4.2: Chemical Reaction Rates
- Define chemical reaction rate
- Derive rate expressions from the balanced equation for a given chemical reaction
- Calculate reaction rates from experimental data
A rate is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time.
The rate of reaction is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution’s conductivity.
The progress of a simple reaction (A → B) is shown in (Figure \(\PageIndex{1}\)); the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in (Figure \(\PageIndex{1}\)). The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time.
Figure \(\PageIndex{1}\): The Progress of a Simple Reaction (A → B). The mixture initially contains only A molecules (purple). Over time, the number of A molecules decreases and more B molecules (green) are formed (top). The graph shows the change in the number of A and B molecules in the reaction as a function of time over a 1 min period (bottom).
\[\text { rate }=\frac{\Delta[\mathrm{B}]}{\Delta t}=-\frac{\Delta[\mathrm{A}]}{\Delta t}\]
Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The formula shows that the rate of the reaction is equal to the negative of the rate of disappearance of A, and is equal to the rate of appearance of B.
For reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. If we measure the concentration of hydrogen peroxide, H 2 O 2 , in an aqueous solution, we find that it changes slowly over time as the H 2 O 2 decomposes, according to the equation:
\[\ce{2H2O2}(aq)⟶\ce{2H2O}(l)+\ce{O2}(g) \nonumber \]
The rate at which the hydrogen peroxide decomposes can be expressed in terms of the rate of change of its concentration, as shown here:
\[\begin{align*}
\ce{rate\: of\: decomposition\: of\: H_2O_2}
&=\mathrm{−\dfrac{change\: in\: concentration\: of\: reactant}{time\: interval}}\\[4pt]
&=−\dfrac{[\ce{H2O2}]_{t_2}−[\ce{H2O2}]_{t_1}}{t_2−t_1}\\[4pt]
&=−\dfrac{Δ[\ce{H2O2}]}{Δt}
\end{align*} \nonumber \]
This mathematical representation of the change in species concentration over time is the rate expression for the reaction. The brackets indicate molar concentrations, and the symbol delta (Δ) indicates “change in.” Thus, \([\ce{H2O2}]_{t_1}\) represents the molar concentration of hydrogen peroxide at some time t 1 ; likewise, \([\ce{H2O2}]_{t_2}\) represents the molar concentration of hydrogen peroxide at a later time t 2 ; and Δ[H 2 O 2 ] represents the change in molar concentration of hydrogen peroxide during the time interval Δ t (that is, t 2 − t 1 ).  Figure \(\PageIndex{1}\) provides an example of data collected during the decomposition of H 2 O 2 .
To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of 40 °C. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period:
\[\dfrac{−Δ[\ce{H2O2}]}{Δt}=\mathrm{\dfrac{−(0.500\: mol/L−1.000\: mol/L)}{(6.00\: h−0.00\: h)}=0.0833\: mol\:L^{−1}\:h^{−1}} \nonumber \]
Notice that the reaction rates vary with time, decreasing as the reaction proceeds. Results for the last 6-hour period yield a reaction rate of:
\[\dfrac{−Δ[\ce{H2O2}]}{Δt}=\mathrm{\dfrac{−(0.0625\:mol/L−0.125\:mol/L)}{(24.00\:h−18.00\:h)}=0.0104\:mol\:L^{−1}\:h^{−1}} \nonumber \]
This behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an average rate for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its instantaneous rate . The instantaneous rate of a reaction at “time zero,” when the reaction commences, is its initial rate . Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle’s initial rate—analogous to the beginning of a chemical reaction—would be the speedometer reading at the moment the driver begins pressing the brakes ( t 0 ). A few moments later, the instantaneous rate at a specific moment—call it t 1 —would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car’s average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop (Δ t ). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates.
The instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. If we plot the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of H 2 O 2 at any time t is given by the slope of a straight line that is tangent to the curve at that time (Figure \(\PageIndex{2}\)). We can use calculus to evaluating the slopes of such tangent lines, but the procedure for doing so is beyond the scope of this chapter.
Physicians often use disposable test strips to measure the amounts of various substances in a patient’s urine (Figure \(\PageIndex{4}\)). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in color upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations.
The test for urinary glucose relies on a two-step process represented by the chemical equations shown here:
\[\ce{C6H12O6 + O2}\underset{\large\textrm{catalyst}}{\xrightarrow{\hspace{45px}}}\ce{C6H10O6 + H2O2} \label{eq1} \]
\[\ce{2H2O2 + 2I-}\underset{\large\textrm{catalyst}}{\xrightarrow{\hspace{45px}}}\ce{I2 + 2H2O + O2} \label{eq2} \]
Equation \(\ref{eq1}\) depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colorless iodide ion to yield brown iodine (Equation \(\ref{eq2}\)), which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct color change.
The two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of catalysis , a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for completion of the color-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a false-negative result). Waiting too long to assess the color change can lead to a false positive due to the slower (not catalyzed) oxidation of iodide ion by other substances found in urine.
Relative Rates of Reaction
The rate of a reaction may be expressed in terms of the change in the amount of any reactant or product, and may be simply derived from the stoichiometry of the reaction.
For a balanced reaction equation:
\[\ce{aA} + {bB} ⟶\ce{cC} + \ce{dD} \nonumber \]
The rate can be expressed in terms of the decrease in reactant concentration over time or increase in product concentration over time using the stoichiometric coefficients:
\[\text { rate }=-\left(\frac{1}{\mathrm{a}}\right)\left(\frac{\Delta \mathrm{[A]}}{\Delta t}\right)=-\left(\frac{1}{\mathrm{~b}}\right)\left(\frac{\Delta \mathrm{[B]}}{\Delta t}\right) = \left(\frac{1}{\mathrm{~c}}\right)\left(\frac{\Delta \mathrm{[C]}}{\Delta t}\right) = \left(\frac{1}{\mathrm{~d}}\right)\left(\frac{\Delta \mathrm{[D]}}{\Delta t}\right)\]
Reaction rate is positive, so rate of reaction is equal to the negative of the rate of disappearance of reactants.
Consider the reaction represented by the following equation:
\[\ce{2NH3}(g)⟶\ce{N2}(g)+\ce{3H2}(g) \nonumber \]
The stoichiometric factors derived from this equation may be used to relate reaction rates in the same manner that they are used to related reactant and product amounts. The relation between the reaction rates expressed in terms of nitrogen production and ammonia consumption, for example, is:
\[\mathrm{−\dfrac{Δmol\: NH_3}{Δ\mathit t}×\dfrac{1\: mol\: N_2}{2\: mol\: NH_3}=\dfrac{Δmol\:N_2}{Δ\mathit t}} \nonumber \]
We can express this more simply without showing the stoichiometric factor’s units:
\[−\dfrac{1}{2}\dfrac{\mathrm{Δmol\:NH_3}}{Δt}=\dfrac{\mathrm{Δmol\:N_2}}{Δt} \nonumber \]
Note that a negative sign has been added to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). If the reactants and products are present in the same solution, the molar amounts may be replaced by concentrations:
\[−\dfrac{1}{2}\dfrac{Δ[\ce{NH3}]}{Δt}=\dfrac{Δ[\ce{N2}]}{Δt} \nonumber \]
Similarly, the rate of formation of H 2 is three times the rate of formation of N 2 because three moles of H 2 form during the time required for the formation of one mole of N 2 :
\[\dfrac{1}{3}\dfrac{Δ[\ce{H2}]}{Δt}=\dfrac{Δ[\ce{N2}]}{Δt} \nonumber \]
Figure \(\PageIndex{3}\) illustrates the change in concentrations over time for the decomposition of ammonia into nitrogen and hydrogen at 1100 °C. We can see from the slopes of the tangents drawn at t = 500 seconds that the instantaneous rates of change in the concentrations of the reactants and products are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production:
\[\dfrac{2.91×10^{−6}\:M/\ce s}{9.71×10^{−6}\:M/\ce s}≈3 \nonumber \]
Figure \(\PageIndex{3}\): This graph shows the changes in concentrations of the reactants and products during the reaction \(\ce{2NH3⟶N2 + 3H2}\). The rates of change of the three concentrations are related by their stoichiometric factors, as shown by the different slopes of the tangents at t = 500 s.
Consider the conversion of dinitrogen tetroxide, N 2 O 4 (g) to nitrogen dioxide, NO 2 (g):
\[ \mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \ ⟶ 2 \mathrm{NO}_2(\mathrm{~g}) \nonumber\]
Write the rate expression for the reaction in terms of disappearance of the reactants and appearance of the products.
Solution
For every one mole of reactant that disappears, two moles of product are formed. The rate of disappearance of N 2 O 4 must be must be half of the rate of appearance of NO 2 . The rate of reaction therefore may be expressed as:
\[\text { rate } = −\dfrac{\mathrm{Δ[\:N_2O_4]}}{Δt}=\dfrac{1}{2}\dfrac{\mathrm{Δ[\:NO_2]}}{Δt} \nonumber \]
The first step in the production of nitric acid is the combustion of ammonia:
\[\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(g) \nonumber \]
Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.
Solution
Considering the stoichiometry of this homogeneous reaction, the rates for the consumption of reactants and formation of products are:
\[−\dfrac{1}{4}\dfrac{Δ[\ce{NH3}]}{Δt}=−\dfrac{1}{5}\dfrac{Δ[\ce{O2}]}{Δt}=\dfrac{1}{4}\dfrac{Δ[\ce{NO}]}{Δt}=\dfrac{1}{6}\dfrac{Δ[\ce{H2O}]}{Δt} \nonumber \]
The rate of formation of Br 2 is 6.0 × 10 −6 mol/L/s in a reaction described by the following net ionic equation:
\[\ce{5Br- + BrO3- + 6H+ ⟶ 3Br2 + 3H2O} \nonumber \]
Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.
- Answer
-
\[−\dfrac{1}{5}\dfrac{Δ[\ce{Br-}]}{Δt}=−\dfrac{Δ[\ce{BrO3-}]}{Δt}=−\dfrac{1}{6}\dfrac{Δ[\ce{H+}]}{Δt}=\dfrac{1}{3}\dfrac{Δ[\ce{Br2}]}{Δt}=\dfrac{1}{3}\dfrac{Δ[\ce{H2O}]}{Δt} \nonumber \]
The graph in Figure \(\PageIndex{3}\) shows the rate of the decomposition of H 2 O 2 over time:
\[\ce{2H2O2 ⟶ 2H2O + O2} \nonumber \]
Based on these data, the instantaneous rate of decomposition of H 2 O 2 at t = 11.1 h is determined to be 3.20 × 10 −2 mol/L/h, that is:
\[−\dfrac{Δ[\ce{H2O2}]}{Δt}=\mathrm{3.20×10^{−2}\:mol\: L^{−1}\:h^{−1}} \nonumber \]
What is the instantaneous rate of production of H 2 O and O 2 ?
Solution
Using the stoichiometry of the reaction, we may determine that:
\[−\dfrac{1}{2}\dfrac{Δ[\ce{H2O2}]}{Δt}=\dfrac{1}{2}\dfrac{Δ[\ce{H2O}]}{Δt}=\dfrac{Δ[\ce{O2}]}{Δt} \nonumber \]
Therefore:
and
\[\dfrac{Δ[\ce{O2}]}{Δt}=\mathrm{1.60×10^{−2}\:mol\:L^{−1}\:h^{−1}} \nonumber \]
If the rate of decomposition of ammonia, NH 3 , at 1150 K is 2.10 × 10 −6 mol/L/s, what is the rate of production of nitrogen and hydrogen?
