12.5: Conic Sections in Polar Coordinates
- Identify a conic in polar form.
- Graph the polar equations of conics.
- Define conics in terms of a focus and a directrix.
Most of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planets’ orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits.
In an elliptical orbit, the periapsis is the point at which the two objects are closest, and the apoapsis is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial body’s gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system.
Identifying a Conic in Polar Form
Any conic may be determined by three characteristics: a single focus , a fixed line called the directrix , and the ratio of the distances of each to a point on the graph. Consider the parabola \(x=2+y^2\) shown in Figure \(\PageIndex{2}\).
We previously learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus \(P(r,\theta)\) at the pole, and a line, the directrix, which is perpendicular to the polar axis.
If \(F\) is a fixed point, the focus, and \(D\) is a fixed line, the directrix, then we can let \(e\) be a fixed positive number, called the eccentricity , which we can define as the ratio of the distances from a point on the graph to the focus and the point on the graph to the directrix. Then the set of all points \(P\) such that \(e=\dfrac{PF}{PD}\) is a conic. In other words, we can define a conic as the set of all points \(P\) with the property that the ratio of the distance from \(P\) to \(F\) to the distance from \(P\) to \(D\) is equal to the constant \(e\).
For a conic with eccentricity \(e\),
- if \(0≤e<1\), the conic is an ellipse
- if \(e=1\), the conic is a parabola
- if \(e>1\), the conic is an hyperbola
With this definition, we may now define a conic in terms of the directrix, \(x=\pm p\), the eccentricity \(e\), and the angle \(\theta\). Thus, each conic may be written as a polar equation , an equation written in terms of \(r\) and \(\theta\).
For a conic with a focus at the origin, if the directrix is \(x=\pm p\), where \(p\) is a positive real number, and the eccentricity is a positive real number \(e\), the conic has a polar equation
\[r=\dfrac{ep}{1\pm e \cos \theta}\]
For a conic with a focus at the origin, if the directrix is \(y=\pm p\), where \(p\) is a positive real number, and the eccentricity is a positive real number \(e\), the conic has a polar equation
\[r=\dfrac{ep}{1\pm e \sin \theta}\]
- Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.
- Identify the eccentricity \(e\) as the coefficient of the trigonometric function in the denominator.
- Compare \(e\) with \(1\) to determine the shape of the conic.
- Determine the directrix as \(x=p\) if cosine is in the denominator and \(y=p\) if sine is in the denominator. Set \(ep\) equal to the numerator in standard form to solve for \(x\) or \(y\).
For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity.
- \(r=\dfrac{6}{3+2 \sin \theta}\)
- \(r=\dfrac{12}{4+5 \cos \theta}\)
- \(r=\dfrac{7}{2−2 \sin \theta}\)
Solution
For each of the three conics, we will rewrite the equation in standard form. Standard form has a \(1\) as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and denominator by the reciprocal of the constant of the original equation, \(\dfrac{1}{c}\), where \(c\) is that constant.
- Multiply the numerator and denominator by \(\dfrac{1}{3}\).
\(r=\dfrac{6}{3+2\sin \theta}⋅\dfrac{\left(\dfrac{1}{3}\right)}{\left(\dfrac{1}{3}\right)}=\dfrac{6\left(\dfrac{1}{3}\right)}{3\left(\dfrac{1}{3}\right)+2\left(\dfrac{1}{3}\right)\sin \theta}=\dfrac{2}{1+\dfrac{2}{3} \sin \theta}\)
Because \(\sin \theta\) is in the denominator, the directrix is \(y=p\). Comparing to standard form, note that \(e=\dfrac{2}{3}\).Therefore, from the numerator,
\[\begin{align*} 2&=ep\\ 2&=\dfrac{2}{3}p\\ \left(\dfrac{3}{2}\right)2&=\left(\dfrac{3}{2}\right)\dfrac{2}{3}p\\ 3&=p \end{align*}\]
Since \(e<1\), the conic is an ellipse . The eccentricity is \(e=\dfrac{2}{3}\) and the directrix is \(y=3\).
- Multiply the numerator and denominator by \(\dfrac{1}{4}\).
\[\begin{align*} r&=\dfrac{12}{4+5 \cos \theta}\cdot \dfrac{\left(\dfrac{1}{4}\right)}{\left(\dfrac{1}{4}\right)}\\ r&=\dfrac{12\left(\dfrac{1}{4}\right)}{4\left(\dfrac{1}{4}\right)+5\left(\dfrac{1}{4}\right)\cos \theta}\\ r&=\dfrac{3}{1+\dfrac{5}{4} \cos \theta} \end{align*}\]
Because \(\cos \theta\) is in the denominator, the directrix is \(x=p\). Comparing to standard form, \(e=\dfrac{5}{4}\). Therefore, from the numerator,
\[\begin{align*} 3&=ep\\ 3&=\dfrac{5}{4}p\\ \left(\dfrac{4}{5}\right)3&=\left(\dfrac{4}{5}\right)\dfrac{5}{4}p\\ \dfrac{12}{5}&=p \end{align*}\]
Since \(e>1\), the conic is a hyperbola . The eccentricity is \(e=\dfrac{5}{4}\) and the directrix is \(x=\dfrac{12}{5}=2.4\).
- Multiply the numerator and denominator by \(\dfrac{1}{2}\).
\[\begin{align*} r&=\dfrac{7}{2-2 \sin \theta}\cdot \dfrac{\left(\dfrac{1}{2}\right)}{\left(\dfrac{1}{2}\right)}\\ r&=\dfrac{7\left(\dfrac{1}{2}\right)}{2\left(\dfrac{1}{2}\right)-2\left(\dfrac{1}{2}\right) \sin \theta}\\ r&=\dfrac{\dfrac{7}{2}}{1-\sin \theta} \end{align*}\]
Because sine is in the denominator, the directrix is \(y=−p\). Comparing to standard form, \(e=1\). Therefore, from the numerator,
\[\begin{align*} \dfrac{7}{2}&=ep\\ \dfrac{7}{2}&=(1)p\\ \dfrac{7}{2}&=p \end{align*}\]
Because \(e=1\), the conic is a parabola. The eccentricity is \(e=1\) and the directrix is \(y=−\dfrac{7}{2}=−3.5\).
Identify the conic with focus at the origin, the directrix, and the eccentricity for \(r=\dfrac{2}{3−\cos \theta}\).
- Answer
-
ellipse; \(e=\dfrac{1}{3}\); \(x=−2\)
Graphing the Polar Equations of Conics
When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with \(1\). This enables us to determine \(e\) and, therefore, the shape of the curve. The next step is to substitute values for \(\theta\) and solve for \(r\) to plot a few key points. Setting \(\theta\) equal to \(0\), \(\dfrac{\pi}{2}\), \(\pi\), and \(\dfrac{3\pi}{2}\) provides the vertices so we can create a rough sketch of the graph.
Graph \(r=\dfrac{5}{3+3 \cos \theta}\).
Solution
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of \(3\), which is \(\dfrac{1}{3}\).
\[\begin{align*} r &= \dfrac{5}{3+3 \cos \theta}=\dfrac{5\left(\dfrac{1}{3}\right)}{3\left(\dfrac{1}{3}\right)+3\left(\dfrac{1}{3}\right)\cos \theta} \\ r &= \dfrac{\dfrac{5}{3}}{1+\cos \theta} \end{align*}\]
Because \(e=1\),we will graph a parabola with a focus at the origin. The function has a \(\cos \theta\), and there is an addition sign in the denominator, so the directrix is \(x=p\).
\[\begin{align*} \dfrac{5}{3}&=ep\\ \dfrac{5}{3}&=(1)p\\ \dfrac{5}{3}&=p \end{align*}\]
The directrix is \(x=\dfrac{5}{3}\).
Plotting a few key points as in Table \(\PageIndex{1}\) will enable us to see the vertices. See Figure \(\PageIndex{3}\).
| A | B | C | D | |
|---|---|---|---|---|
| \(\theta\) | \(0\) | \(\dfrac{\pi}{2}\) | \(\pi\) | \(\dfrac{3\pi}{2}\) |
| \(r=\dfrac{5}{3+3 \cos \theta}\) | \(\dfrac{5}{6}≈0.83\) | \(\dfrac{5}{3}≈1.67\) | undefined | \(\dfrac{5}{3}≈1.67\) |
We can check our result with a graphing utility. See Figure \(\PageIndex{4}\).
Graph \(r=\dfrac{8}{2−3 \sin \theta}\).
Solution
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of \(2\), which is \(\dfrac{1}{2}\).
\[\begin{align*} r &=\dfrac{8}{2−3\sin \theta}=\dfrac{8\left(\dfrac{1}{2}\right)}{2\left(\dfrac{1}{2}\right)−3\left(\dfrac{1}{2}\right)\sin \theta} \\ r &= \dfrac{4}{1−\dfrac{3}{2} \sin \theta} \end{align*}\]
Because \(e=\dfrac{3}{2}\), \(e>1\), so we will graph a hyperbola with a focus at the origin. The function has a \(\sin \theta\) term and there is a subtraction sign in the denominator, so the directrix is \(y=−p\).
\[\begin{align*} 4&=ep\\ 4&=\left(\dfrac{3}{2}\right)p\\ 4\left(\dfrac{2}{3}\right)&=p\\ \dfrac{8}{3}&=p \end{align*}\]
The directrix is \(y=−\dfrac{8}{3}\).
Plotting a few key points as in Table \(\PageIndex{2}\) will enable us to see the vertices. See Figure \(\PageIndex{5}\).
| A | B | C | D | |
|---|---|---|---|---|
| \(\theta\) | \(0\) | \(\dfrac{\pi}{2}\) | \(\pi\) | \(\dfrac{3\pi}{2}\) |
| \(r=\dfrac{8}{2−3\sin \theta}\) | \(4\) | \(−8\) | \(4\) | \(\dfrac{8}{5}=1.6\) |
Graph \(r=\dfrac{10}{5−4 \cos \theta}\).
Solution
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is \(\dfrac{1}{5}\).
\[\begin{align*} r &= \dfrac{10}{5−4\cos \theta}=\dfrac{10\left(\dfrac{1}{5}\right)}{5\left(\dfrac{1}{5}\right)−4\left(\dfrac{1}{5}\right)\cos \theta} \\ r &= \dfrac{2}{1−\dfrac{4}{5} \cos \theta} \end{align*}\]
Because \(e=\dfrac{4}{5}\), \(e<1\), so we will graph an ellipse with a focus at the origin. The function has a \(\cos \theta\), and there is a subtraction sign in the denominator, so the directrix is \(x=−p\).
\[\begin{align*} 2&=ep\\ 2&=\left(\dfrac{4}{5}\right)p\\ 2\left(\dfrac{5}{4}\right)&=p\\ \dfrac{5}{2}&=p \end{align*}\]
The directrix is \(x=−\dfrac{5}{2}\).
Plotting a few key points as in Table \(\PageIndex{3}\) will enable us to see the vertices. See Figure \(\PageIndex{6}\).
| A | B | C | D | |
|---|---|---|---|---|
| \(\theta\) | \(0\) | \(\dfrac{\pi}{2}\) | \(\pi\) | \(\dfrac{3\pi}{2}\) |
| \(r=\dfrac{10}{5−4 \cos \theta}\) | \(10\) | \(2\) | \(\dfrac{10}{9}≈1.1\) | \(2\) |
Analysis
We can check our result using a graphing utility. See Figure \(\PageIndex{7}\).
