left panel shows the anterior view (from the front) and the right panel shows the posterior view (from the back). Labels read (from the top of the skull): skull (cranial portion, facial portion), pectoral shoulder girdle, clavicle, scapula, thoracic cage (sternum, ribs), upper limb (humerus, ulna, radius, carpals, metacarpals, phalanges), vertebral column, pelvic girdle (hip bones), lower limb (femur, patella, tibia, fibula, tarsals, metatarsals, phalanges). [Return to Figure 6.1].
Figure 6.2 image description: This image shows the structure of the vertebral column. The left panel shows the front view of the vertebral column. Labels and the right panel show the side view of the vertebral column. labels read (from top): 7 cervical vertebrae (C1-C7) form cervical curve, 12 thoracic vertebrae (T1-T12) form the thoracic curve, intervertebral disc, 5 lumbar vertebrae (L1-L5) form lumbar curve, Fused vertebrae of sacrum and coccyx form a sacrococcygeal curve, sacrum, coccyx. [Return to Figure 6.2].
Figure 6.3 image description: This figure shows the skeletal structure of the rib cage. The left panel shows the anterior view of the sternum. Labels read (from top): clavicular notch, jugular notch, manubrium, sternal angle, body, xiphoid process. The right panel shows the anterior panel of the sternum including the entire rib cage. Labels read (from top): jugular notch, clavicular notch, clavicle, sternum (manubrium, body, xiphoid process), scapula, sternal angle, costal cartilages, intercostal space. Ribs are numbered 1-12 from the top. [Return to Figure 6.3].
Figure 6.4 image description: This diagram labels the bones of the lower arm (excluding the hands). Labels read (from top): olecranon process, head of radius, radial notch of the ulna, trochlear notch, coronoid process, radial tuberosity, proximal radioulnar joint, neck of radius, radius, interosseous membrane, ulna, ulnar notch of the radius, head of the ulna, distal radioulnar joint, styloid process of ulna, styloid process of radius. [Return to Figure 6.4].
Figure 6.5 image description: This diagram shows an anterior and posterior view of the hands with corresponding labels. Anterior view labels read (from top): middle finger, ring finger, index finger, little finger, thumb, phalanges (distal, proximal), metacarpals, carpals, ulna, radius. Posterior view labels read (from top): Phalanges (distal, middle, proximal), head shaft and base of the proximal phalanx, head shaft and base of the metatarsal, metatarsals 1-5, carpals, ulna, radius. [Return to Figure 6.5].
Figure 6.6 image description: This image shows the structure of the tibia and the fibula. The left panel shows the anterior view. Labels read (from top): lateral condyle, medial condyle, tibial tuberosity, anterior border, interosseous membrane, fibula, tibia, medial malleolus, lateral malleolus, articular surface. The right panel shows the posterior view. Labels read (from top): the articular surface of medial and lateral condyles, medial condyle, head of the fibula, soleal line, interosseous membrane, tibia, fibula, medial malleolus, lateral malleolus, articular surface. [Return to Figure 6.6].
Figure 6.7 image description: This figure shows the bones of the foot. The left panel shows the superior view. Labels read (from toes): distal, proximal phalanges, distal phalange, middle phalange, proximal phalanx, medial cuneiform, intermediate and lateral cuneiforms, navicular, cuboid, talus, trochlea of talus, calcaneus. The top right panel shows the medial view. Labels read (from left to right starting at toe): first metatarsal, medial cuneiform, intermediate cuneiform, navicular, talus, calcaneus, facet for medial malleolus, sustentaculum tali (talar shelf), calcaneal tuberosity. The bottom right panel shows the lateral view. Labels read (from left at the heel, to right): calcaneus, talus, facet for lateral malleolus, cuboid, navicular, intermediate and lateral cuneiforms, fifth metatarsal. [Return to Figure 6.7].
Figure 6.8 image description: This multi-part image shows different types of movements that are possible by different joints in the body. Labels read (from the top, left): a and b angular movements: flexion and extension at the shoulders and knees, c) angular movements: flexion and extension of the neck (arrows pointing left and right to indicate movement). Labels (from the bottom, left) read d) angular movements: flexion and extension of the vertical column, e) angular movements abduction, adduction, and circumduction of the upper limb at the shoulder, f) rotation of the head, neck, and lower limb. [Return to Figure 6.8].
Figure 6.9 image description: This multi-part image shows different types of movements that are possible by different joints in the body. The top left image shows a hand and forearm in the pronation and supination positions. The top middle image shows a foot in the dorsiflexion and plantar flexion positions. The top right image shows a foot in the inversion and eversion positions. The bottom left image shows the retraction and protraction of a man’s mandible. The bottom middle image shows the elevation and depression of a man’s mandible. The bottom right image shows a hand in the opposition position. [Return to Figure 6.9].
Figure 6.10 image description: This image shows the changes to the abnormal curves of the vertebral columns in different diseases. The left panel shows the change in the curve of the vertebral column in scoliosis, the middle panel shows the change in the curve of the vertebral column in kyphosis, and the right panel shows the change in the curve of the vertebral column in lordosis. [Return to Figure 6.10].
Figure 6.11 image description: This figure shows the changes to the spine in osteoporosis. The left panel shows the structure of normal vertebrae and the right panel shows the curved vertebrae in osteoporosis. [Return to Figure 6.11].
Figure 6.12 image description: In this illustration, each type of fracture is shown on the right femur from an anterior view. In the closed fracture, the femur is broken in the middle of the shaft with the upper and lower halves of the bone completely separated. However, the two halves of the bones are still aligned in that the broken edges are still facing each other. In an open fracture, the femur is broken in the middle of the shaft with the upper and lower halves of the bone completely separated. Unlike the closed fracture, in the open fracture, the two bone halves are misaligned. The lower half is turned laterally and it has protruded through the skin of the thigh. The broken ends no longer line up with each other. In a transverse fracture, the bone has a crack entirely through its width, however, the broken ends are not separated. The crack is perpendicular to the long axis of the bone. Arrows indicate that this is usually caused by compression of the bone in a superior-inferior direction. A spiral fracture travels diagonally through the diameter of the bone. In a comminuted fracture, the bone has several connecting cracks at its middle. The bone could splinter into several small pieces at the site of the comminuted fracture. In an impacted fracture, the crack zig zags throughout the width of the bone like a lightning bolt. An arrow indicates that these are usually caused by an impact that pushes the femur up into the body. A greenstick fracture is a small crack that does not extend through the entire width of the bone. The oblique fracture shown here is traveling diagonally through the shaft of the femur at about a thirty degree angle. [Return to Figure 6.12].
Unless otherwise indicated, this chapter contains material adapted from Anatomy and Physiology (on OpenStax ), by Betts et al. and is used under a CC BY 4.0 international license . Download and access this book for free at https://openstax.org/books/anatomy-and-physiology/pages/1-introduction .


9.3: Test Statistics and Sampling Distributions
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At this point we need to start talking specifics about how a hypothesis test is constructed. To that end, let’s return to the ESP example. Let’s ignore the actual data that we obtained, for the moment, and think about the structure of the experiment. Regardless of what the actual numbers are, the form of the data is that X out of N people correctly identified the colour of the hidden card. Moreover, let’s suppose for the moment that the null hypothesis really is true: ESP doesn’t exist, and the true probability that anyone picks the correct colour is exactly θ=0.5. What would we expect the data to look like? Well, obviously, we’d expect the proportion of people who make the correct response to be pretty close to 50%. Or, to phrase this in more mathematical terms, we’d say that X/N is approximately 0.5. Of course, we wouldn’t expect this fraction to be exactly 0.5: if, for example we tested N=100 people, and X=53 of them got the question right, we’d probably be forced to concede that the data are quite consistent with the null hypothesis. On the other hand, if X=99 of our participants got the question right, then we’d feel pretty confident that the null hypothesis is wrong. Similarly, if only X=3 people got the answer right, we’d be similarly confident that the null was wrong. Let’s be a little more technical about this: we have a quantity X that we can calculate by looking at our data; after looking at the value of X, we make a decision about whether to believe that the null hypothesis is correct, or to reject the null hypothesis in favour of the alternative. The name for this thing that we calculate to guide our choices is a test statistic .
Having chosen a test statistic, the next step is to state precisely which values of the test statistic would cause is to reject the null hypothesis, and which values would cause us to keep it. In order to do so, we need to determine what the sampling distribution of the test statistic would be if the null hypothesis were actually true (we talked about sampling distributions earlier in Section 10.3.1). Why do we need this? Because this distribution tells us exactly what values of X our null hypothesis would lead us to expect. And therefore, we can use this distribution as a tool for assessing how closely the null hypothesis agrees with our data.