- Answer
-
1.05 × 10 −6 mol/L/s, N 2 and 3.15 × 10 −6 mol/L/s, H 2 .
Summary
The rate of a reaction can be expressed either in terms of the decrease in the amount of a reactant or the increase in the amount of a product per unit time. Relations between different rate expressions for a given reaction are derived directly from the stoichiometric coefficients of the equation representing the reaction.
Glossary
- average rate
- rate of a chemical reaction computed as the ratio of a measured change in amount or concentration of substance to the time interval over which the change occurred
- initial rate
- instantaneous rate of a chemical reaction at t = 0 s (immediately after the reaction has begun)
- instantaneous rate
- rate of a chemical reaction at any instant in time, determined by the slope of the line tangential to a graph of concentration as a function of time
- rate of reaction
- measure of the speed at which a chemical reaction takes place
- rate expression
- mathematical representation relating reaction rate to changes in amount, concentration, or pressure of reactant or product species per unit time


1.1: Introduction to Developmental Theories
-
- Last updated
- Save as PDF
Learning Objectives
- Define theory.
- Describe Freud’s theory of psychosexual development.
- Identify the parts of the self in Freud’s model.
- List five defense mechanisms.
- Describe five defense mechanisms.
- Appraise the strengths and weaknesses of Freud’s theory.
- List Erikson’s eight stages of psychosocial development.
- Apply Erikson’s stages to examples of people in various stages of the lifespan.
- Appraise the strengths and weaknesses of Erikson’s theory of psychosocial development.
- Compare and contrast Freud and Erikson’s theories of human development.
- Describe the principles of classical conditioning.
- Identify unconditioned stimulus, conditioned stimulus, unconditioned response, and conditioned response in classical conditioning.
- Describe the principles of operant conditioning.
- Identify positive and negative reinforcement, and primary and secondary reinforcement.
- Contrast reinforcement and punishment.
- Contrast classical and operant conditioning and the kinds of behaviors learned in each.
- Describe social learning theory.
- Describe Piaget’s theory of cognitive development.
- Define schema, assimilation, accommodation, and cognitive equilibrium.
- List Piaget’s stages of cognitive development.
- Describe Piaget’s stages of cognitive development.
- Critique Piaget’s theory of cognitive development.
- Describe Vygotsky’s sociocultural theory of cognitive development.
- Explain what is meant by the zone of proximal development.
- Explain guided participation.
- Describe scaffolding.
- Compare Piaget and Vygotsky’s models of cognitive development.
- Describe Bronfenbrenner’s ecological systems model.
What is a Theory?
Students sometimes feel intimidated by theory; even the phrase, “Now we are going to look at some theories…” is met with blank stares and other indications that the audience is now lost. But theories are valuable tools for understanding human behavior; if fact they are proposed explanations for the “how” and “whys” of development.  Theories can help explain these and other occurrences. Developmental theories offer explanations about how we develop, why we change over time and the kinds of influences that impact development.
A theory guides and helps us interpret research findings as well. It provides the researcher with a blueprint or model to be used to help piece together various studies. Think of theories are guidelines much like directions that come with an appliance or other object that required assembly. The instructions can help one piece together smaller parts more easily than if trial and error are used.
Theories can be developed using induction in which a number of single cases are observed and after patterns or similarities are noted, the theorist develops ideas based on these examples. Established theories are then tested through research; however, not all theories are equally suited to scientific investigation. Some theories are difficult to test but are still useful in stimulating debate or providing concepts that have practical application. Keep in mind that theories are not facts; they are guidelines for investigation and practice, and they gain credibility through research that fails to disprove them.
REFERENCES
Bandura, A. (1977). Social learning theory. New York: General Learning Press.
Bandura, A. (1986). Social foundations of thought and action; A social-cognitive theory. Upper Saddle River, NJ: Prentice Hall.
Bandura, A, Ross, D. &. Ross S. (1963). Imitation of film-mediated aggressive models. Journal of Abnormal and social Psychology 66:3-11.
Bronfenbrenner, U. (1979). The ecology of human development: Experiments by nature and design. Cambridge, MA: Harvard University Press.
Erikson, E. H. (1950). Childhood and society. New York: Norton.
Erikson, E. H. (1968). Identity, youth, and crisis. New York: Norton.
O’Grady, D. & Metz, J. (1987). Resilience in children at high risk for psychological disorder. Journal of Pediatric Psychology 12(1):3-23.
Piaget, J. (1929). The child’s conception of the world. NY: Harcourt, Brace Jovanovich.


8.4: Control of the Cell Cycle
By the end of this section, you will be able to do the following:
- Understand how the cell cycle is controlled by mechanisms that are both internal and external to the cell
- Explain how the three internal “control checkpoints” occur at the end of G 1 , at the G 2 /M transition, and during metaphase
- Describe the molecules that control the cell cycle through positive and negative regulation
The length of the cell cycle is highly variable, even within the cells of a single organism. In humans, the frequency of cell turnover ranges from a few hours in early embryonic development, to an average of two to five days for epithelial cells, and to an entire human lifetime spent in G 0 by specialized cells, such as cortical neurons or cardiac muscle cells.
There is also variation in the time that a cell spends in each phase of the cell cycle. When rapidly dividing mammalian cells are grown in a culture (outside the body under optimal growing conditions), the length of the cell cycle is about 24 hours. In rapidly dividing human cells with a 24-hour cell cycle, the G 1 phase lasts approximately nine hours, the S phase lasts 10 hours, the G 2 phase lasts about four and one-half hours, and the M phase lasts approximately one-half hour. By comparison, in fertilized eggs (and early embryos) of fruit flies, the cell cycle is completed in about eight minutes. This is because the nucleus of the fertilized egg divides many times by mitosis but does not go through cytokinesis until a multinucleate “zygote” has been produced, with many nuclei located along the periphery of the cell membrane, thereby shortening the time of the cell division cycle. The timing of events in the cell cycle of both “invertebrates” and “vertebrates” is controlled by mechanisms that are both internal and external to the cell.
Regulation of the Cell Cycle by External Events
Both the initiation and inhibition of cell division are triggered by events external to the cell when it is about to begin the replication process. An event may be as simple as the death of nearby cells or as sweeping as the release of growth-promoting hormones, such as human growth hormone (HGH or hGH) . A lack of HGH can inhibit cell division, resulting in dwarfism, whereas too much HGH can result in gigantism. Crowding of cells can also inhibit cell division. In contrast, a factor that can initiate cell division is the size of the cell: As a cell grows, it becomes physiologically inefficient due to its decreasing surface-to-volume ratio. The solution to this problem is to divide.
Whatever the source of the message, the cell receives the signal, and a series of events within the cell allows it to proceed into interphase. Moving forward from this initiation point, every parameter required during each cell cycle phase must be met or the cycle cannot progress.
Regulation at Internal Checkpoints
It is essential that the daughter cells produced be exact duplicates of the parent cell. Mistakes in the duplication or distribution of the chromosomes lead to mutations that may be passed forward to every new cell produced from an abnormal cell. To prevent a compromised cell from continuing to divide, there are internal control mechanisms that operate at three main cell-cycle checkpoints : A checkpoint is one of several points in the eukaryotic cell cycle at which the progression of a cell to the next stage in the cycle can be halted until conditions are favorable. These checkpoints occur near the end of G 1 , at the G 2 /M transition, and during metaphase (Figure \(\PageIndex{1}\)).
The G 1 Checkpoint
The G 1 checkpoint determines whether all conditions are favorable for cell division to proceed. The G 1 checkpoint, also called the restriction point (in yeast), is a point at which the cell irreversibly commits to the cell division process. External influences, such as growth factors, play a large role in carrying the cell past the G 1 checkpoint. In addition to adequate reserves and cell size, there is a check for genomic DNA damage at the G 1 checkpoint. A cell that does not meet all the requirements will not be allowed to progress into the S phase. The cell can halt the cycle and attempt to remedy the problematic condition, or the cell can advance into G 0 and await further signals when conditions improve.
The G 2 Checkpoint
The G 2 checkpoint bars entry into the mitotic phase if certain conditions are not met. As at the G 1 checkpoint, cell size and protein reserves are assessed. However, the most important role of the G 2 checkpoint is to ensure that all of the chromosomes have been replicated and that the replicated DNA is not damaged. If the checkpoint mechanisms detect problems with the DNA, the cell cycle is halted, and the cell attempts to either complete DNA replication or repair the damaged DNA.
The M Checkpoint
The M checkpoint occurs near the end of the metaphase stage of karyokinesis. The M checkpoint is also known as the spindle checkpoint, because it determines whether all the sister chromatids are correctly attached to the spindle microtubules. Because the separation of the sister chromatids during anaphase is an irreversible step, the cycle will not proceed until the kinetochores of each pair of sister chromatids are firmly anchored to at least two spindle fibers arising from opposite poles of the cell.
Link to Learning
Link to Learning
Watch what occurs at the G 1 , G 2 , and M checkpoints by exploring the HHMI BioInteractive to see an animation of the cell cycle.
Regulator Molecules of the Cell Cycle
In addition to the internally controlled checkpoints, there are two groups of intracellular molecules that regulate the cell cycle. These regulatory molecules either promote progress of the cell to the next phase (positive regulation) or halt the cycle (negative regulation). Regulator molecules may act individually, or they can influence the activity or production of other regulatory proteins. Therefore, the failure of a single regulator may have almost no effect on the cell cycle, especially if more than one mechanism controls the same event. However, the effect of a deficient or non-functioning regulator can be wide-ranging and possibly fatal to the cell if multiple processes are affected.
Positive Regulation of the Cell Cycle
Two groups of proteins, called cyclins and cyclin-dependent kinases (Cdks), are termed positive regulators. They are responsible for the progress of the cell through the various checkpoints. The levels of the four cyclin proteins fluctuate throughout the cell cycle in a predictable pattern (Figure \(\PageIndex{2}\)). Increases in the concentration of cyclin proteins are triggered by both external and internal signals. After the cell moves to the next stage of the cell cycle, the cyclins that were active in the previous stage are degraded by cytoplasmic enzymes, as shown in Figure \(\PageIndex{2}\) below.
Cyclins regulate the cell cycle only when they are tightly bound to Cdks. To be fully active, the Cdk/cyclin complex must also be phosphorylated in specific locations to activate the complex. Like all kinases, Cdks are enzymes ( kinases ) that in turn phosphorylate other proteins. Phosphorylation activates the protein by changing its shape. The proteins phosphorylated by Cdks are involved in advancing the cell to the next phase. (Figure \(\PageIndex{3}\)). The levels of Cdk proteins are relatively stable throughout the cell cycle; however, the concentrations of cyclin fluctuate and determine when Cdk/cyclin complexes form. The different cyclins and Cdks bind at specific points in the cell cycle and thus regulate different checkpoints.
Because the cyclic fluctuations of cyclin levels are largely based on the timing of the cell cycle and not on specific events, regulation of the cell cycle usually occurs by either the Cdk molecules alone or the Cdk/cyclin complexes. Without a specific concentration of fully activated cyclin/Cdk complexes, the cell cycle cannot proceed through the checkpoints.
Although the cyclins are the main regulatory molecules that determine the forward momentum of the cell cycle, there are several other mechanisms that fine-tune the progress of the cycle with negative, rather than positive, effects. These mechanisms essentially block the progression of the cell cycle until problematic conditions are resolved. Molecules that prevent the full activation of Cdks are called Cdk inhibitors. Many of these inhibitor molecules directly or indirectly monitor a particular cell-cycle event. The block placed on Cdks by inhibitor molecules will not be removed until the specific event that the inhibitor monitors is completed.