Graph \(r=\dfrac{2}{4−\cos \theta}\).
- Answer
Defining Conics in Terms of a Focus and a Directrix
So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation.
- Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of \(y\), we use the general polar form in terms of sine. If the directrix is given in terms of \(x\), we use the general polar form in terms of cosine.
- Determine the sign in the denominator. If \(p<0\), use subtraction. If \(p>0\), use addition.
- Write the coefficient of the trigonometric function as the given eccentricity.
- Write the absolute value of \(p\) in the numerator, and simplify the equation.
Find the polar form of the conic given a focus at the origin, \(e=3\) and directrix \(y=−2\).
Solution
The directrix is \(y=−p\), so we know the trigonometric function in the denominator is sine.
Because \(y=−2\), \(–2<0\), so we know there is a subtraction sign in the denominator. We use the standard form of
\(r=\dfrac{ep}{1−e \sin \theta}\)
and \(e=3\) and \(|−2|=2=p\).
Therefore,
\[\begin{align*} r&=\dfrac{(3)(2)}{1-3 \sin \theta}\\ r&=\dfrac{6}{1-3 \sin \theta} \end{align*}\]
Find the polar form of a conic given a focus at the origin, \(e=\dfrac{3}{5}\), and directrix \(x=4\).
Solution
Because the directrix is \(x=p\), we know the function in the denominator is cosine. Because \(x=4\), \(4>0\), so we know there is an addition sign in the denominator. We use the standard form of
\(r=\dfrac{ep}{1+e \cos \theta}\)
and \(e=\dfrac{3}{5}\) and \(|4|=4=p\).
Therefore,
\[\begin{align*} r &= \dfrac{\left(\dfrac{3}{5}\right)(4)}{1+\dfrac{3}{5}\cos\theta} \\ r &= \dfrac{\dfrac{12}{5}}{1+\dfrac{3}{5}\cos\theta} \\ r &=\dfrac{\dfrac{12}{5}}{1\left(\dfrac{5}{5}\right)+\dfrac{3}{5}\cos\theta} \\ r &=\dfrac{\dfrac{12}{5}}{\dfrac{5}{5}+\dfrac{3}{5}\cos\theta} \\ r &= \dfrac{12}{5}⋅\dfrac{5}{5+3\cos\theta} \\ r &=\dfrac{12}{5+3\cos\theta} \end{align*}\]
Find the polar form of the conic given a focus at the origin, \(e=1\), and directrix \(x=−1\).
- Answer
-
\(r=\dfrac{1}{1−\cos\theta}\)
Convert the conic \(r=\dfrac{1}{5−5\sin \theta}\) to rectangular form.
Solution
We will rearrange the formula to use the identities \(r=\sqrt{x^2+y^2}\), \(x=r \cos \theta\),and \(y=r \sin \theta\).
\[\begin{align*} r&=\dfrac{1}{5-5 \sin \theta} \\ r\cdot (5-5 \sin \theta)&=\dfrac{1}{5-5 \sin \theta}\cdot (5-5 \sin \theta)\qquad \text{Eliminate the fraction.} \\ 5r-5r \sin \theta&=1 \qquad \text{Distribute.} \\ 5r&=1+5r \sin \theta \qquad \text{Isolate }5r. \\ 25r^2&={(1+5r \sin \theta)}^2 \qquad \text{Square both sides. } \\ 25(x^2+y^2)&={(1+5y)}^2 \qquad \text{Substitute } r=\sqrt{x^2+y^2} \text{ and }y=r \sin \theta. \\ 25x^2+25y^2&=1+10y+25y^2 \qquad \text{Distribute and use FOIL. } \\ 25x^2-10y&=1 \qquad \text{Rearrange terms and set equal to 1.} \end{align*}\]
Convert the conic \(r=\dfrac{2}{1+2 \cos \theta}\) to rectangular form.
- Answer
-
\(4−8x+3x^2−y^2=0\)
Access these online resources for additional instruction and practice with conics in polar coordinates.
- Polar Equations of Conic Sections
- Graphing Polar Equations of Conics - 1
- Graphing Polar Equations of Conics - 2
Visit this website for additional practice questions from Learningpod.
Key Concepts
- Any conic may be determined by a single focus, the corresponding eccentricity, and the directrix. We can also define a conic in terms of a fixed point, the focus \(P(r,\theta)\) at the pole, and a line, the directrix, which is perpendicular to the polar axis.
- A conic is the set of all points \(e=\dfrac{PF}{PD}\), where eccentricity \(e\) is a positive real number. Each conic may be written in terms of its polar equation. See Example \(\PageIndex{1}\).
- The polar equations of conics can be graphed. See Example \(\PageIndex{2}\), Example \(\PageIndex{3}\), and Example \(\PageIndex{4}\).
- Conics can be defined in terms of a focus, a directrix, and eccentricity. See Example \(\PageIndex{5}\) and Example \(\PageIndex{6}\).
- We can use the identities \(r=\sqrt{x^2+y^2}\), \(x=r \cos \theta\),and \(y=r \sin \theta\) to convert the equation for a conic from polar to rectangular form. See Example \(\PageIndex{7}\).


An allergic reaction is an immune response to a type of antigen called an
allergen
. Allergens can be found in many different items, from peanuts and insect stings to latex and some drugs. Unlike other kinds of antigens, allergens are not necessarily associated with pathogenic microbes, and many allergens provoke no immune response at all in most people.
Allergic responses vary in severity. Some are mild and localized, like hay fever or hives, but others can result in systemic, life-threatening reactions. Anaphylaxis, for example, is a rapidly developing allergic reaction that can cause a dangerous drop in blood pressure and severe swelling of the throat that may close off the airway.
Allergies are just one example of how the immune system—the system normally responsible for preventing disease—can actually cause or mediate disease symptoms. In this chapter, we will further explore allergies and other disorders of the immune system, including hypersensitivity reactions, autoimmune diseases, transplant rejection, and diseases associated with immunodeficiency.
An allergy is an adaptive immune response, sometimes life-threatening, to an allergen. Hypersensitivity reactions are classified by their immune mechanism.
An allergy is an adaptive immune response, sometimes life-threatening, to an allergen. Hypersensitivity reactions are classified by their immune mechanism.
An allergy is an adaptive immune response, sometimes life-threatening, to an allergen. Hypersensitivity reactions are classified by their immune mechanism.
An allergy is an adaptive immune response, sometimes life-threatening, to an allergen. Hypersensitivity reactions are classified by their immune mechanism.
An allergy is an adaptive immune response, sometimes life-threatening, to an allergen. Hypersensitivity reactions are classified by their immune mechanism.
An allergy is an adaptive immune response, sometimes life-threatening, to an allergen. Hypersensitivity reactions are classified by their immune mechanism.
Thumbnail: Allergens in plant pollen, shown here in a colorized electron micrograph, may trigger allergic rhinitis or hay fever in sensitive individuals. (Public Domain/modified from original; Dartmouth Electron Microscope Facility, Dartmouth College via
Wikimedia Commons
).


6.1: Conclusion
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World War I decimated millions and profoundly altered the course of world history. Postwar instabilities led directly toward a global depression and a second world war. The war sparked the Bolshevik Revolution, which led to the Soviet Union and later the Cold War. It created Middle Eastern nations and aggravated ethnic tensions that the United States could never overcome. And the United States had fought on the European mainland as a major power. America’s place in the world was never the same. By whipping up nationalist passions, American attitudes toward radicalism, dissent, and immigration were poisoned. Postwar disillusionment shattered Americans’ hopes for the progress of the modern world.


17: Sound
Sound is an example of a mechanical wave, specifically, a pressure wave: Sound waves travel through the air and other media as oscillations of molecules. Normal human hearing encompasses an impressive range of frequencies from 20 Hz to 20 kHz. Sounds below 20 Hz are called infrasound , whereas those above 20 kHz are called ultrasound . Some animals, like the bat, can hear sounds in the ultrasonic range. Many of the concepts covered in Waves also have applications in the study of sound. For example, when a sound wave encounters an interface between two media with different wave speeds, reflection and transmission of the wave occur. Ultrasound has many uses in science, engineering, and medicine. Ultrasound is used for nondestructive testing in engineering, such as testing the thickness of coating on metal. In medicine, sound waves are far less destructive than X-rays and can be used to image the fetus in a mother’s womb without danger to the fetus or the mother. Later in this chapter, we discuss the Doppler effect, which can be used to determine the velocity of blood in the arteries or wind speed in weather systems.
-
- 17.1: Prelude to Sound
- Sound is an example of a mechanical wave, specifically, a pressure wave: Sound waves travel through the air and other media as oscillations of molecules. Normal human hearing encompasses an impressive range of frequencies from 20 Hz to 20 kHz. Sounds below 20 Hz are called infrasound, whereas those above 20 kHz are called ultrasound. Some animals, like bats, can hear sounds in the ultrasonic range.
-
- 17.2: Sound Waves
- Sound is a disturbance of matter (a pressure wave) that is transmitted from its source outward. Hearing is the perception of sound. Sound can be modeled in terms of pressure or in terms of displacement of molecules. The human ear is sensitive to frequencies between 20 Hz and 20 kHz.
-
- 17.3: Speed of Sound
- The speed of sound depends on the medium and the state of the medium. In a fluid, because the absence of shear forces, sound waves are longitudinal. A solid can support both longitudinal and transverse sound waves. The speed of sound is the same for all frequencies and wavelengths of sound in air.
-
- 17.4: Sound Intensity
- Intensity is the same for a sound wave as was defined for all waves, where P is the power crossing area A. The SI unit for I is watts per meter squared. Sound intensity level in units of decibels (dB) is equal to the logarithm of the ratio between the intensity heard and the threshold intensity for hearing. The perception of frequency is pitch. The perception of intensity is loudness and loudness has units of phons.
-
- 17.5: Normal Modes of a Standing Sound Wave
- Unwanted sound can be reduced using destructive interference. Sound has the same properties of interference and resonance as defined for all waves. In air columns, the lowest-frequency resonance is called the fundamental, whereas all higher resonant frequencies are called overtones. Collectively, they are called harmonics.
-
- 17.6: Sources of Musical Sound
- Some musical instruments can be modeled as pipes that have symmetrical boundary conditions: open/closed at both ends. Others can be modeled as pipes that have anti-symmetrical boundary conditions: closed at one end and open at the other. String instruments produce sound using a vibrating string with nodes at each end. The air around the string oscillates at the string's frequency. The relationship for the frequencies for the string is similar to the symmetrical boundary conditions of the pipe.
-
- 17.9: Shock Waves
- The Mach number is the velocity of a source divided by the speed of sound. When a sound source moves faster than the speed of sound, a shock wave is produced as the sound waves interfere. A sonic boom is the intense sound that occurs as the shock wave moves along the ground. A bow wake is produced when an object moves faster than the speed of a mechanical wave in the medium, such as a boat moving through the water.