How do we actually determine the sampling distribution of the test statistic? For a lot of hypothesis tests this step is actually quite complicated, and later on in the book you’ll see me being slightly evasive about it for some of the tests (some of them I don’t even understand myself). However, sometimes it’s very easy. And, fortunately for us, our ESP example provides us with one of the easiest cases. Our population parameter θ is just the overall probability that people respond correctly when asked the question, and our test statistic X is the count of the number of people who did so, out of a sample size of N. We’ve seen a distribution like this before, in Section 9.4: that’s exactly what the binomial distribution describes! So, to use the notation and terminology that I introduced in that section, we would say that the null hypothesis predicts that X is binomially distributed, which is written
X∼Binomial(θ,N)
Since the null hypothesis states that θ=0.5 and our experiment has N=100 people, we have the sampling distribution we need. This sampling distribution is plotted in Figure 11.1. No surprises really: the null hypothesis says that X=50 is the most likely outcome, and it says that we’re almost certain to see somewhere between 40 and 60 correct responses.


4.2: Analyzing Discrete Random Variables
Learning Objectives
- Draw connections between a relative frequency distribution of a data set and a probability distribution of a random variable
- Define, compute, and interpret the expected value of a discrete random variable
- Define and compute the variance and standard deviation of a discrete random variable
- Interpret the standard deviation of a discrete random variable
Section \(4.2\) Excel File (contains all of the data sets for this section)
Random Variables and Data
As we seek to understand the world around us, there are times when we are unsure what could happen in a particular situation; we do not necessarily know all the possible outcomes or the likelihood of every outcome happening. That is okay. We learn by observation, data collection, and experimentation. When studying a quantitative variable, we collect data by measuring a value for each observation. Once our data is collected, we can analyze the data using relative frequency distributions to show the proportion of the observations in a particular class and provide an empirical estimate of the probability of that class occurring. We realize that our relative frequency distributions are connected with the probability distributions of our random variables (classes corresponding to values the variable takes on and relative frequencies corresponding to probabilities). There is, however, a significant difference to keep in mind. Relative frequency distributions describe data that has been collected. Random variables describe the possibilities and probabilities of what can happen. Relative frequency distributions are descriptive, while random variables are predictive.
To analyze and understand data, we developed methods of visualization, measures of centrality, and measures of dispersion. For these methods with random variables, we use the connection to relative frequency distributions. Recall how we treated these three concepts when given a relative frequency distribution for a discrete quantitative variable.
Visualization: We visualized the relative frequency distribution using bar graphs.
Measures of Centrality: We discovered that the mean of a data set could be calculated by "weighing" our class values \(\left(x_j\right)\) by their relative frequencies \(\left(P(x_j)\right)\)\[ \mu = \frac{\sum \left[ x_j \cdot P(x_j) \right]}{\sum P(x_j) }=\sum \left[ x_j \cdot P(x_j) \right] \nonumber\]
Measures of Dispersion: We discovered that the variance of a population data set could be calculated by "weighing" the squared deviations from the mean \(\left((x_j-\mu)^2\right)\) by their relative frequencies \(\left(P(x_j)\right)\)\[ \sigma^{2} = \frac{\sum \left[\left( x_j - \mu \right)^2 \cdot P(x_j)\right]}{\sum P(x_j) } =\sum{\left[ (x_j - \mu )^2 \cdot P(x_j) \right]} \nonumber\]From the variance, we can also compute the standard deviation of the data set: \(\sigma=\sqrt{\sigma^2}.\)
We will visualize our discrete random variables with bar graphs and develop measures of centrality and dispersion similar to the mean and variance of a population data set.
Analyzing Discrete Random Variables
The primary challenge in connecting our analyses of relative frequency distributions and random variables is interpretation. As mentioned above, relative frequency distributions are constructed from collected data and describe what happened; random variables are constructed from all the possible values resulting from a random experiment and the associated probabilities of these values occurring. Even in this distinction, the best predictor of future behavior is past behavior; we often use relative frequencies as estimates of probabilities (the empirical method of probability). So, how can we understand the measures of centrality and dispersion for a random variable? They would be estimates for the actual measures of centrality and dispersion if we were to repeatedly run the random experiment, collect data, and analyze it. Because of the Law of Large Numbers, we would expect the estimates and computed measures to converge if we conducted the random experiment repeatedly. With this in mind, we will study discrete random variables.
Graphical Visualization: Bar Graphs
The method of constructing bar graphs remains the same for discrete random variables. The values of the random variable are listed on the horizontal axis, and bars are formed with the height indicating the probability of that particular value occurring. The gaps between the bars indicate the discrete nature of the random variable. Here, we produce bar graphs for the random variables \(X\), \(Y\), and \(D\) from our last section (the sum of two fair dice rolled, the max of two fair dice rolled, and the number of days adults exercise in a week, respectively).
Figure \(\PageIndex{1}\): Bar graph representations of the probability distributions of the random variables \(X\), \(Y\), and \(D\) (left to right)
Measure of Centrality: Expected Value
Recall that the mean of a data set is the balancing point of that data set, which we could visualize from our graphs. Given this same manner of representation, the mean of a random variable \(X,\) also referred to as the expected value of \(X,\) \(\text{E}(X),\) is also the balance point of the bar graph. From the symmetry of \(X\) in the figure above, we conclude that \(\text{E}(X)=7.\) Determining the center of mass for \(Y\) and \(D\) is a little more complicated, but by looking at the distributions, we can see that \(\text{E}(Y)>4\) and \(\text{E}(D)<3.\) A more precise calculation is in order. By understanding the relative frequency as an estimate of that particular event's probability, we define the expected value of a discrete random variable \(X.\) \[ \mu =\text{E}(X)= \frac{\sum \left[ x_j \cdot P(X=x_j) \right]}{\sum P(X=x_j) }=\sum \left[ x_j \cdot P(X=x_j) \right] \nonumber\]In computing the expected value of a discrete random variable, we consider each possible value of the random variable and weigh it according to the associated probability; the higher the probability, the heavier the weight. Since the sum of probabilities of all possible values adds to \(1,\) the denominator consisting of the "total" weight simplifies. Let us confirm that the expected value for the sum of two fair dice is indeed \(7.\)
Table \(\PageIndex{1}\): Computation of the expected value of the sum of two fair dice
| \(X=x_j\) | \(P(X=x_j)\) | \(x_j\cdot P(X=x_j)\) |
|---|---|---|
| \(2\) | \(\dfrac{1}{36}\) | \(2\cdot \dfrac{1}{36}=\dfrac{2}{36}\) |
| \(3\) | \(\dfrac{2}{36}\) | \(3\cdot \dfrac{2}{36}=\dfrac{6}{36}\) |
| \(4\) | \(\dfrac{3}{36}\) | \(4\cdot \dfrac{3}{36}=\dfrac{12}{36}\) |
| \(5\) | \(\dfrac{4}{36}\) | \(5\cdot \dfrac{4}{36}=\dfrac{20}{36}\) |
| \(6\) | \(\dfrac{5}{36}\) | \(6\cdot \dfrac{5}{36}=\dfrac{30}{36}\) |
| \(7\) | \(\dfrac{6}{36}\) | \(7\cdot \dfrac{6}{36}=\dfrac{42}{36}\) |
| \(8\) | \(\dfrac{5}{36}\) | \(8\cdot \dfrac{5}{36}=\dfrac{40}{36}\) |
| \(9\) | \(\dfrac{4}{36}\) | \(9\cdot \dfrac{4}{36}=\dfrac{36}{36}\) |
| \(10\) | \(\dfrac{3}{36}\) | \(10\cdot \dfrac{3}{36}=\dfrac{30}{36}\) |
| \(11\) | \(\dfrac{2}{36}\) | \(11\cdot \dfrac{2}{36}=\dfrac{22}{36}\) |
| \(12\) | \(\dfrac{1}{36}\) | \(12\cdot\dfrac{1}{36}=\dfrac{12}{36}\) |
| \(\mu=\text{E}(X)=\dfrac{2}{36}+\dfrac{6}{36}+\ldots+\dfrac{12}{36}=\dfrac{252}{36}=7\) |
After adding all of the entries (\(x_j\cdot P(X=x_j)\)) in the third column, we arrived at a total of \(7,\) confirming our visual estimation. \(\text{E}(X)=7\) means that as we roll a pair of fair dice repeatedly, we expect the mean of all of the sums to be close to \(7.\) Consider rolling two fair dice using a simulator and calculating the mean of the dice sums. We (the authors) ran the simulation with \(20\) trials, producing a data set of \(20\) dice sums, and found the mean to be \(\frac{142}{20}\) \(=7.1;\) this is fairly close to our expected value of \(7.\) Run the simulation yourself and compute the mean. How close to \(7\) is your computed mean?