Negative Regulation of the Cell Cycle
The second group of cell-cycle regulatory molecules are negative regulators , which stop the cell cycle. Remember that in positive regulation, active molecules cause the cycle to progress.
The best understood negative regulatory molecules are retinoblastoma protein (Rb) , p53 , and p21 . Retinoblastoma proteins are a group of tumor-suppressor proteins common in many cells. We should note here that the 53 and 21 designations refer to the functional molecular masses of the proteins (p) in kilodaltons (a dalton is equal to an atomic mass unit , which is equal to one proton or one neutron or 1 g/mol). Much of what is known about cell-cycle regulation comes from research conducted with cells that have lost regulatory control . All three of these regulatory proteins were discovered to be damaged or non-functional in cells that had begun to replicate uncontrollably (i.e., became cancerous). In each case, the main cause of the unchecked progress through the cell cycle was a faulty copy of the regulatory protein.
Rb, p53, and p21 act primarily at the G 1 checkpoint. p53 is a multi-functional protein that has a major impact on the commitment of a cell to division because it acts when there is damaged DNA in cells that are undergoing the preparatory processes during G 1 . If damaged DNA is detected, p53 halts the cell cycle and then recruits specific enzymes to repair the DNA. If the DNA cannot be repaired, p53 can trigger apoptosis, or cell suicide, to prevent the duplication of damaged chromosomes. As p53 levels rise, the production of p21 is triggered. p21 enforces the halt in the cycle dictated by p53 by binding to and inhibiting the activity of the Cdk/cyclin complexes. As a cell is exposed to more stress, higher levels of p53 and p21 accumulate, making it less likely that the cell will move into the S phase.
Rb, which largely monitors cell size, exerts its regulatory influence on other positive regulator proteins. In the active , dephosphorylated state, Rb binds to proteins called transcription factors , most commonly, E2F (Figure \(\PageIndex{4}\)). Transcription factors “turn on” specific genes, allowing the production of proteins encoded by that gene. When Rb is bound to E2F, production of proteins necessary for the G 1 /S transition is blocked. As the cell increases in size, Rb is slowly phosphorylated until it becomes inactivated . Rb releases E2F, which can now turn on the gene that produces the transition protein, and this particular block is removed. For the cell to move past each of the checkpoints, all positive regulators must be “turned on,” and all negative regulators must be “turned off.”
Visual Connection
Visual Connection
Rb and other proteins that negatively regulate the cell cycle are sometimes called tumor suppressors. Why do you think the name tumor suppressor might be appropriate for these proteins?
Contributors and Attributions
Remixed and/or curated from the following works:
Clark, M. A., Douglas, M., & Choi, J. (2018). 2.7.4 Control of the Cell Cycle . In Biology 2e . OpenStax (CC BY 4.0; Access for free at https://openstax.org/books/biology-2e/pages/1-introduction ).


7.4: Order of Operations in Expressions and Formulas
Learning Outcomes
- Use order of operations in statistics formulas.
We have already encountered the order of operations: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). In this section, we will give some additional examples where the order of operations must be used properly to evaluate statistics.
Example \(\PageIndex{1}\)
The formula for the sample standard deviation asks us to add up the squared deviations from the mean, take the square root and divide by one less than the sample size. For example, suppose that there are three data values: 3, 5, 10. The mean of these values is 6. Then the standard deviation is:
\[s=\sqrt{\dfrac{\left(3-6\right)^2+\left(5-6\right)^2+\left(10-6\right)^2}{3-1}}\nonumber\]
Evaluate this number rounded to the nearest hundredth.
Solution
The first thing in the order of operations is to do what is in the parentheses. We must subtract:
\[3-6=-3,\:\:\:5-6\:=\:-1,\:\:\:10-6=4 \nonumber\]
We can substitute the numbers in to get:
\[=\sqrt{\frac{\left(-3\right)^2+\left(-1\right)^2+\left(4\right)^2}{3-1}}\nonumber\]
Next, we exponentiate:
\[\left(-3\right)^2=9,\:\:\:\left(-1\right)^2=1,\:\:\:4^2=16 \nonumber\]
Substitute these in to get:
\[\sqrt{\dfrac{9+1+16}{3-1}} \nonumber\]
We can now perform the addition inside the square root to get:
\[\sqrt{\dfrac{26}{3-1}} \nonumber\]
Next, perform the subtraction of the denominator to get:
\[\sqrt{\dfrac{26}{2}} \nonumber\]
We can divide to get:
\[\sqrt{13} \nonumber\]
We don't want to do this by hand, so in a calculator or computer type in:
\[\sqrt{13} = 3.61 \nonumber\] (rounded to the nearest hundredth)
Example \(\PageIndex{2}\)
When calculating the probability that a value will be less than 4.6 if the value is taken randomly from a uniform distribution between 3 and 7, we have to calculate:
\[\left(4.6-3\right)\times\dfrac{1}{7-3} \nonumber\]
Find this probability.
Solution
We can use a calculator or computer, but we must be very careful about the order of operations. Notice that there are implied parentheses due to the fraction bar. The answer is:
\[\dfrac{(4.6 - 3) \times 1}{7-3} \nonumber\]
Using technology, we get:
\[(4.6 - 3)\times 1/(7 - 3) = 0.4\]
Try It
When finding the upper bound, \(U\), of a confidence interval given the lower bound, \(L\), and the margin of error, \(E\), we use the formula
\[U=\:L+2E \nonumber\]
Find the upper bound of the confidence interval for the proportion of babies that are born preterm if the lower bound is 0.085 and the margin of error is 0.03.
- Answer
-
Plug in \(L\) = 0.085 and \(E\) = 0.03:
\[U=\:L+2E = 0.085 + 2(0.03) = 0.085 + 0.06 = 0.145\nonumber\]
The upper bound of the confidence interval for the proportion of babies that are born preterm is 0.145 or 14.5%.
Additional video resources:


26.2: Growing Diversity of Plants
When plants developed basic tissues and organs and thus became mature enough to survive on land, they started to increase in their diversity. All plants studied in this and following chapters belong to plants\(_2\), or kingdom Vegetabilia which is split into three phyla (Figure 6.1.1): Bryophyta (mosses and relatives), Pteridophyta (ferns and allies), and Spermatophyta (seed plants). The most striking differences between these phyla lay in the organization of their life cycles.
Land plants have a sporic life-cycle (Figure 4.4.7) that begins with a diplont (sporophyte); the mother cell of spores goes through the meiosis and produces haploid spores. These spores develop into haplont which produces female and male gametangia (gamete “homes”). Female is called archegonium , the male— antheridium ; the archegonium produces oocyte which is fertilized by the antheridium’s spermatozoon in the process of oogamy. When this fertilization happens, it forms a diploid zygote which then matures into a young sporophyte growing on a gametophyte . This kind of same species parasitism is almost unique in the living world. Only viviparous animals (like mammals with their pregnancy) could be compared with land plants.
-
- 26.2.1: Bryophyta - the Mosses
- Bryophyta has approximately 20,000 species. They do not have roots, but have long dead cells capable of water absorbency via apoplastic transport, these cells are called rhizoid cells. Their sporophyte is reduced to sporogon, which is simply a sporangium with setamosses: stalk of the sporogon (see) (stalk), and is usually parasitic. Gametophyte of bryophytes starts its development from a protonema, thread of cells.
-
- 26.2.2: Pteridophyta - the Ferns
- Pteridophyta, ferns and allies, have approximately 12,000 species and six classes. They have a sporic life cycle with sporophyte predominance whereas their gametophytes are often reduced to prothallium, small hornwort-like plant. Another frequent variant is the underground, mycoparasitic gametophyte. Pteridophyta (with one exception) have true roots. Most of them have vascular tissues and are homoiohydric. This is why seed plants together with ferns have a name vascular plants.


Skip to main content
-
Revising and editing are the end stages of the writing process in which you improve your work before producing a final draft.
-
When revising, you add, cut, move, or change information in order to improve content.
-
During editing and proofreading, you take a second (and maybe third) look at the words and sentences you used to express your ideas and fix any problems in grammar, punctuation, and sentence structure.
-
Unity in writing means that all the ideas in each paragraph and in the entire essay clearly belong together and are arranged in an order that makes logical sense.
-
Cohesion in writing means that the writer’s wording clearly indicates how one idea leads to another within a paragraph and between paragraphs; transitional words and phrases effectively make writing more cohesive.
-
Writing should be clear and concise, with no unnecessary words, and effective formal writing uses specific, appropriate words and avoids slang, contractions, clichés, and overly general words.
-
Peer reviews, done properly, can give writers objective feedback about their writing. It is the writer’s responsibility to evaluate the results of peer reviews and incorporate only useful feedback.
-
Remember to budget time for careful editing and proofreading. Use all available resources, including editing checklists, peer editing, and your institution’s writing lab, to improve your editing skills.
-
Deal with higher order concerns (changes in content in organization) before dealing with lower order concerns, such as word choice and sentence structure.
-
Along with getting feedback from others, reverse outlining is a good way to check your draft to see if any higher order concern changes need to be made.
-
Be sure to proofread your work (for spelling, punctuation, grammar, and formatting) using one of the proofreading techniques before turning in your paper.


3.6: Distribution and Quantile Functions
As usual, our starting point is a random experiment modeled by a with probability space \((\Omega, \mathscr F, \P)\). So to review, \(\Omega\) is the set of outcomes, \(\mathscr F\) is the collection of events, and \(\P\) is the probability measure on the sample space \((\Omega, \mathscr F)\). In this section, we will study two types of functions that can be used to specify the distribution of a real-valued random variable.
Distribution Functions
Definition
Suppose that \(X\) is a random variable with values in \(\R\). The (cumulative) distribution function of \(X\) is the function \(F: \R \to [0, 1]\) defined by \[ F(x) = \P(X \le x), \quad x \in \R\]
The distribution function is important because it makes sense for any type of random variable, regardless of whether the distribution is discrete, continuous, or even mixed, and because it completely determines the distribution of \(X\). In the picture below, the light shading is intended to represent a continuous distribution of probability, while the darker dots represents points of positive probability; \(F(x)\) is the total probability mass to the left of (and including) \(x\).
Basic Properties
A few basic properties completely characterize distribution functions. Notationally, it will be helpful to abbreviate the limits of \(F\) from the left and right at \(x \in \R\), and at \(\infty\) and \(-\infty\) as follows: \[F(x^+) = \lim_{t \downarrow x} F(t), \; F(x^-) = \lim_{t \uparrow x} F(t), \; F(\infty) = \lim_{t \to \infty} F(t), \; F(-\infty) = \lim_{t \to -\infty} F(t) \]
Suppose that \( F \) is the distribution function of a real-valued random variable \( X \).
- \(F\) is increasing: if \(x \le y\) then \(F(x) \le F(y)\).
- \(F(x^+) = F(x)\) for \(x \in \R\). Thus, \(F\) is continuous from the right .
- \(F(x^-) = \P(X \lt x)\) for \(x \in \R\). Thus, \(F\) has limits from the left .
- \(F(-\infty) = 0\).
- \(F(\infty) = 1\).