Thumbnail Figure 17.1 - Hearing is an important human sense that can detect frequencies of sound, ranging between 20 Hz and 20 kHz. However, other species have very different ranges of hearing. Bats, for example, emit clicks in ultrasound, using frequencies beyond 20 kHz. They can detect nearby insects by hearing the echo of these ultrasonic clicks. Ultrasound is important in several human applications, including probing the interior structures of human bodies, Earth, and the Sun. Ultrasound is also useful in industry for nondestructive testing. (credit: modification of work by Angell Williams)


5.2: Binomial Probability Distribution
Section 5.1 introduced the concept of a probability distribution. The focus of the section was on discrete probability distributions (pdf). To find the pdf for a situation, you usually needed to actually conduct the experiment and collect data. Then you can calculate the experimental probabilities. Normally you cannot calculate the theoretical probabilities instead. However, there are certain types of experiment that allow you to calculate the theoretical probability. One of those types is called a Binomial Experiment .
Properties of a binomial experiment (or Bernoulli trial)
- Fixed number of trials, n , which means that the experiment is repeated a specific number of times.
- The n trials are independent, which means that what happens on one trial does not influence the outcomes of other trials.
- There are only two outcomes, which are called a success and a failure.
- The probability of a success doesn’t change from trial to trial, where p = probability of success and q = probability of failure, q = 1- p .
If you know you have a binomial experiment, then you can calculate binomial probabilities. This is important because binomial probabilities come up often in real life. Examples of binomial experiments are:
- Toss a fair coin ten times, and find the probability of getting two heads.
- Question twenty people in class, and look for the probability of more than half being women?
- Shoot five arrows at a target, and find the probability of hitting it five times?
To develop the process for calculating the probabilities in a binomial experiment, consider Example \(\PageIndex{1}\).
Example \(\PageIndex{1}\): Deriving the Binomial Probability Formula
Suppose you are given a 3 question multiple-choice test. Each question has 4 responses and only one is correct. Suppose you want to find the probability that you can just guess at the answers and get 2 questions right. (Teachers do this all the time when they make up a multiple-choice test to see if students can still pass without studying. In most cases the students can’t.) To help with the idea that you are going to guess, suppose the test is in Martian.
- What is the random variable?
- Is this a binomial experiment?
- What is the probability of getting 2 questions right?
- What is the probability of getting zero right, one right, and all three right?
Solution
a. x = number of correct answers
b.
- There are 3 questions, and each question is a trial, so there are a fixed number of trials. In this case, n = 3.
- Getting the first question right has no affect on getting the second or third question right, thus the trials are independent.
- Either you get the question right or you get it wrong, so there are only two outcomes. In this case, the success is getting the question right.
- The probability of getting a question right is one out of four. This is the same for every trial since each question has 4 responses. In this case, \(p=\dfrac{1}{4} \text { and } q=1-\dfrac{1}{4}=\dfrac{3}{4}\)
This is a binomial experiment, since all of the properties are met.
c. To answer this question, start with the sample space. SS = {RRR, RRW, RWR, WRR, WWR, WRW, RWW, WWW}, where RRW means you get the first question right, the second question right, and the third question wrong. The same is similar for the other outcomes.
Now the event space for getting 2 right is {RRW, RWR, WRR}. What you did in chapter four was just to find three divided by eight. However, this would not be right in this case. That is because the probability of getting a question right is different from getting a question wrong. What else can you do?
Look at just P(RRW) for the moment. Again, that means P(RRW) = P(R on 1st, R on 2nd, and W on 3rd)
Since the trials are independent, then P(RRW) = P(R on 1st, R on 2nd, and W on 3rd) = P(R on 1st) * P(R on 2nd) * P(W on 3rd)
Just multiply p * p * q
\(P(\mathrm{RRW})=\dfrac{1}{4} * \dfrac{1}{4} * \dfrac{3}{4}=\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1}\)
The same is true for P(RWR) and P(WRR). To find the probability of 2 correct answers, just add these three probabilities together. You get
\(\begin{aligned} P(2 \text { correct answers }) &=P(\mathrm{RRW})+P(\mathrm{RWR})+P(\mathrm{WRR}) \\ &=\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1}+\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1}+\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1} \\ &=3\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1} \end{aligned}\)
d. You could go through the same argument that you did above and come up with the following:
| r right | P(r right) |
|---|---|
| 0 right | \(1^{*}\left(\dfrac{1}{4}\right)^{0}\left(\dfrac{3}{4}\right)^{3}\) |
| 1 right | \(3^{*}\left(\dfrac{1}{4}\right)^{1}\left(\dfrac{3}{4}\right)^{2}\) |
| 2 right | \(3 *\left(\dfrac{1}{4}\right)^{2}\left(\dfrac{3}{4}\right)^{1}\) |
| 3 right | \(1^{*}\left(\dfrac{1}{4}\right)^{3}\left(\dfrac{3}{4}\right)^{0}\) |
Hopefully you see the pattern that results. You can now write the general formula for the probabilities for a Binomial experiment
First, the random variable in a binomial experiment is x = number of successes. Be careful, a success is not always a good thing. Sometimes a success is something that is bad, like finding a defect. A success just means you observed the outcome you wanted to see happen.
Definition \(\PageIndex{1}\)
Binomial Formula for the probability of r successes in n trials is
\(P(x=r)=_{n} C_{r} p^{r} q^{n \cdot r} \text { where }_{n} C_{r}=\dfrac{n !}{r !(n-r) !}\)
The \(_{n} C_{r}\) is the number of combinations of n things taking r at a time. It is read “ n choose r ”. Some other common notations for n choose r are \(C_{n, r}\), and \(\left( \begin{array}{l}{n} \\ {r}\end{array}\right)\). n ! means you are multiplying \(n^{*}(n-1)^{*}(n-2)^{*} \dots^{*} 2^{*} 1\). As an example, \(5 !=5^{*} 4^{*} 3^{*} 2^{*} 1=120\).
When solving problems, make sure you define your random variable and state what n, p, q, and r are. Without doing this, the problems are a great deal harder.
Example \(\PageIndex{2}\): Calculating Binomial Probabilities
When looking at a person’s eye color, it turns out that 1% of people in the world has green eyes ("What percentage of," 2013). Consider a group of 20 people.
- State the random variable.
- Argue that this is a binomial experiment.
- Find the probability that none have green eyes.
- Find the probability that nine have green eyes.
- Find the probability that at most three have green eyes.
- Find the probability that at most two have green eyes.
- Find the probability that at least four have green eyes.
- In Europe, four people out of twenty have green eyes. Is this unusual? What does that tell you?
Solution
a. x = number of people with green eyes
b.
- There are 20 people, and each person is a trial, so there are a fixed number of trials. In this case, \(n = 20\).
- If you assume that each person in the group is chosen at random the eye color of one person doesn’t affect the eye color of the next person, thus the trials are independent.
- Either a person has green eyes or they do not have green eyes, so there are only two outcomes. In this case, the success is a person has green eyes.
- The probability of a person having green eyes is 0.01. This is the same for every trial since each person has the same chance of having green eyes. p = 0.01 and q = 1 - 0.01 = 0.99
c. \(P(x=0)=_{20} C_{0}(0.01)^{0}(0.99)^{20-0} \approx 0.818\)
d. \(P(x=9)=_{20} C_{9}(0.01)^{9}(0.99)^{20-9} \approx 1.50 \times 10^{-13} \approx 0.000\)
e. At most three means that three is the highest value you will have. Find the probability of x is less than or equal to three.
\(\begin{aligned} P(x \leq 3) &=P(x=0)+P(x=1)+P(x=2)+P(x=3) \\ &=_{20} C_{0}(0.01)^{0}(0.99)^{20}+_{20} C_{1}(0.01)^{1}(0.99)^{19} \\& +_{20}C_{2}(0.01)^{2}(0.99)^{18}+_{20}C_{3}(0.01)^{3}(0.99)^{17} \\ & \approx 0.818+0.165+0.016+0.001>0.999 \end{aligned}\)
The reason the answer is written as being greater than 0.999 is because the answer is actually 0.9999573791, and when that is rounded to three decimal places you get 1. But 1 means that the event will happen, when in reality there is a slight chance that it won’t happen. It is best to write the answer as greater than 0.999 to represent that the number is very close to 1, but isn’t 1.
f.
\(\begin{aligned} P(x \leq 2) &=P(x=0)+P(x=1)+P(x=2) \\ &=_{20} C_{0}(0.01)^{0}(0.99)^{20}+_{20} C_{1}(0.01)^{1}(0.99)^{19}+_{20} C_{2}(0.01)^{2}(0.99)^{18} \\ & \approx 0.818+0.165+0.016 \approx 0.999 \end{aligned}\)
g. At least four means four or more. Find the probability of x being greater than or equal to four. That would mean adding up all the probabilities from four to twenty. This would take a long time, so it is better to use the idea of complement. The complement of being greater than or equal to four is being less than four. That would mean being less than or equal to three. Part (e) has the answer for the probability of being less than or equal to three. Just subtract that number from 1.
\(P(x \geq 4)=1-P(x \leq 3)=1-0.999=0.001\)
Actually the answer is less than 0.001, but it is fine to write it this way.
h. Since the probability of finding four or more people with green eyes is much less than 0.05, it is unusual to find four people out of twenty with green eyes. That should make you wonder if the proportion of people in Europe with green eyes is more than the 1% for the general population. If this is true, then you may want to ask why Europeans have a higher proportion of green-eyed people. That of course could lead to more questions.
The binomial formula is cumbersome to use, so you can find the probabilities by using technology. On the TI-83/84 calculator, the commands on the TI-83/84 calculators when the number of trials is equal to n and the probability of a success is equal to p are \(\text{binompdf}(n, p, r)\) when you want to find P (x=r) and \(\text{binomcdf}(n, p, r)\) when you want to find \(P(x \leq r)\). If you want to find \(P(x \geq r)\), then you use the property that \(P(x \geq r)=1-P(x \leq r-1)\), since \(x \geq r\) and \(x<r\) or \(x \leq r-1\) are complementary events. Both binompdf and binomcdf commands are found in the DISTR menu. Using R, the commands are \(P(x=r)=\text { dbinom }(r, n, p) \text { and } P(x \leq r)=\text { pbinom }(r, n, p)\).
Example \(\PageIndex{3}\) using the binomial command on the ti-83/84
When looking at a person’s eye color, it turns out that 1% of people in the world has green eyes ("What percentage of," 2013). Consider a group of 20 people.
- State the random variable.
- Find the probability that none have green eyes.
- Find the probability that nine have green eyes.
- Find the probability that at most three have green eyes.
- Find the probability that at most two have green eyes.
- Find the probability that at least four have green eyes.
Solution
a. x = number of people with green eyes
b. You are looking for P (x=0). Since this problem is x=0, you use the binompdf command on the TI-83/84 or dbinom command on R. On the TI83/84, you go to the DISTR menu, select the binompdf, and then type into the parenthesis your n , p , and r values into your calculator, making sure you use the comma to separate the values. The command will look like \(\text{binompdf}(20,.01,0)\) and when you press ENTER you will be given the answer. (If you have the new software on the TI-84, the screen looks a bit different.)