- Recall the discrete random variable \(Y\) which describes the maximum value of two fair dice when rolled. Compute and interpret the expected value \(\text{E}(Y)\) using its probability distribution reproduced below.
Table \(\PageIndex{2}\): Probability distribution for the random variable \(Y\)
| \(Y=y_j\) | \(P(Y=y_j)\) | \(y_j \cdot (P(Y=y_j)\) |
|---|---|---|
| \(1\) | \(\dfrac{1}{36}\) | |
| \(2\) | \(\dfrac{3}{36}\) | |
| \(3\) | \(\dfrac{5}{36}\) | |
| \(4\) | \(\dfrac{7}{36}\) | |
| \(5\) | \(\dfrac{9}{36}\) | |
| \(6\) | \(\dfrac{11}{36}\) |
- Answer
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Table \(\PageIndex{3}\): Table of computation
\(Y=y_j\) \(P(Y=y_j)\) \(y_j \cdot (P(Y=y_j)\) \(1\) \(\dfrac{1}{36}\) \(1 \cdot \dfrac{1}{36} =\dfrac{1}{36}\) \(2\) \(\dfrac{3}{36}\) \(2 \cdot \dfrac{3}{36} =\dfrac{6}{36}\) \(3\) \(\dfrac{5}{36}\) \(3 \cdot \dfrac{5}{36} =\dfrac{15}{36}\) \(4\) \(\dfrac{7}{36}\) \(4 \cdot \dfrac{7}{36} =\dfrac{28}{36}\) \(5\) \(\dfrac{9}{36}\) \(5 \cdot \dfrac{9}{36} =\dfrac{45}{36}\) \(6\) \(\dfrac{11}{36}\) \(6 \cdot \dfrac{11}{36} =\dfrac{66}{36}\) \(\mu=\text{E}(Y)=\dfrac{1}{36}+\dfrac{6}{36}+\ldots+\dfrac{66}{36}=\dfrac{161}{36}=4+\frac{17}{36}\approx4.4722\) The expected value of the random variable \(Y\) is approximately \(4.4722,\) meaning after repeatedly rolling two fair dice and examining the maximum values from each pair, the mean maximum value would be about \(4.4722.\) We can verify this experimentally; see for yourself. Roll two dice, record the highest number, repeat \(20\) times, and then take the mean of those \(20\) numbers. We (the authors) obtained a mean of \(4.7\) when doing this. Notice that this seems less accurate than our empirical estimate obtained for \(X.\) Some random variables require more trials than others to accurately estimate the expected value. Note that our expected value cannot occur for any single trial of our random experiment.
- Recall the discrete random variable \(D\) which describes the number of days adults exercise per week. Compute and interpret the expected value \(\text{E}(D)\) using its probability distribution reproduced below.
Table \(\PageIndex{4}\): Probability distribution of the random variable \(D\)
| \(D=d_j\) | \(P(D=d_j)\) | \(d_j \cdot (P(D=d_j)\) |
|---|---|---|
| \(0\) | \(0.28\) | |
| \(1\) | \(0.11\) | |
| \(2\) | \(0.10\) | |
| \(3\) | \(0.14\) | |
| \(4\) | \(0.10\) | |
| \(5\) | \(0.15\) | |
| \(6\) | \(0.08\) | |
| \(7\) | \(0.04\) |
- Answer
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Table \(\PageIndex{5}\): Table of computation
\(D=d_j\) \(P(D=d_j)\) \(d_j \cdot (P(D=d_j)\) \(0\) \(0.28\) \(0 \cdot 0.28=0\) \(1\) \(0.11\) \(1 \cdot 0.11=0.11\) \(2\) \(0.10\) \(2 \cdot 0.10=0.20\) \(3\) \(0.14\) \(3 \cdot 0.14=0.42\) \(4\) \(0.10\) \(4 \cdot 0.10=0.4\) \(5\) \(0.15\) \(5 \cdot 0.15=0.75\) \(6\) \(0.08\) \(6 \cdot 0.08=0.48\) \(7\) \(0.04\) \(7 \cdot 0.04=0.28\) \(\mu=\text{E}(D)=0+0.11+\ldots+0.28=2.64\) The expected value of the random variable \(D\) is \(2.64\) days, meaning that after repeatedly asking random adults how many days they work out per week, we would predict the mean number of days to be about \(2.64.\)
Measures of Dispersion: Variance and Standard Deviation
A question arises from our discussion of expected value: how much variation should there be as the random experiment is repeated? Will most values be close to the expected value, or will a wide variety of values occur? We can answer these questions intuitively using bar graphs, but the intuition is rarely straightforward. Use Figure \(\PageIndex{1}\) to determine which random variable has the most spread before continuing with your reading.
Using the notion of relative frequency as an estimate for a value's probability and utilizing our expected value as our measure of center, we define the variance of a discrete random variable \(X.\) \[\begin{align*} \sigma^2=\text{Var}(X)&=\frac{\sum \left[ \left( x_j - \text{E}(X) \right)^2 \cdot P(X=x_j)\right]}{\sum P(X=x_j) }=\sum{\left[ (x_j - \text{E}(X) )^2 \cdot P(X=x_j) \right]}\\[5pt]&=\frac{\sum \left[ \left( x_j - \mu \right)^2 \cdot P(X=x_j)\right]}{\sum P(X=x_j) }=\sum{\left[ (x_j - \mu )^2 \cdot P(X=x_j) \right]}\end{align*}\]In computing the variance of a discrete random variable, we "weigh" the values that we are averaging, namely the squared deviations from the mean, by their probabilities. Since the sum of the probabilities of all the possible values is \(1,\) the denominator reduces. As before, we define the standard deviation of a discrete random variable \(X\) as the square root of the random variable's variance. Let us determine the dispersion of the discrete random variables, \(X,\) \(Y,\) and \(D\) by computing their variances and standard deviations.
Table \(\PageIndex{6}\): Table of computation
| \(X=x_j\) | \(P(X=x_j)\) | \(\left(x_j - \mu\right)^2\) | \(\left( x_j - \mu\right)^2 \cdot P(X=x_j)\) |
|---|---|---|---|
| \(2\) | \(\dfrac{1}{36}\) | \(\left(2-7\right)^2=25\) | \(25\cdot \dfrac{1}{36}=\dfrac{25}{36}\) |
| \(3\) | \(\dfrac{2}{36}\) | \(\left(3-7\right)^2=16\) | \(16\cdot \dfrac{2}{36}=\dfrac{32}{36}\) |
| \(4\) | \(\dfrac{3}{36}\) | \(\left(4-7\right)^2=9\) | \(9\cdot \dfrac{3}{36}=\dfrac{27}{36}\) |
| \(5\) | \(\dfrac{4}{36}\) | \(\left(5-7\right)^2=4\) | \(4\cdot \dfrac{4}{36}=\dfrac{16}{36}\) |
| \(6\) | \(\dfrac{5}{36}\) | \(\left(6-7\right)^2=1\) | \(1\cdot \dfrac{5}{36}=\dfrac{5}{36}\) |
| \(7\) | \(\dfrac{6}{36}\) | \(\left(7-7\right)^2=0\) | \(0\cdot \dfrac{6}{36}=\dfrac{0}{36}\) |
| \(8\) | \(\dfrac{5}{36}\) | \(\left(8-7\right)^2=1\) | \(1\cdot \dfrac{5}{36}=\dfrac{5}{36}\) |
| \(9\) | \(\dfrac{4}{36}\) | \(\left(9-7\right)^2=4\) | \(4\cdot \dfrac{4}{36}=\dfrac{16}{36}\) |
| \(10\) | \(\dfrac{3}{36}\) | \(\left(10-7\right)^2=9\) | \(9\cdot \dfrac{3}{36}=\dfrac{27}{36}\) |
| \(11\) | \(\dfrac{2}{36}\) | \(\left(11-7\right)^2=16\) | \(16\cdot \dfrac{2}{36}=\dfrac{32}{36}\) |
| \(12\) | \(\dfrac{1}{36}\) | \(\left(12-7\right)^2=25\) | \(25\cdot\dfrac{1}{36}=\dfrac{25}{36}\) |
| \(\sigma^2=\text{Var}(X)=\dfrac{25}{36}+\dfrac{32}{36}+\ldots+\dfrac{25}{36}=\dfrac{210}{36}=5\frac{5}{6}\approx5.8333\) | |||
| \(\sigma=\sqrt{\text{Var}(X)}=\sqrt{\frac{35}{6}}\approx2.4152\) |
As with the mean, we understand the standard deviation as an estimate for the computed standard deviation as we repeatedly conduct the random experiment. As repetitions increase, the computed standard deviation approaches the expected standard deviation value. We understand the mean of a data set as its center and can loosely understand the standard deviation as the typical distance our observations are from the mean. If we translated this in terms of random variables, we could say that a typical value is within a standard deviation from the expected value. Within the context of summing the rolls of two fair dice, we would expect a typical value to be within \(2.4152\) units from \(7,\) those sums being \(5,\) \(6,\) \(7,\) \(8,\) or \(9.\) If we were playing a board game based on the sum of the rolls of two fair dice, we could expect these \(5\) numbers to occur somewhat "regularly" and plan our strategy accordingly.