Proof
The following result shows how the distribution function can be used to compute the probability that \(X\) is in an interval. Recall that a probability distribution on \(\R\) is completely determined by the probabilities of intervals ; thus, the distribution function determines the distribution of \(X\).
Suppose again that \( F \) is the distribution function of a real-valued random variable \( X \). If \(a, \, b \in \R\) with \(a \lt b\) then
- \(\P(X = a) = F(a) - F(a^-)\)
- \(\P(a \lt X \le b) = F(b) - F(a)\)
- \(\P(a \lt X \lt b) = F(b^-) - F(a)\)
- \(\P(a \le X \le b) = F(b) - F(a^-)\)
- \(\P(a \le X \lt b) = F(b^-) - F(a^-)\)
Proof
These results follow from the definition , the basic properties , and the difference rule: \(\P(B \setminus A) = \P(B) - \P(A) \) if \( A, \, B \) are events and \( A \subseteq B\).
- \(\{X = a\} = \{X \le a\} \setminus \{X \lt a\}\), so \(\P(X = a) = \P(X \le a) - \P(X \lt a) = F(a) - F(a^-)\).
- \(\{a \lt X \le b\} = \{X \le b\} \setminus \{X \le a\}\), so \(\P(a \lt X \le b) = \P(X \le b) - \P(X \le a) = F(b) - F(a)\).
- \(\{a \lt X \lt b\} = \{X \lt b\} \setminus \{X \le a\}\), so \(\P(a \lt X \lt b) = \P(X \lt b) - \P(X \le a) = F(b^-) - F(a)\).
- \(\{a \le X \le b\} = \{X \le b\} \setminus \{X \lt a\}\), so \(\P(a \le X \le b) = \P(X \le b) - \P(X \lt a) = F(b) - F(a^-)\).
- \(\{a \le X \lt b\} = \{X \lt b\} \setminus \{X \lt a\}\), so \(\P(a \le X \lt b) = \P(X \lt b) - \P(X \lt a) = F(b^-) - F(a^-)\).
Conversely, if a Function \(F: \R \to [0, 1]\) satisfies the basic properties , then the formulas above define a probability distribution on \(\R\), with \(F\) as the distribution function. For more on this point, read the section on Existence and Uniqueness.
If \(X\) has a continuous distribution, then the distribution function \(F\) is continuous.
Proof
If \( X \) has a continuous distribution, then by definition, \( \P(X = x) = 0 \) so \( \P(X \lt x) = \P(X \le x) \) for \( x \in \R \). Hence from part (a) of the previous theorem , \( F(x^-) = F(x^+) = F(x) \).
Thus, the two meanings of continuous come together: continuous distribution and continuous function in the calculus sense. Next recall that the distribution of a real-valued random variable \( X \) is symmetric about a point \( a \in \R \) if the distribution of \( X - a \) is the same as the distribution of \( a - X \).
Suppose that \(X\) has a continuous distribution on \(\R\) that is symmetric about a point \(a\). Then the distribution function \(F\) satisfies \(F(a - t) = 1 - F(a + t)\) for \(t \in \R\).
Proof
Since \( X - a \) and \( a - X \) have the same distribution, \[ F(a - t) = \P(X \le a - t) = \P(X - a \le -t) = \P(a - X \le -t) = \P(X \ge a + t) = 1 - F(a + t) \]
Relation to Density Functions
There are simple relationships between the distribution function and the probability density function. Recall that if \(X\) takes value in \(S \subseteq \R\) and has probability density function \(f\), we can extend \(f\) to all of \(\R\) by the convention that \(f(x) = 0\) for \(x \in S^c\). As in Definition (1) , it's customary to define the distribution function \(F\) on all of \(\R\), even if the random variable takes values in a subset.
Suppose that \(X\) has discrete distribution on a countable subset \(S \subseteq \R\). Let \(f\) denote the probability density function and \(F\) the distribution function.
- \(F(x) = \sum_{t \in S, \, t \le x} f(t)\) for \(x \in \R\)
- \(f(x) = F(x) - F(x^-)\) for \(x \in S\)
Proof
- This follows from the definition of the PDF of \( X \), \( f(t) = \P(X = t) \) for \( t \in S \), and the additivity of probability.
- This is a restatement of part (a) of the theorem above .
Thus, \(F\) is a step function with jumps at the points in \(S\); the size of the jump at \(x\) is \(f(x)\).
There is an analogous result for a continuous distribution with a probability density function.
Suppose that \(X\) has a continuous distribution on \(\R\) with probability density function \(f\) and distribution function \(F\).
- \(F(x) = \int_{-\infty}^x f(t) dt\) for \(x \in \R\).
- \(f(x) = F^\prime(x)\) if \(f\) is continuous at \(x\).
Proof
The last result is the basic probabilistic version of the fundamental theorem of calculus. For mixed distributions, we have a combination of the results in the last two theorems.
Suppose that \(X\) has a mixed distribution, with discrete part on a countable subset \(D \subseteq \R\), and continuous part on \(\R \setminus D\). Let \(g\) denote the partial probability density function of the discrete part and assume that the continuous part has partial probability density function \(h\). Let \(F\) denote the distribution function.
- \(F(x) = \sum_{t \in D, \, t \le x} g(t) + \int_{-\infty}^x h(t) dt\) for \(x \in \R\)
- \(g(x) = F(x) - F(x^-)\) for \(x \in D\)
- \(h(x) = F^\prime (x)\) if \(x \notin D\) and \(h\) is continuous at \(x\)
Go back to the graph of a general distribution function. At a point of positive probability, the probability is the size of the jump. At a smooth point of the graph, the continuous probability density is the slope.
Recall that the existence of a probability density function is not guaranteed for a continuous distribution, but of course the distribution function always makes perfect sense. The advanced section on absolute continuity and density functioons has an example of a continuous distribution on the interval \((0, 1)\) that has no probability density function. The distribution function is continuous and strictly increases from 0 to 1 on the interval, but has derivative 0 at almost every point!
Naturally, the distribution function can be defined relative to any of the conditional distributions we have discussed. No new concepts are involved, and all of the results above hold.
Reliability
Suppose again that \(X\) is a real-valued random variable with distribution function \(F\). The function in the following definition clearly gives the same information as \(F\).
The function \(F^c\) defined by \[ F^c(x) = 1 - F(x) = \P(X \gt x), \quad x \in \R\] is the right-tail distribution function of \(X\). Give the mathematical properties of \(F^c\) analogous to the properties of \(F\) in (2) .
Answer
- \( F^c \) is decreasing.
- \( F^c(t) \to F^c(x) \) as \( t \downarrow x \) for \( x \in \R \), so \( F^c \) is continuous from the right.
- \( F^c(t) \to \P(X \ge x) \) as \( t \uparrow x \) for \( x \in \R \), so \( F^c \) has left limits.
- \( F^c(x) \to 0 \) as \( x \to \infty \).
- \( F^c(x) \to 1 \) as \( x \to -\infty \).
So \(F\) might be called the left-tail distribution function . But why have two distribution functions that give essentially the same information? The right-tail distribution function, and related functions, arise naturally in the context of reliability theory . For the remainder of this subsection, suppose that \(T\) is a random variable with values in \( [0, \infty) \) and that \( T \) has a continuous distribution with probability density function \( f \). Here are the important defintions:
Suppose that \( T \) represents the lifetime of a device.
- The right tail distribution function \( F^c \) is the reliability function of \( T \).
- The function \(h\) defined by \( h(t) = f(t) \big/ F^c(t)\) for \(t \ge 0 \) is the failure rate function of \( T \).
To interpret the reliability function, note that \(F^c(t) = \P(T \gt t)\) is the probability that the device lasts at least \(t\) time units. To interpret the failure rate function, note that if \( dt \) is
small
then \[ \P(t \lt T \lt t + dt \mid T \gt t) = \frac{\P(t \lt T \lt t + dt)}{\P(T \gt t)} \approx \frac{f(t) \, dt}{F^c(t)} = h(t) \, dt \] So \(h(t) \, dt\) is the approximate probability that the device will fail in the interval \((t, t + dt)\), given survival up to time \(t\). Moreover, like the distribution function and the reliability function, the failure rate function also completely determines the distribution of \(T\).
The reliability function can be expressed in terms of the failure rate function by \[ F^c(t) = \exp\left(-\int_0^t h(s) \, ds\right), \quad t \ge 0 \]
Proof
At the points of continuity of \( f \) we have \( \frac{d}{dt}F^c(t) = - f(t) \). Hence
\[ \int_0^t h(s) \, ds = \int_0^t \frac{f(s)}{F^c(s)} \, ds = \int_0^t -\frac{\frac{d}{ds}F^c(s)}{F^c(s)} \, ds = -\ln\left[F^c(t)\right] \]The failure rate function \(h\) satisfies the following properties:
- \(h(t) \ge 0\) for \(t \ge 0\)
- \(\int_0^\infty h(t) \, dt = \infty\)
Proof
- This follows from the definition.
- This follows from the previous result and the fact that \( F^c(t) \to 0 \) as \( t \to \infty \).
Conversely, a function that satisfies these properties is the failure rate function for a continuous distribution on \( [0, \infty) \):
Suppose that \(h: [0, \infty) \to [0, \infty) \) is piecewise continuous and \(\int_0^\infty h(t) \, dt = \infty\). Then the function \( G \) defined by \[ F^c(t) = \exp\left(-\int_0^t h(s) \, ds\right), \quad t \ge 0 \] is a reliability function for a continuous distribution on \( [0, \infty) \)
Proof
The function \( F^c \) is continuous, decreasing, and satisfies \( F^c(0) = 1 \) and \( F^c(t) \to 0 \) as \( t \to \infty \). Hence \( F = 1 - F^c \) is the distribution function for a continuous distribution on \( [0, \infty) \).
Multivariate Distribution Functions
Suppose now that \(X\) and \(Y\) are real-valued random variables for an experiment (that is, defined on the same probability space), so that \((X, Y)\) is random vector taking values in a subset of \(\R^2\).
The distribution function of \((X, Y)\) is the function \(F\) defined by \[ F(x, y) = \P(X \le x, Y \le y), \quad (x, y) \in \R^2\]
In the graph above, the light shading is intended to suggest a continuous distribution of probability, while the darker dots represent points of positive probability. Thus, \(F(x, y)\) is the total probability mass below and to the left (that is, southwest) of the point \((x, y)\). As in the single variable case, the distribution function of \((X, Y)\) completely determines the distribution of \((X, Y)\).
Suppose that \(a, \, b, \, c, \, d \in \R\) with \(a \lt b\) and \(c \lt d\). Then \[ \P(a \lt X \le b, c \lt Y \le d) = F(b, d) - F(a, d) - F(b, c) + F(a, c) \]
Proof
Note that \( \{X \le a, Y \le d\} \cup \{X \le b, Y \le c\} \cup \{a \lt X \le b, c \lt Y \le d\} = \{X \le b, Y \le d\} \). The intersection of the first two events is \( \{X \le a, Y \le c\} \) while the first and third events and the second and third events are disjoint. Thus, from the inclusion-exclusion rule we have \[ F(a, d) + F(b, c) + \P(a \lt X \le b, c \lt Y \le d) - F(a, c) = F(b, d) \]
A probability distribution on \( \R^2 \) is completely determined by its values on rectangles of the form \( (a, b] \times (c, d] \), so just as in the single variable case, it follows that the distribution function of \( (X, Y) \) completely determines the distribution of \( (X, Y) \). See the advanced section on existence and uniqueness of positive measures in the chapter on Probability Measures for more details.