Figure \(\PageIndex{1}\): Calculator Results for binompdfOn R, the command would look like dbinom(0, 20, 0.01)
P (x=0) = 0.8179. Thus there is an 81.8% chance that in a group of 20 people none of them will have green eyes.
c. In this case you want to find the P (x=9). Again, you will use the binompdf command or the dbinom command. Following the procedure above, you will have binompdf(20, .01, 9) on the TI-83/84 or dbinom(9,20,0.01) on R. Your answer is \(P(x=9)=1.50 \times 10^{-13}\). (Remember when the calculator gives you \(1.50 E-13\) and R give you \(1.50 e-13\), this is how they display scientific notation.) The probability that out of twenty people, nine of them have green eyes is a very small chance.
d. At most three means that three is the highest value you will have. Find the probability of x being less than or equal to three, which is \(P(x \leq 3)\). This uses the binomcdf command on the TI-83/84 and pbinom command in R. You use the command on the TI-83/84 of binomcdf(20, .01, 3) and the command on R of pbinom(3,20,0.01)
Figure \(\PageIndex{2}\): Calculator Results for binomcdfYour answer is 0.99996. Thus there is a really good chance that in a group of 20 people at most three will have green eyes. (Note: don’t round this to one, since one means that the event will happen, when in reality there is a slight chance that it won’t happen. It is best to write the answer out to enough decimal points so it doesn’t round off to one.
e. You are looking for \(P(x \leq 2)\). Again use binomcdf or pbinom. Following the procedure above you will have \(\text{binomcdf}(20,.01,2)\) on the TI-83/84 and pbinom(2,20,0.01), with \(P(x \leq 2)=0.998996\). Again there is a really good chance that at most two people in the room will have green eyes.
f. At least four means four or more. Find the probability of x being greater than or equal to four. That would mean adding up all the probabilities from four to twenty. This would take a long time, so it is better to use the idea of complement. The complement of being greater than or equal to four is being less than four. That would mean being less than or equal to three. Part (e) has the answer for the probability of being less than or equal to three. Just subtract that number from 1.
\(P(x \geq 4)=1-P(x \leq 3)=1-0.99996=0.00004\) You can also find this answer by doing the following on TI-83/84:
\(P(x \geq 4)=1-P(x \leq 3)=1-\text { binomcdf }(20,.01,3)=1-0.99996=0.00004\) on R:
\(P(x \geq 4)=1-P(x \leq 3)=1-\text { pbinom }(3,20,.01)=1-0.99996=0.0004\) Again, this is very unlikely to happen.
There are other technologies that will compute binomial probabilities.
Example \(\PageIndex{4}\) calculating binomial probabilities
According to the Center for Disease Control (CDC), about 1 in 88 children in the U.S. have been diagnosed with autism ("CDC-data and statistics,," 2013). Suppose you consider a group of 10 children.
- State the random variable.
- Argue that this is a binomial experiment.
- Find the probability that none have autism.
- Find the probability that seven have autism.
- Find the probability that at least five have autism.
- Find the probability that at most two have autism.
- Suppose five children out of ten have autism. Is this unusual? What does that tell you?
Solution
a. x = number of children with autism
b.
- There are 10 children, and each child is a trial, so there are a fixed number of trials. In this case, n = 10.
- If you assume that each child in the group is chosen at random, then whether a child has autism does not affect the chance that the next child has autism. Thus the trials are independent.
- Either a child has autism or they do not have autism, so there are two outcomes. In this case, the success is a child has autism.
- The probability of a child having autism is 1/88. This is the same for every trial since each child has the same chance of having autism. \(p=\dfrac{1}{88}\) and \(q=1-\dfrac{1}{88}=\dfrac{87}{88}\).
c. Using the formula:
\(P(x=0)=_{10} C_{0}\left(\dfrac{1}{88}\right)^{0}\left(\dfrac{87}{88}\right)^{10-0} \approx 0.892\)
Using the TI-83/84 Calculator:
\(P(x=0)=\text { binompdf }(10,1 \div 88,0) \approx 0.892\)
Using R:
\(P(x=0)=\text { pbinom }(0,10,1 / 88) \approx 0.892\)
d. Using the formula:
\(P(x=7)=_{10} C_{7}\left(\dfrac{1}{88}\right)^{7}\left(\dfrac{87}{88}\right)^{10-7} \approx 0.000\)
Using the TI-83/84 Calculator:
\(P(x=7)=\text { binompdf }(10,1 \div 88,7) \approx 2.84 \times 10^{-12}\)
Using R:
\(P(x=7)=\operatorname{dbinom}(7,10,1 / 88) \approx 2.84 \times 10^{-12}\)
e. Using the formula:
\(\begin{aligned} P(x \geq 5) &=P(x=5)+P(x=6)+P(x=7) \\ &+P(x=8)+P(x=9)+P(x=10) \\ &=_{10} C_{5}\left(\dfrac{1}{88}\right)^{5}\left(\dfrac{78}{88}\right)^{10-5}+_{10} C_{6}\left(\dfrac{1}{88}\right)^{6}\left(\dfrac{78}{88}\right)^{10-6} \\ & +_{10}C_{7}\left(\dfrac{1}{88}\right)^{7}\left(\dfrac{78}{88}\right)^{10-7}+_{10}C_{8}\left(\dfrac{1}{88}\right)^{8}\left(\dfrac{78}{88}\right)^{10-8} \\ &+_{10}C_{9}\left(\dfrac{1}{88}\right)^{9}\left(\dfrac{78}{88}\right)^{10-9}+_{10}C_{10}\left(\dfrac{1}{88}\right)^{10}\left(\dfrac{78}{88}\right)^{10-10}\\&=0.000+0.000+0.000+0.000+0.000+0.000 \\ &=0.000 \end{aligned}\)
Using the TI-83/84 Calculator:
To use the calculator you need to use the complement.
\(\begin{aligned} P(x \geq 5) &=1-P(x<5) \\ &=1-P(x \leq 4) \\ &=1-\text { binomcdf }(10,1 \div 88,4) \\ & \approx 1-0.9999999=0.000 \end{aligned}\)
Using R:
To use R you need to use the complement.
\(\begin{aligned} P(x \geq 5) &=1-P(x<5) \\ &=1-P(x \leq 4) \\ &=1-\text { pbinom }(4,10,1 / 88) \\ & \approx 1-0.9999999=0.000 \end{aligned}\)
Notice, the answer is given as 0.000, since the answer is less than 0.000. Don’t write 0, since 0 means that the event is impossible to happen. The event of five or more is improbable, but not impossible.
f. Using the formula:
\(\begin{aligned} P(x \leq 2) &=P(x=0)+P(x=1)+P(x=2) \\ &=_{10} C_{0}\left(\dfrac{1}{88}\right)^{0}\left(\dfrac{78}{88}\right)^{10-0}+_{10} C_{1}\left(\dfrac{1}{88}\right)^{1}\left(\dfrac{78}{88}\right)^{10-1} \\ &+_{10} C_{2}\left(\dfrac{1}{88}\right)^{2}\left(\dfrac{78}{88}\right)^{10-2} \\ &=0.892+0.103+0.005>0.999 \end{aligned}\)
Using the TI-83/84 Calculator:
\(P(x \leq 2)=\text { binomcdf }(10,1 \div 88,2) \approx 0.9998\)
Using R:
\(P(x \leq 2)=\text { pbinom }(2,10,1 / 88) \approx 0.9998\)
g. Since the probability of five or more children in a group of ten having autism is much less than 5%, it is unusual to happen. If this does happen, then one may think that the proportion of children diagnosed with autism is actually more than 1/88.
Homework
Exercise \(\PageIndex{1}\)
-
Suppose a random variable,
x
, arises from a binomial experiment. If
n
= 14, and
p
= 0.13, find the following probabilities using the binomial formula.
- P (x=5)
- P (x=8)
- P (x=12)
- \(P(x \leq 4)\)
- \(P(x \geq 8)\)
- \(P(x \leq 12)\)
-
Suppose a random variable,
x
, arises from a binomial experiment. If
n
= 22, and
p
= 0.85, find the following probabilities using the binomial formula.
- P (x=18)
- P (x=5)
- P (x=20)
- \(P(x \leq 3)\)
- \(P(x \geq 18)\)
- \(P(x \leq 20)\)
-
Suppose a random variable,
x
, arises from a binomial experiment. If
n
= 10, and
p
= 0.70, find the following probabilities using the binomial formula.
- P (x=2)
- P (x=8)
- P (x=7)
- \(P(x \leq 3)\)
- \(P(x \geq 7)\)
- \(P(x \leq 4)\)
-
Suppose a random variable,
x
, arises from a binomial experiment. If
n
= 6, and
p
= 0.30, find the following probabilities using the binomial formula.
- P (x=1)
- P (x=5)
- P (x=3)
- \(P(x \leq 3)\)
- \(P(x \geq 5)\)
- \(P(x \leq 4)\)
-
Suppose a random variable,
x
, arises from a binomial experiment. If
n
= 17, and
p
= 0.63, find the following probabilities using the binomial formula.
- P (x=8)
- P (x=15)
- P (x=14)
- \(P(x \leq 12)\)
- \(P(x \geq 10)\)
- \(P(x \leq 7)\)
-
Suppose a random variable,
x
, arises from a binomial experiment. If
n
= 23, and
p
= 0.22, find the following probabilities using the binomial formula.
- P (x=21)
- P (x=6)
- P (x=12)
- \(P(x \leq 14)\)
- \(P(x \geq 17)\)
- \(P(x \leq 9)\)
-
Approximately 10% of all people are left-handed ("11 little-known facts," 2013). Consider a grouping of fifteen people.
- State the random variable.
- Argue that this is a binomial experiment Find the probability that
- None are left-handed.
- Seven are left-handed.
- At least two are left-handed.
- At most three are left-handed.
- At least seven are left-handed.
- Seven of the last 15 U.S. Presidents were left-handed. Is this unusual? What does that tell you?
-
According to an article in the American Heart Association’s publication Circulation, 24% of patients who had been hospitalized for an acute myocardial infarction did not fill their cardiac medication by the seventh day of being discharged (Ho, Bryson & Rumsfeld, 2009). Suppose there are twelve people who have been hospitalized for an acute myocardial infarction.
- State the random variable.
- Argue that this is a binomial experiment Find the probability that
- All filled their cardiac medication.
- Seven did not fill their cardiac medication.
- None filled their cardiac medication.
- At most two did not fill their cardiac medication.
- At least three did not fill their cardiac medication.
- At least ten did not fill their cardiac medication.
- Suppose of the next twelve patients discharged, ten did not fill their cardiac medication, would this be unusual? What does this tell you?
-
Eyeglassomatic manufactures eyeglasses for different retailers. In March 2010, they tested to see how many defective lenses they made, and there were 16.9% defective lenses due to scratches. Suppose Eyeglassomatic examined twenty eyeglasses.
- State the random variable.
- Argue that this is a binomial experiment Find the probability that
- None are scratched.
- All are scratched.
- At least three are scratched.
- At most five are scratched.
- At least ten are scratched.
- Is it unusual for ten lenses to be scratched? If it turns out that ten lenses out of twenty are scratched, what might that tell you about the manufacturing process?
-
The proportion of brown M&M’s in a milk chocolate packet is approximately 14% (Madison, 2013). Suppose a package of M&M’s typically contains 52 M&M’s.
- State the random variable.