- Determine the probability that the sum of the rolling of two fair dice would be within one standard deviation of the expected value: \(P\left(|X-\mu|<\sigma\right).\)
- Answer
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Since the mean is \(7\) and the standard deviation is about \(2.4152\), we are looking for the probability that our random variable is between the values between \(7-2.4152\) \(=4.5848\) and \(7+2.4152\) \(=9.4152.\) Our question reduces to finding \(P\left(X=5,6,7,8,\text{ or }9\right)\) \(=\dfrac{4}{36}\)\(+\dfrac{5}{36}\)\(+\dfrac{6}{36}\)\(+\dfrac{5}{36}\)\(+\dfrac{4}{36}\) \(=\dfrac{24}{36}\) \(=\dfrac{2}{3}\) \(\approx66.6667\%.\)
- Determine the probability that the sum of the rolling of two fair dice would be within two standard deviations of the expected value: \(P\left(|X-\mu|<2\sigma\right).\)
- Answer
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We are now looking for the probability that our random variable takes on the values between \(7-2\cdot2.4152\) \(=2.1696\) and \(7+\cdot2.4152\) \(=11.8304.\) So our question reduces to finding \(P\left(3\leq X\leq11\right)\) \(=1-P\left(X=2 \text{ or }12\right)\) \(=1-(\dfrac{1}{36}\)\(+\dfrac{1}{36})\) \(=1-\dfrac{2}{36}\) \(=\dfrac{34}{36}\) \(=\dfrac{17}{18}\) \(\approx94.4444\%.\) Notice that all possible values of our random variable lie within \(3\) standard deviations of the expected value.
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Recall the discrete random variable \(Y\) that describes the maximum value of two fair dice when rolled. In text exercise \(\PageIndex{1}.1,\) we computed that \(\text{E}(Y)\) \(=\frac{161}{36}.\)
- Compute the variance and standard deviation of \(Y\) using its probability distribution reproduced below.
- Compute \(P\left(|Y-\mu|<\sigma\right).\)
- Compute \(P\left(|Y-\mu|<2\sigma\right).\)
Table \(\PageIndex{7}\): Probability distribution of the random variable \(Y\)
| \(Y=y_j\) | \(P(Y=y_j)\) | \(\left( y_j - \mu\right)^2 \cdot P(Y=y_j)\) |
|---|---|---|
| \(1\) | \(\dfrac{1}{36}\) | |
| \(2\) | \(\dfrac{3}{36}\) | |
| \(3\) | \(\dfrac{5}{36}\) | |
| \(4\) | \(\dfrac{7}{36}\) | |
| \(5\) | \(\dfrac{9}{36}\) | |
| \(6\) | \(\dfrac{11}{36}\) |
- Answer
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Table \(\PageIndex{8}\): Table of computation
\(Y=y_j\) \(P(Y=y_j)\) \(\left( y_j - \mu\right)^2 \cdot P(Y=y_j)\) \(1\) \(\dfrac{1}{36}\) \(\left(1 - \dfrac{161}{36}\right)^2\cdot\dfrac{1}{36}\approx0.3349\) \(2\) \(\dfrac{3}{36}\) \(\left(2 - \dfrac{161}{36}\right)^2\cdot\dfrac{3}{36}\approx0.5093\) \(3\) \(\dfrac{5}{36}\) \(\left(3 - \dfrac{161}{36}\right)^2\cdot\dfrac{5}{36}\approx0.3010\) \(4\) \(\dfrac{7}{36}\) \(\left(4 - \dfrac{161}{36}\right)^2\cdot\dfrac{7}{36}\approx0.0434\) \(5\) \(\dfrac{9}{36}\) \(\left(5 - \dfrac{161}{36}\right)^2\cdot\dfrac{9}{36}\approx0.0696\) \(6\) \(\dfrac{11}{36}\) \(\left(6 - \dfrac{161}{36}\right)^2\cdot\dfrac{11}{36}\approx0.7132\) \(\sigma^2=\text{Var}(Y)\approx0.3349+0.5093+0.3010+0.0434+0.0696+0.7132\approx1.9715\) \(\sigma=\sqrt{\text{Var}(Y)}\approx\sqrt{1.9715}\approx1.4041\) - Note: when conducting such involved and messy calculations, be sure to round only at the very last stage. We provide intermediate approximations along the way to help facilitate readability, but we do not use the rounded values in the actual computations for final answers. The variance of \(Y\) is approximately \(1.9715\) and the standard deviation is about \(1.4041\).
- We are tasked with determining the probability that the value produced in running the random experiment will fall within one standard deviation of the expected value.\[\begin{align*}P\left(|Y-\mu|<\sigma\right)&=P\left(4.4722-1.4041<Y<4.4722+1.4041\right)=P\left(3.0681<Y<5.8763\right)\\&=P\left(Y=4\text{ or }5\right)\\&=\dfrac{7}{36}+\dfrac{9}{36}=\dfrac{16}{36}\\&=\dfrac{4}{9}\approx44.4444\%\end{align*}\]
- We now determine the probability that the value produced in running the random experiment will fall within two standard deviations of the expected value.\[\begin{align*}P\left(|Y-\mu|<2\sigma\right)&=P\left(4.4722-2\cdot1.4041<Y<4.4722+2\cdot1.4041\right)=P\left(1.664<Y<7.2804\right)\\&=P\left(Y=2,3,4,5,\text{ or }6\right)\\&=1-P\left(Y=1\right)=1-\dfrac{1}{36}\\&=\dfrac{35}{36}\approx97.2222\%\end{align*}\]
-
Recall the discrete random variable \(D\) that describes the number of days adults exercise per week. In text exercise \(\PageIndex{1}.2\), we computed that \(\text{E}(D)=2.64\) days.
- Compute the variance and standard deviation of \(D\) using its probability distribution reproduced below.
- Compute \(P\left(|D-\mu|<\sigma\right)\).
- Compute \(P\left(|D-\mu|<2\sigma\right)\).
Table \(\PageIndex{9}\): Probability distribution of the random variable \(D\)
| \(D=d_j\) | \(P(D=d_j)\) | \(\left( d_j - \mu\right)^2 \cdot P(D=d_j)\) |
|---|---|---|
| \(0\) | \(0.28\) | |
| \(1\) | \(0.11\) | |
| \(2\) | \(0.10\) | |
| \(3\) | \(0.14\) | |
| \(4\) | \(0.10\) | |
| \(5\) | \(0.15\) | |
| \(6\) | \(0.08\) | |
| \(7\) | \(0.04\) |
- Answer
-
Table \(\PageIndex{10}\): Table of computation
\(D=d_j\) \(P(D=d_j)\) \(\left( d_j - \mu\right)^2 \cdot P(D=d_j)\) \(0\) \(0.28\) \(\left(0 - 2.64\right)^2\cdot0.28\approx1.9515\) \(1\) \(0.11\) \(\left(1 - 2.64\right)^2\cdot0.11\approx0.2959\) \(2\) \(0.10\) \(\left(2 - 2.64\right)^2\cdot0.10\approx0.0410\) \(3\) \(0.14\) \(\left(3 - 2.64\right)^2\cdot0.14\approx0.0181\) \(4\) \(0.10\) \(\left(4 - 2.64\right)^2\cdot0.10\approx\0.1850\) \(5\) \(0.15\) \(\left(5 - 2.64\right)^2\cdot0.15\approx0.8354\) \(6\) \(0.08\) \(\left(6 - 2.64\right)^2\cdot0.08\approx0.9032\) \(7\) \(0.04\) \(\left(7 - 2.64\right)^2\cdot0.04\approx0.7604\) \(\sigma^2=\text{Var}(D)\approx1.9515+0.2959+\dots+0.7604\approx4.9904\) \(\sigma=\sqrt{\text{Var}(D)}\approx\sqrt{4.9904}\approx2.2339\) - The variance of \(D\) is approximately \(4.9904\) days\(^2\) and the standard deviation is about \(2.2339\) days.