In the setting of the previous result, give the appropriate formula on the right for all possible combinations of weak and strong inequalities on the left.
The joint distribution function determines the individual (marginal) distribution functions.
Let \(F\) denote the distribution function of \((X, Y)\), and let \(G\) and \(H\) denote the distribution functions of \(X\) and \(Y\), respectively. Then
- \(G(x) = F(x, \infty)\) for \( x \in \R \)
- \(H(y) = F(\infty, y)\) for \( y \in \R \)
Proof
These results follow from the continuity theorem for increasing events. For example, in (a) \[ \P(X \le x) = \P(X \le x, Y \lt \infty) = \lim_{y \to \infty} \P(X \le x, Y \le y) = \lim_{y \to \infty} F(x, y) \]
On the other hand, we cannot recover the distribution function of \( (X, Y) \) from the individual distribution functions, except when the variables are independent.
Random variables \(X\) and \(Y\) are independent if and only if \[ F(x, y) = G(x) H(y), \quad (x, y) \in \R^2\]
Proof
If \( X \) and \( Y \) are independent then \( F(x, y) = \P(X \le x, Y \le y) = \P(X \le x) \P(Y \le y) = G(x) H(y) \) for \( (x, y) \in \R^2 \). Conversely, suppose \( F(x, y) = G(x) H(y) \) for \( (x, y) \in \R^2 \). If \( a, \, b, \, c, \, d \in \R \) with \( a \lt b \) and \( c \lt d \) then from (15) , \begin{align} \P(a \lt X \le b, c \lt Y \le d) & = G(b)H(d) - G(a)H(d) -G(b)H(c) + G(a)H(c) \\ & = [G(b) - G(a)][H(d) - H(c)] = \P(a \lt X \le b) \P(c \lt Y \le d) \end{align} so it follows that \( X \) and \( Y \) are independent. (Recall again that a probability distribution on \( \R^2 \) is completely determined by its values on rectangles.)
All of the results of this subsection generalize in a straightforward way to \(n\)-dimensional random vectors. Only the notation is more complicated.
The Empirical Distribution Function
Suppose now that \( X \) is a real-valued random variable for a basic random experiment and that we repeat the experiment \( n \) times independently. This generates (for the new compound experiment) a sequence of independent variables \( (X_1, X_2, \ldots, X_n) \) each with the same distribution as \( X \). In statistical terms, this sequence is a random sample of size \( n \) from the distribution of \( X \). In statistical inference, the observed values \((x_1, x_2, \ldots, x_n)\) of the random sample form our data .
The empirical distribution function , based on the data \( (x_1, x_2, \ldots, x_n) \), is defined by \[ F_n(x) = \frac{1}{n} \#\left\{i \in \{1, 2, \ldots, n\}: x_i \le x\right\} = \frac{1}{n} \sum_{i=1}^n \bs{1}(x_i \le x), \quad x \in \R\]
Thus, \(F_n(x)\) gives the proportion of values in the data set that are less than or equal to \(x\). The function \(F_n\) is a statistical estimator of \(F\), based on the given data set. This concept is explored in more detail in the section on the sample mean in the chapter on Random Samples. In addition, the empirical distribution function is related to the Brownian bridge stochastic process which is studied in the chapter on Brownian motion.
Quantile Functions
Definitions
Suppose again that \( X \) is a real-valued random variable with distribution function \( F \).
For \(p \in (0, 1)\), a value of \(x\) such that \( F(x^-) = \P(X \lt x) \le p\) and \(F(x) = \P(X \le x) \ge p\) is called a quantile of order \(p\) for the distribution.
Roughly speaking, a quantile of order \(p\) is a value where the graph of the distribution function crosses (or jumps over) \(p\). For example, in the picture below, \(a\) is the unique quantile of order \(p\) and \(b\) is the unique quantile of order \(q\). On the other hand, the quantiles of order \(r\) form the interval \([c, d]\), and moreover, \(d\) is a quantile for all orders in the interval \([r, s]\). Note also that if \( X \) has a continuous distribution (so that \( F \) is continuous) and \( x \) is a quantile of order \( p \in (0, 1) \), then \( F(x) = p \).
Note that there is an inverse relation of sorts between the quantiles and the cumulative distribution values, but the relation is more complicated than that of a function and its ordinary inverse function, because the distribution function is not one-to-one in general. For many purposes, it is helpful to select a specific quantile for each order; to do this requires defining a generalized inverse of the distribution function \( F \).
The quantile function \( F^{-1} \) of \( X \) is defined by \[ F^{-1}(p) = \min\{x \in \R: F(x) \ge p\}, \quad p \in (0, 1)\]
\( F^{-1} \) is well defined
Since \(F\) is right continuous and increasing, \( \{x \in \R: F(x) \ge p\} \) is an interval of the form \( [a, \infty) \). Thus, the minimum of the set is \( a \).
Note that if \(F\) strictly increases from 0 to 1 on an interval \(S\) (so that the underlying distribution is continuous and is supported on \(S\)), then \(F^{-1}\) is the ordinary inverse of \(F\). We do not usually define the quantile function at the endpoints 0 and 1. If we did, note that \(F^{-1}(0)\) would always be \(-\infty\).
Properties
The following exercise justifies the name: \(F^{-1}(p)\) is the minimum of the quantiles of order \(p\).
Let \(p \in (0, 1)\).
- \(F^{-1}(p)\) is a quantile of order \(p\).
- If \(x\) is a quantile of order \(p\) then \(F^{-1}(p) \le x\).
Proof
Let \( y = F^{-1}(p) \).
- Note that \( F(y) \ge p \) by definition, and if \( x \lt y \) then \( F(x) \lt p \). Hence \( F(y^-) \le p \). Therefore \( y \) is a quantile of order \( p \).
- Suppose that \( x \) is a quantile of order \( p \). Then \( F(x) \ge p \) so by definition, \( y \le x \).
Other basic properties of the quantile function are given in the following theorem.
\(F^{-1}\) satisfies the following properties:
- \(F^{-1}\) is increasing on \((0, 1)\).
- \(F^{-1}\left[F(x)\right] \le x\) for any \(x \in \R\) with \(F(x) \lt 1\).
- \(F\left[F^{-1}(p)\right] \ge p\) for any \(p \in (0, 1)\).
- \(F^{-1}\left(p^-\right) = F^{-1}(p)\) for \(p \in (0, 1)\). Thus \(F^{-1}\) is continuous from the left.
- \(F^{-1}\left(p^+\right) = \inf\{x \in \R: F(x) \gt p\}\) for \(p \in (0, 1)\). Thus \(F^{-1}\) has limits from the right.
Proof
- Note that if \( p, \; q \in (0, 1) \) with \( p \le q \), then \( \{x \in \R: F(x) \ge q\} \subseteq \{x \in \R: F(x) \ge p\} \).
- This follows from the definition: \( F^{-1}\left[F(x)\right] \) is the smallest \( y \in \R \) with \( F(y) \ge F(x) \).
- This also follows from the definition: \( F^{-1}(p) \) is a value \( y \in \R \) satisfying \( F(y) \ge p \).
- This follows from the fact that \( F \) is continuous from the right
- This follows from the fact that \( F \) has limits from the left.
As always, the inverse of a function is obtained essentially by reversing the roles of independent and dependent variables. In the graphs below, note that jumps of \(F\) become flat portions of \(F^{-1}\) while flat portions of \(F\) become jumps of \(F^{-1}\). For \( p \in (0, 1) \), the set of quantiles of order \( p \) is the closed, bounded interval \( \left[F^{-1}(p), F^{-1}(p^+)\right] \). Thus, \( F^{-1}(p) \) is the smallest quantile of order \( p \), as we noted earlier, while \( F^{-1}(p^+) \) is the largest quantile of order \( p \).
The following basic property will be useful in simulating random variables, a topic explored in the section on transformations of random variables.
For \(x \in \R\) and \(p \in (0, 1)\), \(F^{-1}(p) \le x\) if and only if \(p \le F(x)\).
Proof
Suppose that \( F^{-1}(p) \le x \). Then, since \( F \) is increasing, \( F\left[F^{-1}(p)\right] \le F(x) \). But \( p \le F\left[F^{-1}(p)\right] \) by part (c) of the previous result, so \( p \le F(x) \). Conversely, suppose that \( p \le F(x) \). Then, since \( F^{-1} \) is increasing, \( F^{-1}(p) \le F^{-1}[F(x)] \). But \( F^{-1}[F(x)] \le x \) by part (b) of the previous result, so \( F^{-1}(p) \le x \).
Special Quantiles
Certain quantiles are important enough to deserve special names.
Suppose that \( X \) is a real-valued random variable.
- A quantile of order \(\frac{1}{4}\) is a first quartile of the distribution.
- A quantile of order \(\frac{1}{2}\) is a median or second quartile of the distribution.
- A quantile of order \(\frac{3}{4}\) is a third quartile of the distribution.
When there is only one median, it is frequently used as a measure of the center of the distribution, since it divides the set of values of \( X \) in half, by probability. More generally, the quartiles can be used to divide the set of values into fourths, by probability.
Assuming uniqueness, let \(q_1\), \(q_2\), and \(q_3\) denote the first, second, and third quartiles of \(X\), respectively, and let \(a = F^{-1}\left(0^+\right)\) and \(b = F^{-1}(1)\).
- The interquartile range is defined to be \( q_3 - q_1 \).
- The five parameters \( (a, q_1, q_2, q_3, b) \) are referred to as the five number summary of the distribution.
Note that the interval \( [q_1, q_3] \) roughly gives the middle half of the distribution, so the interquartile range, the length of the interval, is a natural measure of the dispersion of the distribution about the median. Note also that \(a\) and \(b\) are essentially the minimum and maximum values of \(X\), respectively, although of course, it's possible that \( a = -\infty \) or \( b = \infty \) (or both). Collectively, the five parameters give a great deal of information about the distribution in terms of the center, spread, and skewness. Graphically, the five numbers are often displayed as a
boxplot
or
box and whisker plot
, which consists of a line extending from the minimum value \(a\) to the maximum value \(b\), with a rectangular box from \(q_1\) to \(q_3\), and
whiskers
at \(a\), the median \(q_2\), and \(b\). Roughly speaking, the five numbers separate the set of values of \(X\) into 4 intervals of approximate probability \(\frac{1}{4}\) each.
Suppose that \( X \) has a continuous distribution that is symmetric about a point \(a \in \R\). If \(a + t\) is a quantile of order \(p \in (0, 1) \) then \(a - t\) is a quantile of order \(1 - p\).
Proof
Note that this is the quantile function version of symmetry result for the distribution function. If \( a + t \) is a qantile of order \( p \) then (since \( X \) has a continuous distribution) \( F(a + t) = p \). But then \( F(a - t) = 1 - F(a + t) = 1 - p \) so \( a - t \) is a quantile of order \( 1 - p \).
Examples and Applications
Distributions of Different Types
Let \(F\) be the function defined by \[ F(x) = \begin{cases} 0, & x \lt 1\\ \frac{1}{10}, & 1 \le x \lt \frac{3}{2}\\ \frac{3}{10}, & \frac{3}{2} \le x \lt 2\\ \frac{6}{10}, & 2 \le x \lt \frac{5}{2}\\ \frac{9}{10}, & \frac{5}{2} \le x \lt 3\\ 1, & x \ge 3; \end{cases}\]
- Sketch the graph of \(F\) and show that \(F\) is the distribution function for a discrete distribution.