- Argue that this is a binomial experiment Find the probability that
- Six M&M’s are brown.
- Twenty-five M&M’s are brown.
- All of the M&M’s are brown.
- Would it be unusual for a package to have only brown M&M’s? If this were to happen, what would you think is the reason?
- Answer
-
1. a. P(x=5) = 0.0212, b. P(x=8) = \(1.062 \times 10^{-4}\), c. P(x=12) = \(1.605 \times 10^{-9}\), d. \(P(x \leq 4)=0.973\), e. \(P(x \geq 8)=1.18 \times 10^{-4}\), f. \(P(x \leq 12)=0.99999\)
3. a. \(P(x=2)=0.0014\), b. \(P(x=8)=0.2335\), c. \(P(x=7)=0.2668\), d. \(P(x \leq 3)=0.0106\), e. \(P(x \geq 7)=0.6496\), f. \(P(x \leq 4)=0.0473\)
5. a. \(P(x=8)=0.0784\), b. \(P(x=15)=0.0182\), c. \(P(x=14)=0.0534\), d. \(P(x \leq 12)=0.8142\), e. \(P(x \geq 10)=0.7324\), f. \(P(x \leq 7)=0.0557\)
7. a. See solutions, b. See solutions, c. P(x=0) = 0.2059, d. \(P(x=7)=2.770 \times 10^{-4}\), e. \(P(x \geq 2)=0.4510\), f. \(P(x \leq 3)=0.944\), g.\(P(x \geq 7)=3.106 \times 10^{-4}\), h. See solutions
9. a. See solutions, b. See solutions, c. \(P(x=0)=0.0247\), d. \(P(x=20)=3.612 \times 10^{-16}\), e. \(P(x \geq 3)=0.6812\), f. \(P(x \leq 5)=0.8926\), g. \(P(x \geq 10)=6.711 \times 10^{-4}\), h. See solutions


4.8: Data Mining
-
- Last updated
- Save as PDF
- Ly-Huong T. Pham, Tejal Desai-Naik, Laurie Hammond, & Wael Abdeljabbar
- ASCCC Open Educational Resources Initiative (OERI)
Data mining is the process of sorting through big data (measured in terabytes). In the past, there was a lack of data to analyze. The challenge is an overabundance of data that must be reviewed, which is called data overload. This becomes an issue because the user needs to evaluate which information is useful and which is not. Many businesses do mining to get detailed insight on their customers, products and to optimize business decisions. The analysis is executed with sophisticated programs. The programs can combine multiple databases. The end effect is so complex that companies must find a way to store the data. Data warehouses are needed. The data warehouse is where the information is stored and processed from the data mining. The price for a simple warehouse could start at $10 million.
Companies like Google, Netflix, Amazon, and Facebook are big users of data mining. They seek to find out who their consumer is and how best to keep them and sell them more products. They also review their products. The means used are reviewing data and finding trends, patterns, and associations to make decisions. Generally, data mining is accomplished through automated means against extensive data sets, such as a data warehouse.
Examples of data mining include:
- An analysis of sales from a large grocery chain might determine that milk is purchased more frequently the day after it rains in cities with a population of less than 50,000.
- A bank may find that loan applicants whose bank accounts show particular deposit and withdrawal patterns are not good credit risks.
- A baseball team may find those collegiate baseball players with specific statistics in hitting, pitching, and fielding for more successful major league players.
In some cases, a data-mining project is begun with a hypothetical result in mind. For example, a grocery chain may already have some idea that the buying patterns change after it rains and want to get a deeper understanding of exactly what is happening. In other cases, there are no presuppositions, and a data-mining program is run against large data sets to find patterns and associations.


27.9.2: Thought Questions
Thought Questions
Suppose you observe a star-like object in the sky.
13.
Why don’t any of the methods for establishing distances to galaxies, described in Galaxies (other than Hubble’s law itself), work for quasars?
14.
One of the early hypotheses to explain the high redshifts of quasars was that these objects had been ejected at very high speeds from other galaxies. This idea was rejected, because no quasars with large blueshifts have been found. Explain why we would expect to see quasars with both blueshifted and redshifted lines if they were ejected from nearby galaxies.
15.
A friend of yours who has watched many Star Trek episodes and movies says, “I thought that black holes pulled everything into them. Why then do astronomers think that black holes can explain the great outpouring of energy from quasars?” How would you respond?
16.
Could the Milky Way ever become an active galaxy? Is it likely to ever be as luminous as a quasar?
17.
Why are quasars generally so much more luminous (why do they put out so much more energy) than active galaxies?
18.
Suppose we detect a powerful radio source with a radio telescope. How could we determine whether or not this was a newly discovered quasar and not some nearby radio transmission?
19.
A friend tries to convince you that she can easily see a quasar in her backyard telescope. Would you believe her claim?


13.6: Review Questions
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1.
Public health surveillance can be described primarily as which of the following actions?
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Monitoring occurrences of public health problems
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Developing interventions to control disease outbreaks
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Collecting health-related information from community members
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Tracking individuals who have been exposed to an infectious disease
2.
Which factor related to the agent will the nurse include when discussing the epidemiological triad?
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Pathogenicity
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Susceptibility
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Reservoirs
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Environmental changes
3.
Which disease does the elementary school nurse identify as being spread via airborne transmission?
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Measles
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Influenza
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Pertussis
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Pneumonia
4.
Which of the following factors is a component of the epidemiological triad?
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Susceptibility
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Environment
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Portal of exit
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Pathogenicity
5.
The community health nurse is conducting a health screening of an immigrant family. Which finding is an indication of the need for T-Spot testing instead of Mantoux TST testing?
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Prior vaccination with BCG
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Family member who is 10 years of age
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Symptoms of night sweats, fevers, chills, and hemoptysis
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A negative chest X-ray and negative sputum testing for MTB
6.
Which primary prevention strategy will the nurse include in a community program on the prevention of hepatitis C?
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Advising individuals with risky sexual behaviors to be screened for hepatitis C
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Stressing that handwashing is important after changing diapers or using the bathroom
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Recommending immunizations against the other viral hepatitis infections
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Instructing individuals to avoid sharing needles, syringes, or other equipment contaminated with blood
7.
Which manifestation would the nurse include when teaching about mild to moderate food poisoning at a community health fair?
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Bloody diarrhea
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Diarrhea lasting more than 3 days
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Dehydration
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Fever less than 102°F
8.
Which diagnosis would the nurse anticipate when observing a “bull’s eye” rash on a client’s leg?
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Zika virus
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West Nile virus
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Lyme disease
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Dengue fever
9.
Which surveillance system would the CDC utilize to track and investigate foodborne illnesses?
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System for Enteric Disease Response, Investigation, and Coordination (SEDRIC)
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National Notifiable Disease Surveillance System (NNDSS)
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Global Outbreak Alert and Response Network (GOARN)
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Waterborne Disease and Outbreak Surveillance System (WBDOSS)
10.
Which action by the nurse during an infectious disease outbreak is an example of a secondary prevention measure?
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Administering immunizations
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Engaging in contact tracing
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Providing direct care to infected individuals
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Educating the public about disease transmission


5: Population Genetics
Deoxyribonucleic acid, or DNA-a large double-stranded, helical molecule, with rungs made from the four base pairs adenine (A), cytosine (C), thymine (T) and guanine \((G)\) -carries inherited genetic information. The ordering of the base pairs \(A, C\) , \(\mathrm{T}\) and \(\mathrm{G}\) determines the DNA sequence. A gene is a particular DNA sequence that is the fundamental unit of heredity for a particular trait. Some species develop as diploids, carrying two copies of every gene, one from each parent, and some species develop as haploids with only one copy. There are even species that develop as both diploids and haploids.
Consider the pea plant, which develops as a diploid. When we say there is a gene for pea color, say, we mean there is a particular DNA sequence that may vary in a pea plant population, and that there are at least two subtypes, called alleles, where plants with two copies of the yellow-color allele have yellow peas, those with two copies of the green-color allele, green peas. A plant with two copies of the same allele is homozygous for that particular gene (or a homozygote), while a plant carrying two different alleles is heterozygous (or a heterozygote). For the pea color gene, a plant carrying both a yellow- and green-color allele has yellow peas. We say that the green color is a recessive trait (or the green-color allele is recessive), and the yellow color is a dominant trait (or the yellow-color allele is dominant). The combination of alleles carried by the plant is called its genotype, while the actual trait (green or yellow peas) is called its phenotype. A gene that has more than one allele in a population is called polymorphic, and we say the population has a polymorphism for that particular gene.
Population genetics can be defined as the mathematical modeling of the evolution and maintenance of polymorphism in a population. Population genetics together with Charles Darwin’s theory of evolution by natural selection and Gregor Mendel’s theory of biological inheritance forms the modern evolutionary synthesis (sometimes called the modern synthesis, the evolutionary synthesis, the neoDarwinian synthesis, or neo-Darwinism). The primary founders in the early twentieth century of population genetics were Sewall Wright, J. B. S. Haldane and Ronald Fisher.
Allele frequencies in a population can change due to the influence of four primary evolutionary forces: natural selection, genetic drift, mutation, and migration. Here, we mainly focus on natural selection and mutation. Genetic drift is the study of stochastic effects, and it is important in small populations. Migration typically requires consideration of the spatial distribution of a population, and it is usually modeled mathematically by partial differential equations.
The simplified models we will consider assume infinite population sizes (neglecting stochastic effects except in \(\$ 5.5\) ), well-mixed populations (neglecting any spatial distribution), and discrete generations (neglecting any age-structure). Our main purpose is to illustrate the fundamental ways that a genetic polymorphism can be maintained in a population.
| genotype | \(A\) | \(a\) |
| number | \(n_{A}\) | \(n_{a}\) |
| viability fitness | \(g_{A}\) | \(g_{a}\) |
| fertility fitness | \(f_{A}\) | \(f_{a}\) |


5.4.6: BF₃
The case of boron trifluoride (\(\ce{BF_3}\)) is an example of a molecule with one more layer of complexity than the other examples we have seen in previous sections in this chapter. (\(\ce{BF_3}\)) is more complex than previous examples because it is the first case in which there are multiple types of valence orbitals on the pendant atoms. (\(\ce{BF_3}\)) possesses \(s\) and \(p\) orbitals on both the central atom and all of the pendant atoms. We can follow the same steps that we have previously to derive other molecular orbital diagrams; however, there is one important difference: we will treat each type of pendant orbital as an individual set of SALCs.
Step 1. Find the point group of the molecule and assign Cartesian coordinates so that z is the principal axis.
The BH\(_3\) molecule is trigonal planar, and its point group is \(D_{3h}\). The \(z\) axis is colinear with the principle axis, the \(C_3\) axis.
Step 2. Identify and count the pendant atoms' valence orbitals.
Each of the three pendant fluorine atoms has four valence orbitals: one \(2s\) and three \(2p\) orbitals. Thus, we can expect a total of twelve SALCs from the three atoms. However, we will treat each type of orbital from the F atoms as its own set; each will have its own set of coordinates, with the \(y\) axis along the M-L bond. Thus we expect the following:
- One set of \(2s\) SALCs: there is one 2s orbital on each of three atoms. This set will have three SALCs.