- We determine the probability that the value produced in running the random experiment will fall within one standard deviation of the expected value.\[\begin{align*}P\left(|D-\mu|<\sigma\right)&=P\left(2.64-2.2339<D<2.64-2.2339\right)=P\left(0.4061<D<4.8739\right)\\&=P\left(D=1,2,3,\text{ or }4\right)\\&=0.11+0.10+0.14+0.10=45\%\end{align*}\]
- We determine the probability that the value produced in running the random experiment will fall within two standard deviations of the expected value.\[\begin{align*}P\left(|D-\mu|<2\sigma\right)&=P\left(2.64-2\cdot2.2339<D<2.64-2\cdot2.2339\right)=P\left(-1.8278<D<7.1078\right)\\&=P\left(D=0,1,2,3,4,5,6\text{ or }7\right)\\&=100\%\end{align*}\]
As we have seen with the last several text exercises, the probability that a discrete random variable is within a standard deviation or two of the expected value varies depending on the distribution. This, hopefully, does not come as a surprise (if it did, recall our discussion in section \(2.7\) about Chebyshev's Inequality and the Empirical Rule ). For this reason, we must take our understanding of a typical value rather loosely at this stage. It will get better. We end this section with text exercises exploring games of chance with cash prizes.
We may further reduce the formula for the variance of a random variable and produce an equivalent formulation with some computational benefits. The two formulas produce the same values; they are equivalent. The first formula emphasizes the underlying logic of what the measure means; we, therefore, recommend its use primarily, but we provide this second formula regardless.\[\sigma^2=\text{Var}(X)=\left(\sum {\left[ x_j^2 \cdot P(X=x_j)\right]}\right)-\mu^2\nonumber \]
- Optional derivation for the mathematically inclined
-
\[\begin{align*}\sigma^2=\text{Var}(X)&=\sum{\left[ (x_j - \mu )^2 \cdot P(X=x_j) \right]}\\&=\sum{\left[ (x_j^2 -2x_j\mu+\mu^2 )\cdot P(X=x_j) \right]}\\&=\sum{\left[x_j^2\cdot P(X=x_j) -2x_j\mu\cdot P(X=x_j)+ \mu^2\cdot P(X=x_j) \right]}\\&=\sum{\left[x_j^2\cdot P(X=x_j)\right]}-\sum{\left[2x_j\mu\cdot P(X=x_j)\right]}+ \sum{\left[\mu^2\cdot P(X=x_j) \right]}\\&=\sum{\left[x_j^2\cdot P(X=x_j)\right]}-2\mu\sum{\left[x_j\cdot P(X=x_j)\right]}+ \mu^2\sum{P(X=x_j)}\\&=\sum{\left[x_j^2\cdot P(X=x_j)\right]}-2\mu^2+ \mu^2\\&=\left(\sum {\left[ x_j^2 \cdot P(X=x_j)\right]}\right)-\mu^2\end{align*}\]
Games of Chance
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Consider entering a raffle with a single cash prize of \(\$1,000.\) Each player may only purchase one ticket for \(\$25,\) and only \(100\) raffle tickets will be sold.
- Represent our financial net gains or losses with the discrete random variable \(R\) and construct its probability distribution.
- Compute and interpret \(\text{E}(R).\)
- Answer
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The random experiment is entering the raffle; either we win or we lose. If we lose, we are out the \(\$25.\) If we win, our net gain is \(\$975\) because we spent \(\$25\) on the raffle ticket. Our random variable \(R\) takes on two values: \(-25\) and \(975.\) Each ticket is typically equally likely to be drawn. Therefore, \(P\left(X=-25\right)\)(the probability that we lose), is \(\frac{99}{100}.\) Its complement, the probability that we win, is \(\frac{1}{100}.\)
Table \(\PageIndex{11}\): Probability distribution for the random variable \(R\)
\(R=r_j\) \(P(R=r_j)\) \(-25\) \(\dfrac{99}{100}\) \(975\) \(\dfrac{1}{100}\) -
Table \(\PageIndex{12}\): Table of computation
We have found that the expected value for this raffle is \(-\$15.\) When we play, we either lose \(\$25\) or gain \(\$975,\) but on average we should expect to lose \(\$15.\) It is important to note that a negative expected value does not necessarily mean we will lose money. Someone will win the raffle. What it means is that, if we played this raffle many times, we would expect to lose money in the long run. Imagine if we did the raffle \(100\) times; we might expect to win \(1\) time and lose \(99\) times. In such a situation, we won \(\$1000,\) but we also lost \(\$25 \cdot 100\) \(= \$2500\) in purchasing tickets. Hence, we see a net loss of \(\$1500\), for an average loss of \(\$15\) each time.\(R=r_j\) \(P(R=r_j)\) \(r_j\cdot P(R=r_j)\) \(-25\) \(\dfrac{99}{100}\) \(-25\cdot\dfrac{99}{100}=-\dfrac{99}{4}\) \(975\) \(\dfrac{1}{100}\) \(975\cdot\dfrac{1}{100}=\dfrac{39}{4}\) \(\mu=\text{E}(R)=-\dfrac{99}{4}+\dfrac{39}{4}=-\dfrac{60}{4}=-15\)
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The random experiment is entering the raffle; either we win or we lose. If we lose, we are out the \(\$25.\) If we win, our net gain is \(\$975\) because we spent \(\$25\) on the raffle ticket. Our random variable \(R\) takes on two values: \(-25\) and \(975.\) Each ticket is typically equally likely to be drawn. Therefore, \(P\left(X=-25\right)\)(the probability that we lose), is \(\frac{99}{100}.\) Its complement, the probability that we win, is \(\frac{1}{100}.\)
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Consider entering a raffle with one \(\$1,000\) cash prize and one \(\$500\) cash prize. Each player may only purchase one ticket for \(\$50,\) and \(200\) raffle tickets will be sold.
- Represent our financial net gains or losses with the discrete random variable \(R\) and construct its probability distribution.
- Compute and interpret \(\text{E}(R).\)
- Answer
-
-
We will set up our random variable similarly. However, this time, there are three possible values since there are three different outcomes: winning the big prize, winning the small prize, and losing altogether.
Table \(\PageIndex{13}\): Probability distribution for the random variable \(R\)
\(R=r_j\) \(P(R=r_j)\) \(-50\) \(\dfrac{198}{200}\) \(450\) \(\dfrac{1}{200}\) \(950\) \(\dfrac{1}{200}\) -
Table \(\PageIndex{14}\): Table of computation
We have found that the expected value for this raffle is \(\-$42.50.\) Despite more prize money being available, the higher price of entering the raffle and the larger number of tickets sold made the expected value decrease starkly.\(R=r_j\) \(P(R=r_j)\) \(r_j\cdot P(R=r_j)\) \(-50\) \(\dfrac{198}{200}\) \(-50\cdot\dfrac{198}{200}=-\dfrac{198}{4}=\) \(450\) \(\dfrac{1}{200}\) \(450\cdot\dfrac{1}{200}=\dfrac{9}{4}\) \(950\) \(\dfrac{1}{200}\) \(950\cdot\dfrac{1}{200}=\dfrac{19}{4}\) \(\mu=\text{E}(R)=-\dfrac{198}{4}+\dfrac{9}{4}+\dfrac{19}{4}=-\dfrac{170}{4}=-42.50\)
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We will set up our random variable similarly. However, this time, there are three possible values since there are three different outcomes: winning the big prize, winning the small prize, and losing altogether.
- A family is hosting an extended family reunion and thought having a raffle with a single cash prize would be fun. The dad does not want to make or lose any money in running the raffle, but his wife would like the raffle to help cover the cost of hosting the reunion and wants \(\$5\) per ticket to help cover the costs. If \(80\) family members signed up for the raffle at a ticket price of \(\$15,\) determine the cash prize and compute the expected value when
- the dad gets his desire.
- the mom gets her desire.
- Answer
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Since the dad does not want the raffle to make or lose any money on the raffle, the cash prize will need to be all the ticket revenue. Since there are \(80\) tickets being sold at \(\$15\) a ticket. The total ticket revenue is \(\$1200.\) We can construct a random variable similar to the variables above representing the game from the player's perspective.