- Find the corresponding probability density function \(f\) and sketch the graph.
- Find \(\P(2 \le X \lt 3)\) where \(X\) has this distribution.
- Find the quantile function and sketch the graph.
- Find the five number summary and sketch the boxplot.
Answer
- Note that \( F \) increases from 0 to 1, is a step function, and is right continuous.
- \(f(x) = \begin{cases} \frac{1}{10}, & x = 1 \\ \frac{1}{5}, & x = \frac{3}{2} \\ \frac{3}{10}, & x = 2 \\ \frac{3}{10}, & x = \frac{5}{2} \\ \frac{1}{10}, & x = 3 \end{cases}\)
- \(\P(2 \le X \lt 3) = \frac{3}{5}\)
- \(F^{-1}(p) = \begin{cases} 1, & 0 \lt p \le \frac{1}{10} \\ \frac{3}{2}, & \frac{1}{10} \lt p \le \frac{3}{10} \\ 2, & \frac{3}{10} \lt p \le \frac{6}{10} \\ \frac{5}{2}, & \frac{6}{10} \lt p \le \frac{9}{10} \\ 3, & \frac{9}{10} \lt p \le 1 \end{cases}\)
- \(\left(1, \frac{3}{2}, 2, \frac{5}{2}, 3\right)\)
Let \(F\) be the function defined by
\[F(x) = \begin{cases} 0, & x \lt 0\\ \frac{x}{x + 1}, & x \ge 0 \end{cases}\]- Sketch the graph of \(F\) and show that \(F\) is the distribution function for a continuous distribution.
- Find the corresponding probability density function \(f\) and sketch the graph.
- Find \(\P(2 \le X \lt 3)\) where \(X\) has this distribution.
- Find the quantile function and sketch the graph.
- Find the five number summary and sketch the boxplot.
Answer
- Note that \( F \) is continuous and increases from 0 to 1.
- \(f(x) = \frac{1}{(x + 1)^2}, \quad x \gt 0\)
- \(\P(2 \le X \lt 3) = \frac{1}{12}\)
- \(F^{-1}(p) = \frac{p}{1 - p}, \quad 0 \lt p \lt 1\)
- \(\left(0, \frac{1}{3}, 1, 3, \infty\right)\)
The expression \( \frac{p}{1 - p} \) that occurs in the quantile function in the last exercise is known as the odds ratio associated with \( p \), particularly in the context of gambling.
Let \(F\) be the function defined by
\[ F(x) = \begin{cases} 0, & x \lt 0\\ \frac{1}{4} x, & 0 \le x \lt 1\\ \frac{1}{3} + \frac{1}{4} (x - 1)^2, & 1 \le x \lt 2\\ \frac{2}{3} + \frac{1}{4} (x - 2)^3, & 2 \le x \lt 3\\ 1, & x \ge 3 \end{cases}\]- Sketch the graph of \(F\) and show that \(F\) is the distribution function of a mixed distribution.
- Find the partial probability density function of the discrete part and sketch the graph.
- Find the partial probability density function of the continuous part and sketch the graph.
- Find \(\P(2 \le X \lt 3)\) where \(X\) has this distribution.
- Find the quantile function and sketch the graph.
- Find the five number summary and sketch the boxplot.
Answer
- Note that \( F \) is piece-wise continuous, increases from 0 to 1, and is right continuous.
- \(g(1) = g(2) = g(3) = \frac{1}{12}\)
- \(h(x) = \begin{cases} \frac{1}{4}, & 0 \lt x \lt 1 \\ \frac{1}{2}(x - 1), & 1 \lt x \lt 2 \\ \frac{3}{4}(x - 2)^2, & 2 \lt x \lt 3 \end{cases}\)
- \(\P(2 \le X \lt 3) = \frac{1}{3}\)
- \(F^{-1}(p) = \begin{cases} 4 p, & 0 \lt p \le \frac{1}{4} \\ 1, & \frac{1}{4} \lt p \le \frac{1}{3} \\ 1 + \sqrt{4(p - \frac{1}{3})}, & \frac{1}{3} \lt p \le \frac{7}{12} \\ 2, & \frac{7}{12} \lt p \le \frac{2}{3} \\ 2 + \sqrt[3]{4 (p - \frac{2}{3})}, & \frac{2}{3} \lt p \le \frac{11}{12} \\ 3, & \frac{11}{12} \lt p \le 1 \end{cases}\)
- \(\left(0, 1, 1 + \sqrt{\frac{2}{3}}, 2 + \sqrt[3]{\frac{1}{3}}, 3\right)\)
The Uniform Distribution
Suppose that \(X\) has probability density function \(f(x) = \frac{1}{b - a}\) for \(x \in [a, b]\), where \(a, \, b \in \R\) and \(a \lt b\).
- Find the distribution function and sketch the graph.
- Find the quantile function and sketch the graph.
- Compute the five-number summary.
- Sketch the graph of the probability density function with the boxplot on the horizontal axis.
Answer
- \(F(x) = \frac{x - a}{b - a}, \quad a \le x \lt b\)
- \(F^{-1}(p) = a + (b - a) p, \quad 0 \le p \le 1\)
- \(\left(a, \frac{3 a + b}{4}, \frac{a + b}{2}, \frac{a + 3 b}{4}, b\right)\)
The distribution in the last exercise is the
uniform distribution
on the interval \( [a, b] \). The left endpoint \( a \) is the
location parameter
and the length of the interval \( w = b - a \) is the
scale parameter.
The uniform distribution models a point chose
at random
from the interval, and is studied in more detail in the chapter on Special Distributions.
In the special distribution calculator, select the continuous uniform distribution. Vary the location and scale parameters and note the shape of the probability density function and the distribution function.
The Exponential Distribution
Suppose that \(T\) has probability density function \(f(t) = r e^{-r t}\) for \(0 \le t \lt \infty\), where \(r \gt 0\) is a parameter.
- Find the distribution function and sketch the graph.
- Find the reliability function and sketch the graph.
- Find the failure rate function and sketch the graph.
- Find the quantile function and sketch the graph.
- Compute the five-number summary.
- Sketch the graph of the probability density function with the boxplot on the horizontal axis.
Answer
- \(F(t) = 1 - e^{-r t}, \quad 0 \le t \lt \infty\)
- \(F^c(t) = e^{-r t}, \quad 0 \le t \lt \infty\)
- \(h(t) = r, \quad 0 \le t \lt \infty\)
- \(F^{-1}(p) = -\frac{1}{r} \ln(1 - p), \quad 0 \le p \lt 1\)
- \(\left(0, \frac{1}{r}[\ln 4 - \ln 3], \frac{1}{r} \ln 2, \frac{1}{r} \ln 4 , \infty\right)\)
The distribution in the last exercise is the exponential distribution with rate parameter \(r\). Note that this distribution is characterized by the fact that it has constant failure rate (and this is the reason for referring to \( r \) as the rate parameter). The reciprocal of the rate parameter is the scale parameter . The exponential distribution is used to model failure times and other random times under certain conditions, and is studied in detail in the chapter on The Poisson Process.
In the special distribution calculator, select the exponential distribution. Vary the scale parameter \(b\) and note the shape of the probability density function and the distribution function.
The Pareto Distribution
Suppose that \(X\) has probability density function \(f(x) = \frac{a}{x^{a+1}}\) for \(1 \le x \lt \infty\) where \(a \gt 0\) is a parameter.
- Find the distribution function.
- Find the reliability function.
- Find the failure rate function.
- Find the quantile function.
- Compute the five-number summary.
- In the case \(a = 2\), sketch the graph of the probability density function with the boxplot on the horizontal axis.
Answer
- \(F(x) = 1 - \frac{1}{x^a}, \quad 1 \le x \lt \infty\)
- \(F^c(x) = \frac{1}{x^a}, \quad 1 \le x \lt \infty\)
- \(h(x) = \frac{a}{x}, \quad 1 \le x \lt \infty\)
- \(F^{-1}(p) = (1 - p)^{-1/a}, \quad 0 \le p \lt 1\)
- \(\left(1, \left(\frac{3}{4}\right)^{-1 / a}, \left(\frac{1}{2}\right)^{-1/a}, \left(\frac{1}{4}\right)^{-1/a}, \infty \right)\)
The distribution in the last exercise is the Pareto distribution with shape parameter \(a\), named after Vilfredo Pareto. The Pareto distribution is a heavy-tailed distribution that is sometimes used to model income and certain other economic variables. It is studied in detail in the chapter on Special Distributions.
In the special distribution calculator, select the Pareto distribution. Keep the default value for the scale parameter, but vary the shape parameter and note the shape of the density function and the distribution function.
The Cauchy Distribution
Suppose that \(X\) has probability density function \( f(x) = \frac{1}{\pi (1 + x^2)} \) for \(x \in \R\).
- Find the distribution function and sketch the graph.
- Find the quantile function and sketch the graph.
- Compute the five-number summary and the interquartile range.
- Sketch the graph of the probability density function with the boxplot on the horizontal axis.
Answer
- \(F(x) = \frac{1}{2} + \frac{1}{\pi} \arctan x, \quad x \in \R\)
- \(F^{-1}(p) = \tan\left[\pi\left(p - \frac{1}{2}\right)\right], \quad 0 \lt p \lt 1\)
- \((-\infty, -1, 0, 1, \infty)\), \(\text{IQR} = 2\)
The distribution in the last exercise is the Cauchy distribution , named after Augustin Cauchy. The Cauchy distribution is studied in more generality in the chapter on Special Distributions.
In the special distribution calculator, select the Cauchy distribution and keep the default parameter values. Note the shape of the density function and the distribution function.
The Weibull Distribution
Let \(h(t) = k t^{k - 1}\) for \(0 \lt t \lt \infty\) where \(k \gt 0\) is a parameter.
- Sketch the graph of \(h\) in the cases \(0 \lt k \lt 1\), \(k = 1\), \(1 \lt k \lt 2\), \( k = 2 \), and \( k \gt 2 \).
- Show that \(h\) is a failure rate function.
- Find the reliability function and sketch the graph.
- Find the distribution function and sketch the graph.
- Find the probability density function and sketch the graph.
- Find the quantile function and sketch the graph.
- Compute the five-number summary.
Answer
- \( h \) is decreasing and concave upward if \( 0 \lt k \lt 1 \); \( h = 1 \) (constant) if \( k = 1 \); \( h \) is increasing and concave downward if \( 1 \lt k \lt 2 \); \( h(t) = t \) (linear) if \( k = 2 \); \( h \) is increasing and concave upward if \( k \gt 2 \);
- \( h(t) \gt 0 \) for \( 0 \lt t \lt \infty \) and \( \int_0^\infty h(t) \, dt = \infty \)
- \(F^c(t) = \exp\left(-t^k\right), \quad 0 \le t \lt \infty\)
- \(F(t) = 1 - \exp\left(-t^k\right), \quad 0 \le t \lt \infty\)
- \(f(t) = k t^{k-1} \exp\left(-t^k\right), \quad 0 \le t \lt \infty\)
- \(F^{-1}(p) = [-\ln(1 - p)]^{1/k}, \quad 0 \le p \lt 1\)
- \(\left(0, [\ln 4 - \ln 3]^{1/k}, [\ln 2]^{1/k}, [\ln 4]^{1/k}, \infty\right)\)
The distribution in the previous exercise is the Weibull distributions with shape parameter \(k\), named after Walodi Weibull. The Weibull distribution is studied in detail in the chapter on Special Distributions. Since this family includes increasing, decreasing, and constant failure rates, it is widely used to model the lifetimes of various types of devices.