- One set of \(2p_x\) SALCs: there is one \(2p_x\) orbital on each of three atoms. Each of these individual orbitals is perpendicular to the M-L bond and coplanar with the molecule. This set will have three SALCs.
- One set of \(2p_y\) SALCs: there is one \(2p_y\) orbital on each of three atoms. Each of these individual orbitals is colinear with the M-L bond. This set will have three SALCs.
- One set of \(2p_z\) SALCs: there is one \(2p_z\) orbital on each of three atoms. Each of these individual orbitals is perpendicular to the M-L bond and parallel with the principal axis of the molecule. This set will have three SALCs.
Step 3. Generate the \(\Gamma\)'s for each set of SALCs
Use the \(D_{3h}\) character table to generate one reducible representation (\(\Gamma\)) for each of the four sets of pendant atomic orbitals shown in Figure \(\PageIndex{2}\); in this case we need four \(\Gamma\) because there are four types of valence orbital (the \(s, p_x, p_y, p_z\)).
- For the set of \(s\) orbitals: perform the operation for each class in the character table. Each \(s\) orbital is assigned a value depending on whether it moves or not: assign a value of 1 if it remains in place during the operation, or 0 (zero) if it moves out of its original place.
- for each set of \(p\) orbitals: Now there are phases (signs of the wavefunction), so a negative value is possible. Assign a value of 1 to each \(p\) orbital if it remains in place and in phase during the operation; assign a -1 if the atom's position remains, while the phase of the orbital is inverted (the positive lobe moves to the position of the negative lobe and vice versa); assign a 0 if it moves out of its original position.
The \(\Gamma\) for each type is given below:
\[\begin{array}{|c|cccccc|} \hline \bf{D_{3h}} & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_v \\ \hline
\bf{\Gamma_{2s}} & 3 & 0 & 1 & 3 & 0 & 1 \\
\bf{\Gamma_{2p_x}} & 3 & 0 & -1 & 3 & 0 & -1 \\
\bf{\Gamma_{2p_y}} & 3 & 0 & 1 & 3 & 0 & 1 \\
\bf{\Gamma_{2p_z}} & 3 & 0 & -1 & -3 & 0 & 1 \\
\hline \end{array} \nonumber \]
Step 4. Break \(\Gamma\)'s into irreducible representations for individual SALCs
Reduce each \(\Gamma\) into its component irreducible representations. Using either of the processes described previously, we find that each of the \(\Gamma\) reduces to two irreducible representations under the \(D_{3h}\) point group; in each case one is singly degenerate (\(A\)) and the other doubly degenerate (\(E\)). This equates to three SALCs for each \(\Gamma\) (orbital type) and a total of twelve SALCs. This should give us confidence! We were in fact expecting 12 SALCs because we found that there were twelve pendant atomic orbitals to start with.
\[\begin{array}{|c|cccccc|} \hline {D_{3h}} & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_v \\ \hline
\bf{\Gamma_{2s}} & \bf 3 & \bf 0 & \bf 1 & \bf 3 & \bf 0 & \bf 1 \\
A_{1}' & 1 & 1 & 1 & 1 & 1 & 1 \\
E' & 2 & -1 & 0 & 2 & -1 & 0 \\
\hline \end{array} \nonumber \]
\[\begin{array}{|c|cccccc|} \hline {D_{3h}} & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_v \\ \hline
\bf{\Gamma_{2p_x}} &\bf 3 &\bf 0 &\bf -1 &\bf 3 &\bf 0 &\bf -1 \\
A_{2}' & 1 & 1 & -1 & 1 & 1 & -1 \\
E' & 2 & -1 & 0 & 2 & -1 & 0 \\
\hline \end{array} \nonumber \]
\[\begin{array}{|c|cccccc|} \hline {D_{3h}} & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_v \\ \hline
\bf{\Gamma_{2p_y}} &\bf 3 &\bf 0 &\bf 1 &\bf 3 &\bf 0 &\bf 1 \\
A_{1}' & 1 & 1 & 1 & 1 & 1 & 1 \\
E' & 2 & -1 & 0 & 2 & -1 & 0 \\
\hline \end{array} \nonumber \]
\[\begin{array}{|c|cccccc|} \hline {D_{3h}} & E & 2C_3 & 3C_2 & \sigma_h & 2S_3 & 3\sigma_v \\ \hline
\bf{\Gamma_{2p_z}} &\bf 3 &\bf 0 &\bf -1 &\bf -3 &\bf 0 &\bf 1 \\
A_{2}" & 1 & 1 & -1 & -1 & -1 & 1 \\
E" & 2 & -1 & 0 & -2 & 1 & 0 \\
\hline \end{array} \nonumber \]
Step 5: Sketch the SALCs
We'll skip this step for now to make this problem simpler.
Step 6: Draw the MO diagram by combining SALCs with AO’s of like symmetry.
We now know the symmetries of different pendant orbital SALCs - and it is helpful if we also know their relative energies, and the energies of the valence orbitals on the central boron atom. We can make predictions based on periodic trends, or we can use a table of ionization energies as a guide. The symmetries of the boron \(2s\) and \(2p\) orbitals can be found from the \(D_{3h}\) character table.


11.2: The Process of Meiosis - Meiosis I
- Describe the stages and results of meiosis I
Meiosis I
Meiosis is preceded by an interphase consisting of three stages. The G 1 phase (also called the first gap phase) initiates this stage and is focused on cell growth. The S phase is next, during which the DNA of the chromosomes is replicated. This replication produces two identical copies, called sister chromatids, that are held together at the centromere by cohesin proteins. The centrosomes, which are the structures that organize the microtubules of the meiotic spindle, also replicate. Finally, during the G 2 phase (also called the second gap phase), the cell undergoes the final preparations for meiosis.
Prophase I
During prophase I, chromosomes condense and become visible inside the nucleus. As the nuclear envelope begins to break down, homologous chromosomes move closer together. The synaptonemal complex, a lattice of proteins between the homologous chromosomes, forms at specific locations, spreading to cover the entire length of the chromosomes. The tight pairing of the homologous chromosomes is called synapsis. In synapsis, the genes on the chromatids of the homologous chromosomes are aligned with each other. The synaptonemal complex also supports the exchange of chromosomal segments between non-sister homologous chromatids in a process called crossing over. The crossover events are the first source of genetic variation produced by meiosis. A single crossover event between homologous non-sister chromatids leads to an exchange of DNA between chromosomes. Following crossover, the synaptonemal complex breaks down and the cohesin connection between homologous pairs is also removed. At the end of prophase I, the pairs are held together only at the chiasmata; they are called tetrads because the four sister chromatids of each pair of homologous chromosomes are now visible.
Prometaphase I
The key event in prometaphase I is the formation of the spindle fiber apparatus where spindle fiber microtubules attach to the kinetochore proteins at the centromeres. Microtubules grow from centrosomes placed at opposite poles of the cell. The microtubules move toward the middle of the cell and attach to one of the two fused homologous chromosomes at the kinetochores. At the end of prometaphase I, each tetrad is attached to microtubules from both poles, with one homologous chromosome facing each pole. In addition, the nuclear membrane has broken down entirely.
Metaphase I
During metaphase I, the tetrads move to the metaphase plate with kinetochores facing opposite poles. The homologous pairs orient themselves randomly at the equator. This event is the second mechanism that introduces variation into the gametes or spores. In each cell that undergoes meiosis, the arrangement of the tetrads is different. The number of variations is dependent on the number of chromosomes making up a set. There are two possibilities for orientation at the metaphase plate. The possible number of alignments, therefore, equals 2n, where n is the number of chromosomes per set. Given these two mechanisms, it is highly unlikely that any two haploid cells resulting from meiosis will have the same genetic composition.
Anaphase I
In anaphase I, the microtubules pull the attached chromosomes apart. The sister chromatids remain tightly bound together at the centromere. The chiasmata are broken in anaphase I as the microtubules attached to the fused kinetochores pull the homologous chromosomes apart.
Telophase I and Cytokinesis
In telophase I, the separated chromosomes arrive at opposite poles. In some organisms, the chromosomes decondense and nuclear envelopes form around the chromatids in telophase I. Then cytokinesis, the physical separation of the cytoplasmic components into two daughter cells, occurs without reformation of the nuclei. In nearly all species of animals and some fungi, cytokinesis separates the cell contents via a cleavage furrow (constriction of the actin ring that leads to cytoplasmic division). In plants, a cell plate is formed during cell cytokinesis by Golgi vesicles fusing at the metaphase plate. This cell plate will ultimately lead to the formation of cell walls that separate the two daughter cells.
Two haploid cells are the end result of the first meiotic division. The cells are haploid because at each pole there is just one of each pair of the homologous chromosomes. Therefore, only one full set of the chromosomes is present. Although there is only one chromosome set, each homolog still consists of two sister chromatids.
Key Points
- Meiosis is preceded by interphase which consists of the G 1 phase (growth), the S phase ( DNA replication), and the G 2 phase.
- During prophase I, the homologous chromosomes condense and become visible as the x shape we know, pair up to form a tetrad, and exchange genetic material by crossing over.
- During prometaphase I, microtubules attach at the chromosomes’ kinetochores and the nuclear envelope breaks down.
- In metaphase I, the tetrads line themselves up at the metaphase plate and homologous pairs orient themselves randomly.
- In anaphase I, centromeres break down and homologous chromosomes separate.
- In telophase I, chromosomes move to opposite poles; during cytokinesis the cell separates into two haploid cells.
Key Terms
- crossing over : the exchange of genetic material between homologous chromosomes that results in recombinant chromosomes
- tetrad : two pairs of sister chromatids (a dyad pair) aligned in a certain way and often on the equatorial plane during the meiosis process
- chromatid : either of the two strands of a chromosome that separate during meiosis


34: Earlier American Literature in Film Last updated Save as PDF Page ID 40689 Robin DeRosa, Abby Goode et al. Plymouth State University/ 34.1: “You Can’t Hide What’s In Your Heart” 34.2: Patrick Batemen is Ralph Waldo Emerson 34.3: GTFO 34.4: It’s Time to “Get Out” Your Antiquated Views on Race 34.5: Twelve Years…in the Life of a Slave Girl 34.6: I’m Not the Right Person to Be Talking About This 34.7: Life Imitates Art…or Art Imitates Life? 34.8: Madison Unchained 34.9: Into Self-Reliance 34.10: Down the Rabbit Hole 34.11: The Self-Reliant Baby Driver (Spoiler Alert!) – Where is American Literature Now? 34.12: Avatar is just Pocahontas with Blue People in Space 34.13: SHOOK- Django is Madison Washington.. with a trigger happy finger. 34.14: Why Does it ALWAYS Take a White Character to be Able to Save those of Color? 34.15: Wakanda Forever? 34.16: Equality? I don’t think so. 34.17: The H8teful 8 34.18: The Culture With The Attention Span Of A Goldfish 34.19: Harriet Jacobs and William Peter Blatty- Interpretations of Slavery, Maternity, and Weaponized Religion 34.20: American Literature Today 34.21: Technology’s Impact on Literature 34.22: The Dance of American Literature


29: Harriet Beecher Stowe Last updated Save as PDF Page ID 40684 Robin DeRosa, Abby Goode et al. Plymouth State University/ 29.1: Introduction (2017) 29.2: Introduction (2016) 29.3: Uncle Tom’s Cabin (Chapters 1-3, 7) 29.4: Lesson Plan for Harriet Beecher Stowe’s Uncle Tom’s Cabin (Group Anthology Contribution 2019) 29.5: When Sympathy Goes Beyond the Pages 29.6: All The Slave Texts Thus Far 29.7: “Slaves Are Human Too…When They Act White and Look Attractive” 29.8: Untitled 29.9: How to Depict a Black Character… by Depicting Them as “Not all Black” 29.10: Uncle Tom’s Cabin versus Benito Cereno versus The Heroic Slave 29.11: “Hot People Are White-Looking and Good, Ugly People Are Evil” 29.12: Using Religion to Internalize Whiteness 29.13: Stowe was White 29.14: The Depiction of Race in Uncle Tom’s Cabin 29.15: The Mother, The Father, and the Holy American Man 29.16: The Book That Started the Civil War 29.17: Things To Stop Romanticizing- The Stranger Things Kids and 19th-Century Slaves 29.18: Gentle Giants 29.19: Uncle Tom’s Cabin- The Good, The Bad, and the Ugly 29.20: Racism’s Prevalence in Early American Literature


3.7: Shaft Work
Shaft work refers to energy transferred across the boundary by a rotating shaft.