Table \(\PageIndex{15}\): Probability distribution for the random variable \(R\) along with computation
\(R=r_j\) \(P(R=r_j)\) \(r_j\cdot P(R=r_j)\) \(-15\) \(\dfrac{79}{80}\) \(-15\cdot\dfrac{79}{80}\) \(1185\) \(\dfrac{1}{80}\\) \(1185\cdot\dfrac{1}{80}\) \(\mu=\text{E}(R)=-\dfrac{15\cdot79}{80}+\dfrac{1185}{80}=\dfrac{-1185+1185}{80}=0\) If the dad gets his desire and all the ticket revenue is given as the cash prize, the expected value would \(\$0.\)
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If the mom gets her desire to help cover the cost of hosting the reunion, \(\$10\) of every ticket will go to the cash prize, and \(\$5\) will go towards the reunion. With \(80\) tickets sold, the cash prize would be \(80\cdot 10\) \(=800\) dollars.
Table \(\PageIndex{16}\): Probability distribution for the random variable \(R\) along with computation
\(R=r_j\) \(P(R=r_j)\) \(r_j\cdot P(R=r_j)\) \(-15\) \(\dfrac{79}{80}\) \(-15\cdot\dfrac{79}{80}\) \(785\) \(\dfrac{1}{80}\\) \(785\cdot\dfrac{1}{80}\) \(\mu=\text{E}(R)=-\dfrac{15\cdot79}{80}+\dfrac{785}{80}=\dfrac{-1185+785}{80}=-\dfrac{400}{80}=-5\) If the mom gets her desire, the expected value would \(-\$5.\) We notice that the dad did not want to make any money off the raffle and the expected value was \(\$0,\) while the mom wanted to make \(\$5\) per ticket and the expected value was \(-\$5.\) In general, if the host wanted to make \(\$x\) off of each ticket, then the expected value from the participant's perspective would be \(-\$x.\)
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Since the dad does not want the raffle to make or lose any money on the raffle, the cash prize will need to be all the ticket revenue. Since there are \(80\) tickets being sold at \(\$15\) a ticket. The total ticket revenue is \(\$1200.\) We can construct a random variable similar to the variables above representing the game from the player's perspective.
Having worked through these text exercises, we understand the expected value as a measure of who the game of chance favors: the players or the house/host. Under the mom's desires, the host family making \(\$5\) a ticket is represented in the expected value being \(-\$5.\) If we then apply this idea to the earlier raffles, where the expected values were \(-\$15\) and \(-\$42.50,\) we understand that the hosts were raising money at \(\$15\) and \(\$42.50\) a ticket. This is reasonable because raffles are generally run as fundraisers for some organizations.
Over the years, there have been big lottery prizes, as in millions and billions of dollars. Consider a simplified version of the Powerball game. While in reality, there are \(9\) different ways to win a cash prize from purchasing a \(\$2\) ticket, we will only consider winning the grand prize without any additional options and assume there is only one winner.   The grand prize increases until a winner is found. Determine how large the grand prize must be for the expected value (from the player's perspective) to be nonnegative.
- Answer
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We are trying to find the smallest grand prize so that the expected value is nonnegative. Let us refer to the grand prize amount as \(p\) and set up a discrete random variable \(R\) modeling this simplified version of the Powerball. Refer to this previous text exercise for additional aid in determining the probabilities.
Table \(\PageIndex{17}\): Probability distribution for the random variable \(R\) along with computation
\(R=r_j\) \(P(R=r_j)\) \(r_j\cdot P(R=r_j)\) \(-2\) \(1-\dfrac{1}{292,201,338}\) \(-2+\dfrac{2}{292,201,338}\) \(p-2\) \(\dfrac{1}{292,201,338}\) (\dfrac{p-2}{292,201,338}\) \(\mu=\text{E}(R)=-2+\dfrac{2}{292,201,338}+\dfrac{p}{292,201,338}-\dfrac{2}{292,201,338}=-2+\dfrac{p}{292,201,338}\) We want the expected value to be nonnegative, so we are looking to solve \(\text{E}(R)\geq 0.\) \begin{align*}\text{E}(R)&\geq 0 \\-2+\dfrac{p}{292,201,338}&\geq 0\\ \dfrac{p}{292,201,338}&\geq 2 \\ p&\geq 2\cdot292,201,338=584,402,676\end{align*}Only after the grand prize grows beyond \(\$584,402,676\) will the expected value of the Powerball be positive. Note that this analysis does not consider taxation on the winnings, which is quite hefty.
Given the fact that expected value is negative (unless the pot is enormous), we know that playing the lottery is not a good way to make money.
- Consider a biased die, weighted so that rolling a one is \(5\) times as likely as the other sides, but each of the other sides are equally probable. Determine the probability distribution for the discrete random variable \(Z\) defined to be the value landing up after rolling this biased die.
- Answer
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We know that \(P(Z=2)\) \(=P(Z=3)\) \(=P(Z=4)\) \(=P(Z=5)\) \(=P(Z=6)\), \(P(Z=1)\) \(=5P(Z=2),\) and that the sum of all the probabilities needs to be \(1.\) We have \(1\) \(=P(Z=1)\) \(+P(Z=2)\) \(+P(Z=3)\) \(+P(Z=4)\) \(+P(Z=5)\) \(+P(Z=6)\) \(=5P(Z=2)\) \(+P(Z=2)\) \(+P(Z=2)\) \(+P(Z=2)\) \(+P(Z=2)\) \(+P(Z=2)\) = \(10P(Z=2).\) Thus \(P(Z=2)=\frac{1}{10}\) and \(P(Z=1)=\frac{5}{10}\) \(=\frac{1}{2}.\)
Table \(\PageIndex{18}\): Probability distribution for the random variable \(Z\)
\(Z=z_j\) \(P(Z=z_j)\) \(1\) \(\dfrac{1}{2}\) \(2\) \(\dfrac{1}{10}\) \(3\) \(\dfrac{1}{10}\) \(4\) \(\dfrac{1}{10}\) \(5\) \(\dfrac{1}{10}\) \(6\) \(\dfrac{1}{10}\)
- Compare the expected value and standard deviation of \(Z\) with those of rolling a fair die.
- Answer
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Table \(\PageIndex{19}\): Table of computation for the two random variables
\(Z=z_j\) \(P(Z=z_j)\) \(z_j\cdot P(Z=z_j)\) \((z_j-\mu)^2\cdot P(Z=z_j)\) \(F=f_j\) \(P(F=f_j)\) \(f_j\cdot P(F=f_j)\) \((f_j-\mu)^2\cdot P(F=f_j)\) \(1\) \(\dfrac{1}{2}\) \(\dfrac{1}{2}\) \(\left(1-\dfrac{5}{2}\right)^2\cdot \dfrac{1}{2}=\dfrac{9}{8}\) \(1\) \(\dfrac{1}{6}\) \(\dfrac{1}{6}\) \(\left(1-\dfrac{7}{2}\right)^2\cdot \dfrac{1}{6}=\dfrac{25}{24}\) \(2\) \(\dfrac{1}{10}\) \(\dfrac{1}{5}\) \(\left(2-\dfrac{5}{2}\right)^2\cdot \dfrac{1}{10}=\dfrac{1}{40}\) \(2\) \(\dfrac{1}{6}\) \(\dfrac{1}{3}\) \(\left(2-\dfrac{7}{2}\right)^2\cdot \dfrac{1}{6}=\dfrac{3}{8}\) \(3\) \(\dfrac{1}{10}\) \(\dfrac{3}{10}\) \(\left(3-\dfrac{5}{2}\right)^2\cdot \dfrac{1}{10}=\dfrac{1}{40}\) \(3\) \(\dfrac{1}{6}\) \(\dfrac{1}{2}\) \(\left(3-\dfrac{7}{2}\right)^2\cdot \dfrac{1}{6}=\dfrac{1}{24}\) \(4\) \(\dfrac{1}{10}\) \(\dfrac{2}{5}\) \(\left(4-\dfrac{5}{2}\right)^2\cdot \dfrac{1}{10}=\dfrac{9}{40}\) \(4\) \(\dfrac{1}{6}\) \(\dfrac{2}{3}\) \(\left(4-\dfrac{7}{2}\right)^2\cdot \dfrac{1}{6}=\dfrac{1}{24}\) \(5\) \(\dfrac{1}{10}\) \(\dfrac{1}{2}\) \(\left(5-\dfrac{5}{2}\right)^2\cdot \dfrac{1}{10}=\dfrac{5}{8}\) \(5\) \(\dfrac{1}{6}\) \(\dfrac{5}{6}\) \(\left(5-\dfrac{7}{2}\right)^2\cdot \dfrac{1}{6}=\dfrac{3}{8}\) \(6\) \(\dfrac{1}{10}\) \(\dfrac{3}{5}\) \(\left(6-\dfrac{5}{2}\right)^2\cdot \dfrac{1}{10}=\dfrac{49}{40}\) \(6\) \(\dfrac{1}{6}\) \(1\) \(\left(6-\dfrac{7}{2}\right)^2\cdot \dfrac{1}{6}=\dfrac{25}{24}\) \(\mu=\text{E}(Z)=\dfrac{1}{2}+\dfrac{1}{5}+\ldots+\dfrac{3}{5}=\dfrac{25}{10}=\dfrac{5}{2}=2.5\) \(\mu=\text{E}(F)=\dfrac{1}{6}+\dfrac{1}{3}+\ldots+1=\dfrac{21}{6}=\dfrac{7}{2}=3.5\) \(\sigma^2=\text{Var}(Z)=\dfrac{9}{8}+\dfrac{1}{40}+\dots+\dfrac{49}{40}=\dfrac{130}{40}=\dfrac{13}{4}=3.25\) \(\sigma^2=\text{Var}(F)=\dfrac{25}{24}+\dfrac{3}{8}+\dots+\dfrac{25}{24}=\dfrac{70}{24}=\dfrac{35}{12}\approx2.9167\) \(\sigma=\sqrt{3.25}\approx1.8028\) \(\sigma=\sqrt{\dfrac{35}{12}}\approx1.7078\) In comparing the two random variables, we notice that the bias brought the expected value closer to the biased value. This is relatively intuitive because one would expect the biased value to occur more often. The bias also increased the standard deviation a little. This may be less intuitive because it seems natural that the spread would decrease. After all, we expect ones to appear quite frequently. With a heavy enough bias, this can happen. The degree to which the expected value is pulled towards the biased value and the bias's impact on the standard deviation depends on the amount of bias and the probabilities of the other values.