In the special distribution calculator, select the Weibull distribution. Keep the default scale parameter, but vary the shape parameter and note the shape of the density function and the distribution function.
Beta Distributions
Suppose that \(X\) has probability density function \(f(x) = 12 x^2 (1 - x)\) for \(0 \le x \le 1\).
- Find the distribution function of \(X\) and sketch the graph.
- Find \( \P\left(\frac{1}{4} \le X \le \frac{1}{2}\right) \).
- Compute the five number summary and the interquartile range. You will have to approximate the quantiles.
- Sketch the graph of the density function with the boxplot on the horizontal axis.
Answer
- \(F(x) = 4 x^3 - 3 x^4, \quad 0 \le x \le 1\)
- \(\P\left(\frac{1}{4} \le X \le \frac{1}{2}\right) = \frac{67}{256}\)
- \((0, 0.4563, 0.6413, 0.7570, 1)\), \(\text{IQR} = 0.3007\)
Suppose that \(X\) has probability density function \(f(x) = \frac{1}{\pi \sqrt{x (1 - x)}}\) for \(0 \lt x \lt 1\).
- Find the distribution function of \(X\) and sketch the graph.
- Compute \( \P\left(\frac{1}{3} \le X \le \frac{2}{3}\right) \).
- Find the quantile function and sketch the graph.
- Compute the five number summary and the interquartile range.
- Sketch the graph of the probability density function with the boxplot on the horizontal axis.
Answer
- \(F(x) = \frac{2}{\pi} \arcsin\left(\sqrt{x}\right), \quad 0 \le x \le 1\)
- \(\P\left(\frac{1}{3} \le X \le \frac{2}{3}\right) = 0.2163\)
- \(F^{-1}(p) = \sin^2\left(\frac{\pi}{2} p\right), \quad 0 \lt p \lt 1\)
- \(\left(0, \frac{1}{2} - \frac{\sqrt{2}}{4}, \frac{1}{2}, \frac{1}{2} + \frac{\sqrt{2}}{4}, 1\right)\), \(\text{IQR} = \frac{\sqrt{2}}{2}\)
The distributions in the last two exercises are examples of beta distributions . The particular beta distribution in the last exercise is also known as the arcsine distribution ; the distribution function explains the name. Beta distributions are used to model random proportions and probabilities, and certain other types of random variables, and are studied in detail in the chapter on Special Distributions.
In the special distribution calculator, select the beta distribution. For each of the following parameter values, note the location and shape of the density function and the distribution function.
- \(a = 3\), \(b = 2\). This gives the first beta distribution above.
- \(a = b = \frac{1}{2}\). This gives the arcsine distribution above
Logistic Distribution
Let \(F(x) = \frac{e^x}{1 + e^x}\) for \(x \in \R\).
- Show that \(F\) is a distribution function for a continuous distribution, and sketch the graph.
- Compute \( \P(-1 \le X \le 1) \) where \(X\) is a random variable with distribution function \(F\).
- Find the quantile function and sketch the graph.
- Compute the five-number summary and the interquartile range.
- Find the probability density function and sketch the graph with the boxplot on the horizontal axis.
Answer
- Note that \( F \) is continuous, and increases from 0 to 1.
- \(\P(-1 \le X \le 1) = 0.4621\)
- \(F^{-1}(p) = \ln \left(\frac{p}{1 - p}\right), \quad 0 \lt p \lt 1\)
- \((-\infty, -\ln 3, 0, \ln 3, \infty)\)
- \(f(x) = \frac{e^x}{(1 + e^x)^2}, \quad x \in \R\)
The distribution in the last exercise is an logistic distribution and the quantile function is known as the logit function . The logistic distribution is studied in detail in the chapter on Special Distributions.
In the special distribution calculator, select the logistic distribution and keep the default parameter values. Note the shape of the probability density function and the distribution function.
Extreme Value Distribution
Let \(F(x) = e^{-e^{-x}}\) for \(x \in \R\).
- Show that \(F\) is a distribution function for a continuous distribution, and sketch the graph.
- Compute \(\P(-1 \le X \le 1)\) where \(X\) is a random variable with distribution function \(F\).
- Find the quantile function and sketch the graph.
- Compute the five-number summary.
- Find the probability density function and sketch the graph with the boxplot on the horizontal axis.
Answer
- Note that \( F \) is continuous, and increases from 0 to 1.
- \(\P(-1 \le X \le 1) = 0.6262\)
- \(F^{-1}(p) = -\ln(-\ln p), \quad 0 \lt p \lt 1\)
- \(\left(-\infty, -\ln(\ln 4), -\ln(\ln 2), -\ln(\ln 4 - \ln 3), \infty\right)\)
- \(f(x) = e^{-e^{-x}} e^{-x}, \quad x \in \R\)
The distribution in the last exercise is the type 1 extreme value distribution , also known as the Gumbel distribution in honor of Emil Gumbel. Extreme value distributions are studied in detail in the chapter on Special Distributions.
In the special distribution calculator, select the extreme value distribution and keep the default parameter values. Note the shape and location of the probability density function and the distribution function.
The Standard Normal Distribution
Recall that the standard normal distribution has probability density function \( \phi \) given by \[ \phi(z) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} z^2}, \quad z \in \R\] This distribution models physical measurements of all sorts subject to small, random errors, and is one of the most important distributions in probability. The normal distribution is studied in more detail in the chapter on Special Distributions. The distribution function \( \Phi \), of course, can be expressed as \[ \Phi(z) = \int_{-\infty}^z \phi(x) \, dx, \quad z \in \R \] but \( \Phi \) and the quantile function \( \Phi^{-1} \) cannot be expressed, in closed from, in terms of elementary functions. Because of the importance of the normal distribution \( \Phi \) and \( \Phi^{-1} \) are themselves considered special functions , like \( \sin \), \( \ln \), and many others. Approximate values of these functions can be computed using most mathematical and statistical software packages. Because the distribution is symmetric about 0, \( \Phi(-z) = 1 - \Phi(z) \) for \( z \in \R \), and equivalently, \( \Phi^{-1}(1 - p) = -\Phi^{-1}(p)\). In particular, the median is 0.
Open the sepcial distribution calculator and choose the normal distribution. Keep the default parameter values and select CDF view. Note the shape and location of the distribution/quantile function. Compute each of the following:
- The first and third quartiles
- The quantiles of order 0.9 and 0.1
- The quantiles of order 0.95 and 0.05
Miscellaneous Exercises
Suppose that \(X\) has probability density function \(f(x) = -\ln x\) for \(0 \lt x \le 1\).
- Sketch the graph of \(f\).
- Find the distribution function \(F\) and sketch the graph.
- Find \( \P\left(\frac{1}{3} \le X \le \frac{1}{2}\right) \).
Answer
- \(F(x) = x - x \ln x, \quad 0 \lt x \lt 1\)
- \(\P(\frac{1}{3} \le X \le \frac{1}{2}) = \frac{1}{6} + \frac{1}{2} \ln 2 - \frac{1}{3} \ln 3\)
Suppose that a pair of fair dice are rolled and the sequence of scores \((X_1, X_2)\) is recorded.
- Find the distribution function of \(Y = X_1 + X_2\), the sum of the scores.
- Find the distribution function of \( V = \max \{X_1, X_2\} \), the maximum score.
- Find the conditional distribution function of \(Y\) given \(V = 5\).
Answer
The random variables are discrete, so the CDFs are step functions, with jumps at the values of the variables. The following tables give the values of the CDFs at the values of the random variables.
-
\(y\) 2 3 4 5 6 7 8 9 10 11 12 \(\P(Y \le y)\) \(\frac{1}{36}\) \(\frac{3}{36}\) \(\frac{6}{36}\) \(\frac{10}{36}\) \(\frac{15}{36}\) \(\frac{21}{36}\) \(\frac{26}{36}\) \(\frac{30}{36}\) \(\frac{33}{36}\) \(\frac{35}{36}\) 1 -
\(v\) 1 2 3 4 5 6 \(\P(V \le v)\) \(\frac{1}{36}\) \(\frac{4}{36}\) \(\frac{9}{36}\) \(\frac{16}{36}\) \(\frac{25}{36}\) 1 -
\(y\) 6 7 8 9 10 \(\P(Y \le y \mid V = 5)\) \(\frac{2}{9}\) \(\frac{4}{9}\) \(\frac{6}{9}\) \(\frac{8}{9}\) 1
Suppose that \((X, Y)\) has probability density function \(f(x, y) = x + y\) for \(0 \le x \le 1\), \(0 \le y \le 1\).
- Find the distribution function of \(X, Y)\).
- Compute \(\P\left(\frac{1}{4} \le X \le \frac{1}{2}, \frac{1}{3} \le Y \le \frac{2}{3}\right)\).
- Find the distribution function of \(X\).
- Find the distribution function of \(Y\).
- Find the conditional distribution function of \(X\) given \(Y = y\) for \(0 \lt y \lt 1 \).
- Find the conditional distribution function of \(Y\) given \(X = x\) for \(0 \lt x \lt 1\).
- Are \(X\) and \(Y\) independent?
Answer
- \(F(x, y) = \frac{1}{2}\left(x y^2 + y x^2\right); \quad 0 \lt x \lt 1, \; 0 \lt y \lt 1\)
- \(\P\left(\frac{1}{4} \le X \le \frac{1}{2}, \frac{1}{3} \le Y \le \frac{2}{3}\right) = \frac{7}{96}\)
- \(G(x) = \frac{1}{2}\left(x + x^2\right), \quad 0 \lt x \lt 1\)
- \(H(y) = \frac{1}{2}\left(y + y^2\right), \quad 0 \lt y \lt 1\)
- \(G(x \mid y) = \frac{x^2 / 2 + x y}{y + 1/2}; \quad 0 \lt x \lt 1, \; 0 \lt y \lt 1\)
- \(H(y \mid x) = \frac{y^2 / 2 + x y}{x + 1/2}; \quad 0 \lt x \lt 1, \; 0 \lt y \lt 1\)
Statistical Exercises
For the M&M data, compute the empirical distribution function of the total number of candies.
Answer
Let \(N\) denote the total number of candies. The empirical distribution function of \(N\) is a step function; the following table gives the values of the function at the jump points.
| \(n\) | 50 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 | 61 |
|---|---|---|---|---|---|---|---|---|---|---|
| \(\P(N \le n)\) | \(\frac{1}{30}\) | \(\frac{2}{30}\) | \(\frac{3}{30}\) | \(\frac{7}{30}\) | \(\frac{11}{30}\) | \(\frac{14}{30}\) | \(\frac{23}{30}\) | \(\frac{26}{30}\) | \(\frac{28}{30}\) | 1 |
For the cicada data, let \(BL\) denotes body length and let \(G\) denote gender. Compute the empirical distribution function of the following variables:
- \(BL\)
- \(BL\) given \(G = 1\) (male)
- \(BL\) given \(G = 0\) (female).
- Do you believe that \(BL\) and \(G\) are independent?
For statistical versions of some of the topics in this section, see the chapter on Random Samples, and in particular, the sections on empirical distributions and order statistics.