The complete apparatus is depicted in Fig. 3.13. In use, two lead weights sank and caused the paddle wheel to rotate. Joule evaluated the stirring work done on the system (the vessel, its contents, and the lid) from the change of the vertical position \(h\) of the weights. To a first approximation, this work is the negative of the change of the weights’ potential energy: \(w = -mg\Del h\) where \(m\) is the combined mass of the two weights. Joule made corrections for the kinetic energy gained by the weights, the friction in the connecting strings and pulley bearings, the elasticity of the strings, and the heat gain from the air surrounding the system.
A typical experiment performed by Joule is described in Prob. 3.10. His results for the mechanical equivalent of heat, based on 40 such experiments at average temperatures in the range \(13{\thinspace}\degC\)–\(16{\thinspace}\degC\) and expressed as the work needed to increase the temperature of one gram of water by one kelvin, was \(4.165\units{J}\). This value is close to the modern value of \(4.1855\units{J}\) for the “\(15{\thinspace}\degC\) calorie,” the energy needed to raise the temperature of one gram of water from \(14.5{\thinspace}\degC\) to \(15.5{\thinspace}\degC\).
The thermochemical calorie (cal), often used as an energy unit in the older literature, is defined as \(4.184\units{J}\). Thus \(1\units{kcal} = 4.184\units{kJ}\).


5.3: Classification and distribution of Languages
This second section will facilitate your understanding of the dimensions of language across geographic areas and cultural landscapes. Three main questions are addressed in this section:
- How are languages classified with respect to issues of national identity and genealogical considerations?
- What are the major language families of the world and how many speakers make use of the respective languages?
- How does language use vary in the United States with respect to dialects of English and multilingualism?
5.3.1 Diffusion of Languages
Language, like any other cultural phenomenon, has an inherent spatiality, and all languages have a history of diffusion. As our ancestors moved from place to place, they brought their languages with them. As people have conquered other places, expanded demographically, or converted others to new religions, languages have moved across space. Writing systems that were developed by one people were adapted and used by others. Indo-European, the largest language family, spread across a large expanse of Europe and Asia through a mechanism that is still being debated. Later, European expansion produced much of the current linguistic map by spreading English, French, Spanish, Portuguese, and Russian far from their native European homelands.
Language is disseminated through diffusion, but in complex ways. Relocation diffusion is associated with settler colonies and conquest, but in many places, hierarchical diffusion is the form that best explains the predominant languages. People may be compelled to adopt a dominant language for social, political or economic mobility. Contagious diffusion is also seen in languages, particularly in the adoption of new expressions in a language. One of the most obvious examples has been in the current convergence of British and American English. The British press has published books 1 and articles 2 decrying the Americanization of British English, while the American press has done the same thing in reverse 3 . In reality, languages borrow bits and pieces from other languages continuously.
The establishment of official languages is often related to the linguistic power differential within countries. Russification and Arabization are just two implementations of processes that use political power to favor one language over another.
5.3.2 Classification of Languages
There is no precise figure as to the total number of languages spoken in the world today. Estimates vary between 5,000 and 7,000, and the accurate number depends partly on the arbitrary distinction between languages and dialects. Dialects (variants of the same language) reflect differences along regional and ethnic lines. In the case of English, most native speakers will agree that they are speakers of English even though differences in pronunciation, vocabulary and sentence structure clearly exist. English speakers from England, Canada, Australia, New Zealand and United States of America will generally agree that they speak English, and this is also confirmed with the use of a standard written form of the language and a common literary heritage. However, there are many other cases in which speakers will not agree when the question of national identity and mutual intelligibility do not coincide.
The most common situation is when similar spoken language varieties are mutually understandable, but for political and historical reasons, they are regarded as different languages as in the case of Scandinavian languages. While Swedes, Danes and Norwegians can communicate with each other in most instances, each national group admits speaking a different language: Swedish, Danish, Norwegian and Icelandic. There are other cases in which political, ethnic, religious, literary and other factors force a distinction between similar language varieties: Hindi vs. Urdu, Flemish vs. Dutch, Serbian vs. Croatian, Gallego vs. Portuguese, Xhosa vs Zulu. An opposite situation occurs when spoken language varieties are not mutually understood, but for political, historical or cultural motives, they are regarded as the same language as in the case of Lapp and Chinese dialects.
Languages are usually classified according to membership in a language family (a group of related languages) which share common linguistic features (pronunciation, vocabulary, grammar) and have evolved from a common ancestor ( proto-language ). This type of linguistic classification is known as the genetic or genealogical approach. Languages can also be classified according to sentence structure (S)ubject+(V)erb+(O)bject, S+O+V, V+S+O). This type of classification is known as typological classification , and is based on a comparison of the formal similarities (pronunciation, grammar or vocabulary) which exist among languages.
Language families around the world reflect centuries of geographic movement and interaction among different groups of people. The Indo-European family of languages, for example, represents nearly half of the world’s population. The language family dominates nearly all of Europe, significant areas of Asia, including Russia and India, North and South America, Caribbean islands, Australia, New Zealand, and parts of South Africa. The Indo-European family of languages consists of various language branches (a collection of languages within a family with a common ancestral language) and numerous language subgroups (a collection of languages within a branch that share a common origin in the relative recent past and exhibit many similarities in vocabulary and grammar.
Indo-European Language Branches and Language Subgroups
Germanic Branch
Western Germanic Group (Dutch, German, Frisian, English)
Northern Germanic Group (Danish, Swedish, Norwegian, Icelandic, Faeroese)
Romance Branch
French, Portuguese, Spanish, Catalan, Provençal, Romansh, Italian, Romanian)
Slavic Branch
West Slavic Group (Polish, Slovak, Czech, Sorbian)
Eastern Slavic Group (Russian, Ukrainian, Belorussian)
Southern Slavic Group (Slovene, Serbo-Croatian, Macedonian, Bulgarian)
Celtic Branch
Britannic Group (Breton, Welsh)
Gaulish Group (Irish Gaelic, Scots Gaelic)
Baltic-Slavonic Branch
Latvian, Lithuanian
Hellenic Branch
Greek
Thracian-Illyrian Branch
Albanian
Armenian Branch
Armenian
Iranian Branch
Kurdish, Persian, Baluchi, Pashto, Tadzhik
Indo-Iranian (Indic) Branch
Northwestern Group (Panjabi, Sindhi, Pahari, Dardic)
Eastern Group (Assamese, Bengali, Oriya)
Midland Group (Rajasthani, Hindi/Urdu, Bihari)
West and Southwestern Group (Gujarati, Marathi, Konda, Maldivian, Sinhalese)
Other languages spoken in Europe, but not belonging to the Indo-European family are subsumed in these other families: Finno-Ugric (Estonian, Hungarian, Karelian, Saami, Altaic (Turkish, Azerbaijani, Uzbek) and Basque. Some of the language branches listed above are represented by only one principal language (Albanian, Armenian, Basque, Greek), while others are spoken by diverse groups in some geographic regions (Northern and Western Germanic languages, Western and Eastern Slavic languages, Midland and Southwestern Indian languages).
Major Language Families of the World by Geographic Region
Europe
Caucasian Family
Abkhaz-Adyghe Group (Circassian, Adyghe, Abkhaz) Nakho-Dagestanian Group (Avar, Kuri, Dargwa) Kartvelian Group (Kartvelian, Georgian, Zan, Mingrelian)
Africa
Afro-Asiatic Family (Arabic, Hebrew, Tigrinya, Amharic) Niger-Congo Family (Benue-Congo, Adamawa, Kwa) Nilo-Saharan Family (Chari-Nile, Nilo-Hamitic, Nara) Khoisan Family (Sandawe, Hatsa)
Asia
Sino-Tibetan Family (Chinese, Tibetan, Burmese)
Tai Family (Laotian, Shan, Yuan)
Austro-Asiatic Family (Vietnamese, Indonesian, Dayak, Malayo-Polynesian) Japanese (an example of an isolated language)
Pacific
Austronesian Family (Malagasy, Malay, Javanese, Palauan, Fijian) Indo-Pacific Family (Tagalog, Maori, Tongan, Samoan)
Americas
Eskimo-Aleut Family (Eskimo-Aleut, Greenlandic Eskimo)
Athabaskan Family (Navaho. Apache)
Algonquian Family (Arapaho, Blackfoot, Cheyenne, Cree, Mohican, Choctaw) Macro-Siouan Family (Cherokee, Dakota, Mohawk, Pawnee)
Aztec-Tanoan Family (Comanche, Hopi, Pima-Papago, Nahuatl, Tarahumara) Mayan Family (Maya, Mam, Quekchi, Quiche)
Oto-Manguean Family (Otomi, Mixtec, Zapotec)
Macro-Chibchan Family (Guaymi, Cuna, Waica, Epera)
Andean-Equatorial Family (Guahibo, Aymara, Quechua, Guarani)
The number of language families distributed around the world is sizable. The linguistic situation of specific member groups of the language family might be influenced by diverse, interacting factors: settlement history (migration, conquest, colonialism, territorial agreements), ways of living (farming, fishing, hunting, trading) and demographic strength and vitality of the speaker groups. Some languages might converge (many local varieties becoming one main language), while others might diverge (one principal language evolves into many other speech varieties). When different linguistic groups come into contact, a pidgin type of language may be the result. A pidgin is a composite language with a simplified grammatical system and a limited vocabulary, typically borrowed from the linguistic groups involved in trade and commerce activities.
Tok Pisin is an example of a pidgin spoken in Papua New Guinea and derived mainly from English. A pidgin may become a creole language when the size of the vocabulary increases, grammatical structures become more complex and children learn it as their native language or mother tongue. There are cases in which one existing language gains the status of a lingua franca . A lingua franca may not necessarily be the mother tongue of any one speaker group, but it serves as the medium of communication and commerce among diverse language groups. Swahili, for instance, serves as a lingua franca for much of East Africa, where individuals speak other local and regional languages.