1.1.2: Discovery of Subatomic Particles
Subatomic Particles
Dalton’s atomic theory stated that different elements were made of different atoms, but did not explain what made the atoms different. The answer is that atoms are composed of subatomic particles, protons, neutron, and electrons, and the number of the protons and electrons in an atom defines the element it represents. So after all, atoms are not so indivisible. However, an atom can still be seen as the smallest particle that represents the full properties of an element. If an atoms gets divided further into subatomic particles then these properties are being lost. To understand atoms further, we must therefore understand the subatomic particles they are made of. The first subatomic particle that was discovered was the electron. It was found through the investigation of so-called cathode rays which were a mysterious phenomenon at the time. Cathode rays occur when a high voltage of about 10-20 kV is applied to an evacuated tube (0.0001 mm Hg). You can see such a tube with a cathode ray going through in Figure 1.1.9. The big question was: What causes these rays?
The answer was found by John Joseph Thomson. He applied electric and magnetic fields to these rays and found them deflected. When an electric field was applied the beam was deflected toward the positive pole. When an equally strong electric and magnetic field was applied, and the magnetic field was perpendicular to the electric field then these fields canceled out, and the rays were no longer deflected. Thomson concluded that these rays must consist of negatively charged particles emitted by the negatively charged electrode because they were attracted by the positive pole of the electric field: Electrons. Knowing the forces associated with the electrical and magnetic fields, he was able to calculate a mass to charge ratio of the electrons which is 5.6857 x 10E-9 g/C. After Thomson, Robert Millikan performed another experiment, the so-called oil drop experiment, that allowed to determine the charge and the mass of the electron. The charge of an electron is -1.6 x 10E-19 C, and its mass is 9.1 x 10E-28 g.
Thomson’s plum pudding atomic model
The discovery of the electron led Thomson to the development of a first atomic model that would include a subatomic particle. It is named plum pudding model (Fig. 1.1.10). In the Thomson atom model electrons are embedded as little particles in a positively charged mass like raisins are embedded in a cake. This atom model is still far from the atom model that we accept today, but it represented an important step forward, because it introduced for the first time the idea that atoms are not indivisible, but contain subatomic particles.
Rutherford's Atom Model
Ernest Rutherford (Fig. 1.1.11) wanted to test Thomson’s plum pudding atom model by experiment. To do so, he placed a photographic film around a gold foil (Figure 1.1.12). In the following he bombarded the gold foil with alpha radiation. Alpha radiation is a form of radioactivity and consists of helium nuclei, also called alpha-particles. It was known at this time the alpha-particles were positively charged and had a large mass compared to an electron. Rutherford’s hypothesis was that if Thomson’s atom model was true, then the alpha particles should all go straight through the gold foil like a bullet going through a plum pudding. Therefore, a blackening of the photographic film, which served as the alpha radiation detector, should only occur directly behind the gold film.
However, what Rutherford observed was that while most of the alpha-particles went straight through the gold foil, some of them were deflected, and a few of them were even reflected. Rutherford concluded that Thomson’s plum pudding model must be wrong. A bullet shot at a plum pudding would just never be reflected by the plum pudding and come back at you.
He suggested that the atom would be made of a positively charged nucleus where almost the entire mass of the atom would be concentrated. In the event of a collision between the alpha particle and that nucleus, the alpha particle would be reflected. In the event of an alpha particle passing by the nucleus close, the alpha particle would be deflected. In the event of an alpha particle passing the nucleus in larger distance, the alpha particle would pass the atom practically non-deflected. The observation that most of the alpha particles were not deflected led Rutherford to the conclusion that the overall atom must be much larger than the nucleus, in fact, the data allowed him to calculate that the radius of the atom would be about 10,000 times larger than the radius of the nucleus. This implies that the atoms would be mostly empty space. To explain the empty space, Rutherford assumed that the electrons would move in orbits around the nucleus like planets around the sun. This would be called the planetary model of the atom (Fig. 1.1.13).


2.2: Purchasing
The purchasing process is an essential part of every food service operation. All competent cooks should be skilled in buying the appropriate ingredients, in accurate amounts, at the right time, and at the best price.
Every kitchen operation has different purchasing procedures. But there is one rule that should always be followed:
Buy only as much as it is anticipated will be needed until the next delivery.
This will ensure that foods stay fresh and will create a high inventory turnover. All foods deteriorate in time, some more quickly than others. It is the job of the purchaser to ensure that only those quantities that will be used immediately or in the near future are purchased.
Market Sourcing
Sources of supply vary considerably from location to location. Large cities have a greater number and variety of suppliers than do small towns and isolated communities. Purchasers should establish contact with available suppliers such as wholesalers, local producers and packers, retailers, cooperative associations, and food importers. In most instances, the person in charge of buying will contact several suppliers to obtain the necessary foods. Some wholesalers diversify their product lines in order to meet all food-related kitchen needs.
Food products are obtained from various sources of supply. For example, a packing house supplies meat and meat products, while a food wholesaler supplies dry goods. Once business is established with a supplier, all transactions should be well documented and kept readily available on file.
There are two major food categories: perishables and non-perishables.
Perishables
Perishable items include fruits, vegetables, fresh fish and shellfish, fresh meats, poultry, and dairy products. As a rule, perishables are bought frequently to ensure freshness. Frozen foods, such as vegetables, fish and meat products, have a longer lifespan and can be ordered less frequently and stored in a freezer.
Non-perishables
Non-perishable items include dry goods, flour, cereals, and miscellaneous items such as olives, pickles, and other condiments. These can be ordered on a weekly or monthly basis.
Keep in mind that just because something does not go bad isn’t a reason to buy it in quantities larger than you need. Every item in your inventory is equal to a dollar amount that you could be saving or spending on something else. Consider that a case of 1000 sheets of parchment paper may cost $250. If you have a case and a half sitting in your inventory, but only use a few sheets a day, that is a lot of money sitting in your storeroom.
Factors That Impact Prices
Food products in particular fluctuate in price over the year, due to many factors:
- Seasonality: When food is in season, there is more of it available in the local food supply, bringing prices down. Additionally, foods in season are usually of higher quality and have longer shelf life than those that are out of season and need to be transported long distances to market.
- Weather: Severe weather can have a huge impact on the cost of food. Drought, flooding, and unseasonable frost have all affected major produce-supplying areas of the world in recent years, causing a rise in prices for many items.
- Costs of transportation: If the cost of fuel or transportation rises, so does the cost of food that needs to travel to market.
- Commodity prices: A number of foods are traded on the commodity market, such as meats and grains. These prices fluctuate as buyers who trade in these products in large volumes buy and sell, much like the stock market.
Before purchasing any food items, ask the following questions.