9: The Gaseous State
Some of the first real breakthroughs in the study of chemistry happened in the study of the gaseous state. In gases, the volume of the actual gas particles is but a tiny fraction of the total volume that the gas occupies. This allowed early chemists to relate parameters such as volume and the number of gas particles, leading to the development of the mole concept. As we have seen in previous chapters, the notion of a chemical mole allows us to do quantitative chemistry and lead us to the point where we can routinely address reaction stoichiometry, etc. In this chapter, we will visit some of the early observations that lead to our current understand of gasses and how they behave. We will see how the relationships between pressure and volume; volume and temperature and volume and moles lead to the ideal gas laws and how these simple rules can allow us to do quantitative calculations in the gas phase.


11.5: Power
The measure of a person's ability to control the environment around them including the behavior of other people. If a human being's environment includes citizens then their power is measured by how much control they have over the masses.
Types of Power
1. Money
Money, in many countries, is the foremost source of power. Those that are more wealthy can often use it as power by using their wealth to their own benefit or to the benefit of their community. Today there are over 178 currencies in use (CIA World Factbook), with most countries using the currency they produce themselves. Money in any form of currency can be exchanged for other types of currency, thus making the power of money worldwide. This tool also has the power to determine the monetary status of certain countries. Depending on how much money is in circulation, the value of that country's money either goes up or down. If there is a lot of money, the value of that country's coin would be a lot less than if there wasn't as much money in circulation. This can influence the exchange rate between countries, making tourists more or less likely to travel there, based on how expensive that country is. In extreme cases, hyperinflation can breakdown a nations monetary system, as seen in Germany during 1923 when hyperinflation got so high that prices were raised by 2500%. http://www.investopedia.com/university/inflation/inflation1.asp
Money can also be used to help persuade politics in a country. In the United States, Super PACs are legal, which allows companies and unions, even sometimes individuals, to anonymously donate as much money as they want to politicians.
2. Social Class
Social class is the hierarchy among members of a society. Often people are born into it, or it is gained through money, education, or career. In some cultures people must stay within their social class through life (ascribed status), and in other cultures it is allowable or even respected if people work their way up the social ladder (achieved status). The class a person belongs to is often associated with an identity or subculture within society. People of a higher class associate and have similar lives as people within that class, and the same goes for people of the lower or middle classes. One extreme example of class is the caste system in India which divides people into five different groups within society.
According to Karl Heinrich Marx, there is no freedom for individuals in the social class. Human beings are motivated by social, economic power such as money and status. These desires that people originally have make the hierarchy. He also argues that for these people tend to believe in religion strongly so they can repress their resentment which emerges because of the social situation.
3. Physical Force
Physical force is using physical coercion as a means to gain power and control over others. Such forces have been used by civilizations for thousands of years in order to survive. Stateless societies typically had this form of power employed, where locals feared other powerful locals. A person in high position of power might use force to persuade others do perform a task, or make them stop practicing certain rituals [14]. Force usually involves violence and inflicts fear. It is used in a variety ways and seen utilized in many places across the globe. Many people try to resist force and try to retaliate, which can often lead to harm being done to one or both parties involved. This is also known as the power of free agency; or "the freedom of self contained individuals to pursue their own interests above everything else and to challenge another for dominance."
A specific example of physical force being used for free agency is the Gulf War. Iraq used their large army to conquer Kuwait so they could gain oil, in an effort to benefit financially.[9] Oil, however, is the lifeblood of many western nations, and for the first time in its history, the United Nations formed a coalition to make sure that the U.S. and Western European Nations would continue to receive cheap gasoline throughout the remainder of the 20 th century. This is free agency because Iraq was pursuing its own interests without worrying about how the world or Kuwait would respond. The world then responded in an efficient manner and evicted Iraq from Kuwait, basically telling Iraq that they couldn't do whatever they want just because they are bigger than another country.
In new era of nuclear weapons and WMD, the threat of violence or physical force can be just as effective as actually perpetuating the violence. We see examples of this during the Cold War, when extreme nuclear proliferation, lead to an arms race between the United States and the USSR. Nuclear proliferation lead to the theory of mutually assured destruction, which occurs when conflict would cause the complete annihilation of both the attacker and the defender. Neither side has any incentive to initiate a conflict or to disarm, and it therefor serves as a preventative measure. We saw the height of this kind of strategic theory, during the Cuban Missile Crisis, which was the closest the world has ever gotten to all out nuclear holocaust.
4. Persuasion
Persuasion: Power based on verbal argument (Schultz-Lavenda, 244) . Persuasion is the act of influencing others into adopting an idea. This is usually done through speech and rarely through physical acts. The ability to persuade is highly coveted and is often associated with powerful people in many societies, because those who can persuade can be trusted to speak for their people. This is why persuasive speakers tend to be the ones to speak for those in their societies. For example, the Huichol People of Northern Mexico have shaman that see themselves as being able to speak for all of the Huichol and are often involved in negotiations with other societies because of their persuasive abilities. (Shultz-Lavenda 244-245). Blackmail is a tool of persuasion with more sinister intentions. It is usually in the form of using a threat, whether it is a way of tarnishing one's reputation or performing an act that can do harm on a person's way of life. And in exchange of not pursuing this threat, needs or satisfactions are met.
CITATION: Some tips taken from: http://mindprod.com/ethics/persuasion.html
5. Fame
Combine the two main sources of power and what do you get? Fame. Fame is based on money and a high social status. Many western cultures look to those who are famous almost as idols. Money and a high social class, whether gained or born into, are closely related to celebrities and socialites. Those who gain or inherit fame are given power almost automatically, even though they do not necessarily have the right to have this power.
An ethnographic example of the power that comes with fame would be celebrities. In the United States, celebrities are fawned over by their fans, and in some cases receive special treatment over non-celebrities. For example, in the case of imprisonment, celebrities are able to get out of prison early for no apparent reason. One example of a celebrity who was able to get out of prison before the end of their sentence was Nicole Richie. She was sentenced for driving the wrong way down a road while drunk, for which she served a total of eighty-two minutes instead of four days. [15]
6. Tradition
The power of Tradition in a culture can be defined as the possession of control or command over others through a long established way of thinking. This type of power is usually asserted through means of religion, cultural beliefs and workforce. Religions have long histories, which inevitably create traditional customs, laws, beliefs and ways of thinking or processing. Certain cultures have traditional beliefs that grant power logically to one sex over the other, such as in patriarchal or matriarchal cultures.
In an ethnographic attempt to further explain the power of tradition, examine the status of women in Islamic religion. In Farnaz Fassihi’s book of her reporting in Iraq post Saddam Hussein’s fall, she states how her gender is a reoccurring problem when Iraq’s policies are being greatly influenced by its Islamic traditions and its Islamic religious leader, Grand Ayatollah Sistani. Fassihi explains, “I never had to tiptoe around my gender the way I do now...at offices of political parties or clerics, I am required to stand outside under the sweltering sun because the waiting room is designated for men...the entrances to the holy shrines in Karbala and Najaf now have segregated entries with a...police officer checking the attire of visiting female” (Fassihi 116). Fassihi is appalled at the way women are thought of and are treated, but it is the power of tradition that allows men to carry on this way. The power of tradition creates customs within cultures and religions. In the Islamic faith, it is customary for women to dress modestly, in this sense women wear head-covers (see Women in Islam). Tradition can furthermore create beliefs, such as the belief that women should not look men in the eye if outside of their immediate family.
Kinds of Social Power
- Interpersonal Power : The ability of one individual to impose his or her own will on another individuals (Schultz-Lavenda, 233).
In its broadest sense, interpersonal power refers to the cause of any change in the behavior of one actor, B, which can be attributed to the effect of another actor, A. It can refer to the capacity and usage of that capacity to cause such change (Weber [1918] 1968), (Dahl 1957; Simon 1953) but always to overcoming the "resistance" of B (Weber [1918] 1968), hence causing B to do something B would not otherwise do (Dahl 1957). Interpersonal power is therefore the power of one individual "over" another as opposed to an individual's power to do something, the capacity of an actor to attain some goal (IPES, BookRags).
- Organizational Power : Highlights how individuals or social units can limit the actions of other individuals in particular social settings (Schultz-Lavenda, 233).
Organizational Power Politics is about how individuals can achieve their objectives in organizational work groups. Office politics or organizational politics, is a significant part of the life of everyone who works with others in formal or informal groups. These relationships are power-tinged, and success can be attained only as we use power effectively. Understanding what power is and how it can be used to gain personal or group objectives is the focus of the book. It provides readers with specific recommendations about the situations in which power use can be effective, and it identifies those tactics most effective in leading subordinates and superiors toward the achievement of our goals. This work will be of interest to scholars and practicing managers seeking information on how better to use organizational politics to attain personal and organizational goals. It provides insight into power theory, as well as a practical model for power use, strategic orientation, and operational tactics (Choice, Greenwood).
- Structural Power : Organizes social settings themselves and controls the allocation of social labor (Schultz-Lavenda, 233).
Winter and Stewart (1978) have provided a useful taxonomy of power-related constructs linking the organizational and individual levels. Power as an attribute of particular social roles (e.g., jobs) locates individuals in organizational roles that legitimize or require actual power behavior (actions affecting the behavior and emotions of other people) from the individual for effective role performance. The enjoyment of power satisfaction, regardless of social role, requires both feeling powerful as a result of successful power behavior and the capacity to find that feeling gratifying. Thus, power as a source of job satisfaction depends on opportunities for power behavior, frequent successful outcomes of power behavior, and the experience of feeling powerful. This should be more likely in jobs that provide structural power as an attribute of the occupational role (Bnet).
The Role of the State
Many early political anthropologists assumed that in order for a civilization to be socially civilized, a state was necessary. This idea was rooted in the western idea that without a state, disorder and anarchy would erupt. Anthropologists such as Lewis Henry Morgan proved that successful societies where an actual state is not present exist. In these societies, it was common that various roles were given to different people, thus distributing power among the people. Order within a social group in the absence of a state can be maintained so long as the group has a system in which they organize themselves.


Skip to main content
-
-
6.1: Introduction to the Skeleton System
-
The skeletal system forms the rigid internal framework of the body. It consists of the bones, cartilages, and ligaments. Bones support the weight of the body, allow for body movements, and protect internal organs. Each bone of the body serves a particular function, and therefore bones vary in size, shape, and strength based on these functions.
-
-
6.2: Divisions of the Skeletal System
-
The skeleton is subdivided into two major divisions—the axial and appendicular. The axial skeleton forms a vertical axis and the appendicular skeleton includes all of the bones of the upper and lower limbs plus the bones that anchor each limb to the axial skeleton
-
-
6.3: The Skull
-
In the adult, the skull consists of 22 individual bones, 21 of which are immobile and united into a single unit. The skull is the skeletal structure of the head that supports the face and protects the brain. It is subdivided into the facial bones and the brain case.
-
-
6.4: The Vertebral Column
-
The adult vertebral column consists of 24 vertebrae, plus the sacrum and coccyx. The vertebrae are divided into three regions: cervical vertebrae, thoracic vertebrae, and lumbar vertebrae.
-
-
6.5: The Thoracic Cage
-
The thoracic cage (rib cage) forms the thorax (chest) portion of the body. It consists of the 12 pairs of ribs with their costal cartilages and the sternum. The ribs are anchored posteriorly to the 12 thoracic vertebrae (T1–T12). The thoracic cage protects the heart and lungs.
-
-
6.6: Embryonic Development of the Axial Skeleton
-
The axial skeleton begins to form during early embryonic development. However, growth, remodeling, and ossification (bone formation) continue for several decades after birth before the adult skeleton is fully formed. Knowledge of the developmental processes that give rise to the skeleton is important for understanding the abnormalities that may arise in skeletal structures.