With increased globalization and interdependence among nations, English is rapidly acquiring the status of lingua franca for much of the world. In Europe, Africa and India and other geographic regions, English serves as a lingua franca across many national-state boundaries. The linguistic consequence results in countless numbers of speaker groups who must become bilingual (the ability to use two languages with varying degrees of fluency) to participate more fully in society.
Some continents have more spoken languages than others. Asia leads with an estimated 2,300 languages, followed by Africa with 2,138. In the Pacific area, there are about 1,300 languages spoken and in North and South America about 1,064 languages have been identified. Europe, even with its many nation-states, is at the bottom of the list with about 286 languages.
|
Language |
Family |
Speakers in Millions |
Main Areas Where Spoken |
|---|---|---|---|
|
Chinese |
Sino-Tibetan |
1197 |
China, Taiwan, Singapore |
|
Spanish |
Indo-European |
406 |
Spain, Latin America, Southwestern United States |
|
English |
Indo-European |
335 |
British Isles, United States, Canada, Caribbean, Australia, New Zealand, South Africa, Philippines, former British colonies in Asia and Africa |
|
Hindi |
Indo-European |
260 |
Northern India, Pakistan |
|
Arabic |
Afro-Asiatic |
223 |
Middle East, North Africa |
|
Portuguese |
Indo-European |
202 |
Portugal, Brazil, southern Africa |
|
Bengali |
Indo-European |
193 |
Bangladesh, eastern India |
|
Russian |
Indo-European |
162 |
Russia, Kazakhstan, part of Ukraine, 0ther former Soviet Republics |
|
Japanese |
Japanese |
122 |
Japan |
|
Javanese |
Austronesian |
84.3 |
Indonesia |
Ten Major Languages of the World in the Number of Native Speakers 5
Other important languages and related dialects, whose total number includes both native speakers and second language users, consist of following: Korean (78 million), Wu/Chinese (71 million), Telugu (75 million), Tamil (74 million), Yue/ Chinese (71 million), Marathi (71 million), Vietnamese (68 million) and Turkish (61 million).
Language Spread and Language Loss
Of the top 20 languages of the world, all these languages have their origin in south or east Asia or in Europe. There is not one from the Americas, Oceania or Africa. The absence of a major world language in these regions seems to be precisely where most of the linguistic diversity is concentrated.
- English, French and Spanish are among the world’s most widespread languages due to the imperial history of the home countries from where they originated.
- Two-thirds (66%) of the world’s population speak 12 of the major languages around the globe
- About 3 percent of the world’s population accounts for 96 percent of all the languages spoken today. Of the current living languages in the world, about 2,000 have less than 1,000 native speakers.
- Nearly half of the world’s spoken languages will disappear by the end of this century. Linguistic extinction ( language death ) will affect some countries and regions more than others.
- In the United States many endangered languages are spoken by Native American groups who reside in reservations. Many languages will be lost in Amazon rain forest, sub-Saharan Africa, Oceania, aboriginal Australia and Southeast Asia.
- English is used as an official language in at least 35 countries, including a number of countries in Africa (Botswana, Kenya, Namibia, Sudan, Tanzania, Uganda among others), Asia (India, Pakistan, Philippines), Pacific Region (Fiji, Solomon Islands, Vanuatu, New Zealand), Caribbean (Puerto Rico, Belize, Guyana, Jamaica), Ireland and Canada.
- English is not by law ( de jure ) the official language in the United Kingdom, United States and Australia. English does enjoy the status of “national language” in these countries due to its power and prestige in institutions and society.
-
English does not have the highest number of native speakers, but it
is the world’s most commonly studied language. More people learn English than French, Spanish, Italian, Japanese, German and Chinese combined. 6
Dialects of English in the United States
At the time of the American Revolution, three principal dialects of English were spoken. ThesevarietiesofEnglishcorrespondedtodifferencesamongtheoriginal setters who populated the East Coast.
Northern English
These settlements in this area were established and populated almost entirely by English settlers. Nearly two-thirds of the colonists in New England were Puritans from East Anglia in southwestern England. The region consists of the following states: Massachusetts, New Hampshire, Maine, Connecticut, Rhode Island, Vermont, New York and New Jersey.
Southern English
About half of the speakers came from southeast England. Some of them came from diverse socialclass backgrounds, including deported prisoners, indentured servants, political and religious persecuted groups. The following states comprise the region: Virginia, Delaware, North Carolina, South Carolina and Georgia.
Midlands English
The settlers of this region included immigrants from diverse backgrounds. Those who settled in Pennsylvania were predominantly Quakers from northern England. Some individuals from Scotland and Ireland also settled in Pennsylvania as well as in New Jersey and Delaware. Immigrants from Germany, Holland and Sweden also migrated to this region and learned their English from local English-speaking settlers. This region is formed by the following areas/states: Upper Ohio Valley, Pennsylvania, Maryland, West Virginia, western areas North and South Carolina.
Dialects of American English have continued to evolve over time and place. Regional differences in pronunciation, vocabulary and grammar do not suggest that a type of linguistic convergence is underway, resulting in some type of “national dialect” of American English. Even with the homogenizing influences of radio, television, internet, and social media, many distinctive varieties of English can be identified. Robert Delaney (2000) has outlined a dialect map for the United States which features at least 24 distinctive dialects of English. Dialect boundaries are established using diverse criteria: language features (differences in pronunciation, vocabulary and grammar) settlement history, ethnic diversity, educational levels and languages in contact (Spanish/English in the American Southwest). The dialect map does not represent the English varieties spoken in Alaska or Hawaii. However, it does include some urban and social (ethnolinguistic) dialects.
General Northern English, spoken by nearly two-thirds of the country.
New England Varieties
- Eastern New England
- Boston Urban
- Western New England
- Hudson Valley
- New York City
- Bonac (Long Island)
- Inland Northern English Varieties
- San Francisco Urban
- Upper Midwestern
- Chicago Urban
Midland English Varieties
- North Midland (Pennsylvania)
- Pennsylvania German-English
Western English Varieties
- Rocky Mountain
- Pacific Northwest
- Pacific Southwest
- Southwest English
- South Midland Varieties
- Ozark
- Southern Appalachian (Smoky Mountain English)
General Southern English Varieties
Southern
- Virginia Piedmont
- Coastal South
- Gullah (coastal Georgia and South Carolina)
- Gulf Southern
- Louisiana (Cajun French and Cajun English) 7
Multilingualism in the United States
Language diversity existed in what is now the United States long before the arrival of the Europeans.
It is estimated that there were between 500 and 1,000 Native American languages spoken around the fifteenth century and that there was widespread language contact and bilingualism among the Indian nations. With the arrival of the Europeans, seven colonial languages established themselves in different regions of the territory:
- English along the Eastern seaboard, Atlantic coast
- Spanish in the South from Florida to California
- French in Louisiana and northern Maine
- German in Pennsylvania
- Dutch in New York (New Amsterdam)
- Swedish in Delaware
- Russian in Alaska
Dutch, Swedish and Russian survived only for a short period, but the other four languages continue to be spoken to the present day. In the 1920’s, six major minority languages were spoken in significant numbers partly to due to massive immigration and territorial histories. The “big six” minority languages of the 1940’s include German, Italian, Polish, Yiddish, Spanish and French. Of the six minority languages, only Spanish and French have shown any gains over time, Spanish because of continued immigration and French because of increased “language consciousness” among individuals from Louisiana and Franco-Americans in the Northeast.
The 2015 Census data for the United States reveals valuable geographic information regarding the top 10 states with the extensive language diversity.
- California: 45 percent of the inhabitants speak a language other than English at home; the major languages include Spanish, Chinese, Korean, Vietnamese, Arabic, Armenian and Tagalog.
- Texas: 35 percent of the residents speak a language other than English at home; Spanish
- is widely used among bilinguals; Chinese, German and Vietnamese are also spoken.
- New Mexico: 34 percent of the state’s population speak another language; most speak Spanish but a fair number speak Navajo and other Native American languages.
- New York: 31 percent of the residents speak a second language; Chinese, Italian, Russian, Spanish and Yiddish; some of these languages can be found within the same city block.
- New Jersey: 31 percent of the state’s residents speak a second language in addition to English; some of the languages spoken include Chinese, Gujarati, Portuguese, Spanish and Italian.
- Nevada: 30 percent of the population is bilingual; Chinese, German and Tagalog are used along with Spanish, the predominant second language of the Southwest.
- Florida: 29 percent of the residents speak a second language, including French (Haitian Creole), German and Italian
- Arizona: 27 percent of the residents claim to be bilingual; most speak Spanish as in New Mexico while others use Native American languages.
- Hawaii: 26 percent of the population claims to be bilingual; Japanese, Chinese, Korean and Tagalog are spoken along with Hawaiian, the state’s second official language.
- Illinois and Massachusetts: 23 percent of their respective populations speak a second language at home; residents of Illinois speak Chinese, German, Spanish and Polish, especially in Chicago; residents of Massachusetts speak Spanish, Haitian Creole, Chinese, Portuguese, Vietnamese and French. 8
Top Ten Languages Spoken in U.S. Homes Other Than English
Data from the 2015 American Community Survey ranks the top ten languages spoken in U.S. homes other than English. The data highlight the size of the speaker population, bilingual proficiency (fluency in the home language and English) and degree of English proficiency (LEP, limited English proficiency).
|
Rank |
Language Spoken at Home |
Total |
Bilingualism % |
Limited English % |
|---|---|---|---|---|
|
1. |
Spanish |
64,716,000 |
60.0 |
40.0 |
|
2. |
Chinese |
40,046,000 |
59.0 |
41.0 |
|
3. |
Tagalog |
3,334.000 |
44.3 |
55.7 |
|
4. |
Vietnamese |
1,737,000 |
67.6 |
32.4 |
|
5. |
French |
1,266,000 |
79.9 |
20.1 |
|
6. |
Arabic |
1,157,000 |
62.8 |
37.2 |
|
7. |
Korean |
1,109,000 |
46.8 |
53.2 |
|
8. |
German |
933,000 |
85.1 |
14.9 |
|
9. |
Russian |
905,000 |
56.0 |
44.0 |
|
10. |
French Creole |
863,000 |
58.8 |
41.2 |
Chinese includes Mandarin and Cantonese. French also comprises Haitian and Cajun varieties. German encompasses Pennsylvania Dutch. 9
While a record number of persons speak a language at home other than English, a substantial figure within each immigrant group claimed an elevated command of English. Overall, some 60 percent of the speaker groups using a second language at home were also highly fluent in English. Limited fluency in English among young children ranged from a high of 55.7 percent in the Tagalog speaker group to a low of 14.9 percent in the German group which included Pennsylvania Dutch users. 10
Most immigrant language groups have tended to follow an intergenerational language shift in the United States. This first generation is basically monolingual, speaking the native language of the group. The second generation is bilingual, speaking both the home language and English. By the third generation, the cultural group is essentially monolingual, speaking only English in most communicative situations.
More recently, some immigrant groups, particularly those with advanced training and degrees in professional fields (technology, health sciences and business), come to the United States with a high degree of fluency in English. At the same time, the variety of English these persons speak is usually marked by the country of origin (India, Philippines, Singapore among others). With globalization “new Englishes” have emerged (Indian English, Filipino English, Nigerian English) which challenge the notion of a Standard English variety (British or American) for use around the world.