- When is the item to be used?
- Which supplier has the best price and the best quality? Where an item is purchased should be determined by the price and the quality of the available supplies. When ordering supplies, it is advisable to get prices from at least three sources, then purchase from the supplier who quotes the best price for comparable quality.
- When will the item be delivered? Depending on the distance of the food service establishment from the supplier, delivery may take hours or days. Remember, it is extremely difficult to maintain food quality and consistency if you do not know when your order will be delivered. For this reason, menu planning and a running inventory are two of the most important aspects of purchasing procedures.
Specifications
Meat, seafood, poultry, processed fruits and vegetables, and fresh fruits and vegetables can be ordered under different specification s. For example,
- Meats can be ordered by grade, cut, weight/thickness, fat limitation, age, whether fresh or frozen, and type of packaging.
- Seafood can be ordered by type (e.g., fin fish/shellfish), species, market form, condition, grade, place of origin, whether fresh or frozen, count, size, and packaging,
- Poultry can be ordered by type, grade, class (e.g., broiler, fryer), style (e.g., breasts, wings), size, whether fresh or frozen, and packaging.
- Processed fruits and vegetables can be ordered by grade (sometimes), variety, packaging size and type, drained weight, count per case, packing medium, and whether canned or frozen.
- Fresh fruits and vegetables can be ordered by grade (sometimes), variety, size, weight per container, growing area, and count per container,
Figure 4 shows an example of a purchasing specification sheet that might be kept in a commercial kitchen or receiving area.
Figure 4: Purchasing Specifications
| Beef | Grade | Weight, Size, and Cut Specifications |
|---|---|---|
| Prime rib | Grade AA | 7 kg, fully trimmed |
| New York strip | Grade AAA | 6 kg, bone out, fully trimmed, max. 15 cm width, min. 5 cm depth |
| Tenderloin | Grade AAA | 3 kg, fully trimmed to silverside |
| Roast sirloin | Grade A | 7 kg, boneless butt |
| Short loins | Grade AAA | 6 kg, fully trimmed, 5 cm from eye |
| Pork | Grade | Weight, Size, and Cut Specifications |
|---|---|---|
| Pork leg | Fresh—Canada #1 | 6 kg, oven ready, lean |
| Pork loin | Fresh—Canada #1 | 5-6 kg, trimmed, lean |
| Ham | 6-8 kg, fully cooked, lean, bone in |
| Poultry | Grade | Weight, Size, and Cut Specifications |
|---|---|---|
| Chicken—Frying | Fancy, Eviscerated | 1.5 kg, always fresh |
| Turkey | Fancy, Eviscerated | 9-13 kg |
| Lamb | Grade | Weight, Size, and Cut Specifications |
|---|---|---|
| Legs | Fresh—Canada #1 | 3-5 kg, bone in |
| Lamb loin | 2-3 kg, trimmed with all fat removed |
| Seafood | Grade | Weight, Size, and Cut Specifications |
|---|---|---|
| Shrimp | Jumbo | 24-30/kg, fresh |
| Oysters | Canada #1 | 35/L |
Contract Buying
Some restaurants and hotels, particularly those belonging to chains, will have contracts in place for the purchasing of all products or for certain items. This may mean that the property can only purchase from a specific supplier, but in return it will have negotiated set pricing for the duration of the contract. This has advantages and disadvantages. On the positive side, the contract price remains stable and the job of managing food costs becomes more consistent since there are no price fluctuations. On the negative side, contract buying takes away the opportunity to compare prices between suppliers and take advantage of specials that may be offered.
Additional Resources
- The CFIA (Canadian Food Inspection Agency) has specifications for food labelling, packaging, an so forth
-
These books are great resources for purchase specifications:
- The Visual Food Encyclopedia
- The Visual Food Lover’s Guide : Includes essential information on how to buy, prepare, and store over 1000 types of food
- Chef’s Book of Formulas, Yields and Sizes
Purchasing Procedures
In most kitchens, purchasing and ordering are done by the chef and sous-chefs, although in larger hotels there may be purchasing departments assigned this responsibility. Most kitchens will have a list of suppliers, contacts, delivery dates and schedules, and order sheets with par stock levels to make purchasing easier. For a special function or event, such as a banquet, it may also be necessary to determine the required supplies for that function alone.
Portion Control Chart
To calculate the quantities of food items to be ordered for any size banquet, a portion control chart must be consulted first. Most establishments will have a portion control chart similar to the one shown in Figure 5. The chart indicates the portions to be used per person for any given menu item.
Figure 5: Portion control chart
| Food Item | Menu Item | Portion Size |
|---|---|---|
| Shrimp | Shrimp cocktail | 80 g (2.82 oz.) |
| Lemon | Shrimp cocktail | 1 wedge (6/lemon) |
| Cocktail sauce | Shrimp cocktail | 60 mL (2.11 oz.) |
| Head lettuce | Tossed salad | 1/4 head |
| Tomato | Tossed salad | 1/2 each |
| Dressing | Tossed salad | 60 mL (2.11 oz.) |
| Prime rib, raw, trimmed ready | Prime rib | 500 g (17.6 oz.) |
| Potato | Baked potato | 1 each (100 count) |
| Green beans | Green beans | 80 g (2.82 oz.) |
| Carrots | Carrots | 80 g (2.82 oz.) |
| Strawberries | Fresh strawberries | 100 g (3.52 oz.) |
| Whipping cream | Berries and cream | 60 mL (2.11 oz.) |
| Coffee | Coffee | 500 g (17.6 oz.) for 75 people |
| Coffee cream | Coffee | 60 mL (2.11 oz.) |
One use for a portion control chart is to estimate the quantity of major ingredients and supplies needed to produce a predicted number of menu servings.
You need to prepare shrimp cocktails and prime rib for a 100-person banquet. Using the portion control chart in Figure 5, you can quickly determine what amounts of major ingredients (Figure 6).
Figure 6L Calculating purchase amounts
| Required Servings | Amount to Order |
|---|---|
| 100 x 80 g shrimp | 8000 g or 8 kg (17.6 lbs.) shrimp |
| 100 x 1 wedge of lemon | 100 wedges = 17 lemons (6 wedges per lemon) |
| 100 x 1/4 head of lettuce | 25 heads lettuce |
| 100 x 500 g prime rib raw oven ready | 50 kg (110 lbs.) prime rib |
Purchase Order Chart with Par Levels
The primary purpose for using a purchasing standard is to ensure that sufficient quantities of all food are on hand to meet daily requirements. To establish and maintain these standards, food inventory must become a daily routine. Having set par levels (the amount you should have on hand to get through to the next order) will help in this regard.
There are three main things you need to know:
- Amount required (par level)
- Amount on hand
- Amount to order
To find the amount to order, subtract the amount on hand from the amount required (Figure 7). In some cases, you may have to order a minimum amount based on the package size, so will need to round your quantity up (such as the whole tub of garlic and full cases of mushrooms, apples, and lettuce in Figure 7).
Figure 7: Purchase order chart
| Meats | Amount Required (Par Level) | Amount on Hand | Amount to Order | Actual Order |
|---|---|---|---|---|
|
10 kg | 2 kg | 8 kg | 8 kg |
|
20 kg | 5 kg | 15 kg | 15 kg |
|
10 kg | – | 10 kg | 10 kg |
|
5 kg | 500 g | 4.5 kg | 4.5 kg |
|
10 kg | 3 kg | 7 kg | 7 kg |
| Fish | Amount Required (Par Level) | Amount on Hand | Amount to Order | Actual Order |
|---|---|---|---|---|
|
25 kg | 5 kg | 20 kg | 20 kg |
| Vegetables | Amount Required (Par Level) | Amount on Hand | Amount to Order | Actual Order |
|---|---|---|---|---|
|
2 kg tub | 250 g | 1.750 kg | 2 kg tub |
|
5 kg case | 500 g | 4.5 kg | 5 kg case |
|
2 cases (24/case) | 12 (1/2 case) | 1 1/2 cases | 2 cases |
| Fruits | Amount Required (Par Level) | Amount on Hand | Amount to Order | Actual Order |
|---|---|---|---|---|
|
2 cases | 1/2 case | 1 1/2 cases | 2 cases |
|
10 kg | – | 10 kg | |
|
1 case | 2 cases | – | – |
Integrating these par levels into your regular ordering sheets or your ordering system will make it very easy to manage inventory coming in.
More and more suppliers are moving to online ordering systems, which have current prices, case sizes, and often your purchase history available to you when placing an order. Online ordering can often be more convenient as the person placing the order does not have to make calls into an order desk during regular office hours.