Figure 7.1 Albert Einstein (1879–1955), best known for his formula E = MC² and for his statement “God does not play dice with the world,” was awarded the Nobel Prize in Physics in 1921.

Preview

“Lest men suspect your tale untrue, Keep probability in view.”*

John Gay, English poet and dramatist (1685–1732)

At the beginning of the nineteenth century, the prevailing scientific view was a clockwork universe in which everything was determinable, if not actually determined. An asteroid hitting Earth was not a chance event; rather, it could have been anticipated if the position and velocity of all asteroids in the solar system were known. This was the view of Pierre-Simon Laplace (1749–1827) whose methods on how to compute future positions of planets and comets from observations of their past positions were published in a five-volume treatise titled Méchanique céleste. A century later, the clockwork view of classical mechanics was shattered by twentieth century quantum mechanics, built on Werner Heisenberg's (1901–1976) uncertainty principle. This principle implies that the precision with which both the position and velocity of any object can be known at a given point in time is limited. Einstein's difficulty in accepting this principle is encapsulated in one of his most quoted statements (translated from the German): “God does not play dice with the world.” Whether we believe that the time of death of an individual is preordained by God or is subject to the throw of a cosmic die is not relevant. In either case, we know that many kinds of computations in biology are intrinsically predictions of the frequencies of the occurrence of particular events. Think of flipping coins: heads can be predicted to occur only 50% of the time.

The mathematical theory of chance started when a French writer, Chevalier de Méré, with a penchant for gambling and an interest in mathematics, challenged the French mathematician Blaise Pascal (1623–1662) to solve a betting problem. Pascal teamed up with another French mathematician, Pierre de Fermat (1601–1665), to solve the problem; in the process, they laid the foundations of probability theory. It is no exaggeration to say that without probability theory, the biological sciences would not exist as we know them today. All concepts, ideas, and calculations relating to the theory and practice of mathematical statistics and its indispensable application to experimental biology would not exist. Thus, students in the biological sciences need to become immersed in ideas that relate to chance and probabilities as early as possible and have these ideas reinforced as often as possible. Certain foundational aspects of probability theory rely on integration; this chapter provides an elementary introduction to these aspects of probability.

We begin with organizing data into histograms, a topic reminiscent of how we used discrete rectangles to approximate the area under a continuous curve. Integration allows us to make the transition from histograms to continuous probability density functions and cumulative distribution functions. These functions often have infinite domains (e.g., heights theoretically can take on any positive value) and, consequently, require improper integrals. Next we consider means and variances of these functions, fundamental concepts for all sciences. We then focus on applications: the probability of getting infected as a disease sweeps through a population, identification of underweight children, time to extinction of marine species, life expectancy of humans and dinosaurs, and probabilities of tumor regrowth after medical treatment.

7.1 Histograms, PDFs, and CDFs

Histograms and probabilities

In previous chapters we saw how gathering data can lead to uncovering and ultimately understanding many interesting phenomena. In the simplest case, data represent single values, each associated with an object or an individual. Some examples of sets of data we considered (or, in the case of a coin or die, familiar to us all) are listed in Table 7.1.

Table 7.1 Examples of random variables with expected ranges

One way of visualizing large data sets is to organize values by the proportion that fall into subintervals, known as bins. For example, consider the fourth data set listed in Table 7.1: If we have data on 100 animals no larger than a cow, we can ask what proportion of these 100 individuals require, respectively, (0, 1], (1, 2], ..., and (9, 10] thousand kilocalories per day. We can then plot the proportions as a bar graph, referred to as a histogram.

Given a data set, various types of histograms can be created, depending on how we divide up the interval over which the data are defined. The most common type is obtained by splitting the range of the data into equal-sized bins, as illustrated in the following example.

As we have seen in Example 1, histograms provide a sense of whether there is a center to a data set (i.e., where most of the data points lie), how much spread there is to the data set (i.e., is it even over the whole range, peaked in more than one region, or concentrated around a center), and how skewed the data set is (i.e., whether there are more data points to the left or right of the center). For example, in Figure 7.2c, the data are centered around the bin of 25–30 species and the histogram is slightly skewed to the right.

One interpretation of the area of a rectangle in a histogram is the proportion of the data in the interval. An alternative interpretation is in terms of random variables. A random variable, usually written X, is a variable whose possible values are outcomes of a random phenomenon, such as the rolling of a die, this evening's winning lottery numbers, or the height of a randomly chosen individual. Two important types of random variables are discrete random variables, which we discuss next, and continuous random variables, which we examine later in this section. In the discussion of random variables, we provide only simple, intuitive definitions to facilitate our presentation of probabilistic applications of integration. Introductory statistics and probability classes provide more in-depth studies of these concepts.

When discrete random variables take on an infinite number of values, the infinite sum p₁ + p₂ + p₃ + ··· corresponds to a infinite series, which is well-defined only if p₁ + p₂ + ··· + p_n exists. As with the improper integrals discussed in Section 7.2, understanding when these limits are well-defined is a delicate issue. However, all of our examples only deal with finite number of values.

To see how random variables relate to data sets and their histograms, imagine writing down each data value on a slip of paper, putting all the slips in a hat, drawing one slip at random from the hat, and calling this value X. For an interval of data values [a, b), we define

For a histogram of this data set, the area of the rectangle above the interval [a, b) equals P(a ≤ X < b).

Probability density functions

Some data sets are naturally discrete, such as the distribution of litter size among female cats of a particular age. Other data sets involving physical measurements—such as height, weight, or time—can take on a continuum of values. In the latter case, when sufficiently many measurements are taken, the histogram may be well approximated by a continuous function, as seen in the next example.

The continuous function approximating the histogram in Figure 7.4 is an example of a probability density function. The graphs of these functions are the continuous analogues of histograms. Since probability density functions are used to describe distributions from data sets, they need to satisfy certain natural properties.

Since PDFs are used to approximate histograms of data sets, we can use them to estimate the fraction of data lying in any interval.

In our approximation of the data in the interval [a, b], we included the right endpoint b, contrary to our convention with histograms. For data sets whose distributions are well approximated by a PDF, the fraction of data taking on exactly the value b is 0 or very small. Hence, the effect of including or excluding an endpoint b is negligible.

An alternative interpretation of the area under a PDF is in terms of continuous random variables. Imagine we have a hat that contains an infinite number of (infinitely thin) slips of paper, each with different numbers such that the proportion of slips with numbers in the interval [a, b] is given by

where f is a PDF. Now shake this hat and, with your eyes closed, grab a slip of paper. Let X denote the value on this slip. Since X can assume any real value, it is called a continuous random variable with PDF f(x), where the probability that X takes on a value in the interval [a, b] equals f(x)dx.

The simplest continuous random variable is a uniform random variable that places equal weight on all values in a given interval. For example, one might expect the birth time of babies to be approximately uniformly distributed over the year, as illustrated in Figure 7.5. However, some interesting patterns emerge, including high peaks in September and late December (could the latter be a possible synergism of tax considerations and induced/Cesarean births?) and a low trough in late April/early May. The peaks and troughs are surely climate and culture dependent.

Figure 7.5 Fraction of births each week in 1978 in the United States

The PDF in Example 8 is an example of the following general class of density functions:

Cumulative distribution functions

Another way of describing a discrete or continuous random variable is with a cumulative distribution function.

If X is a continuous random variable with PDF f(x), then F(x) corresponds to the area under f over the interval (−∞, x]. Formally, we write this as follows:

If there exists an a such that f(t) = 0 for t ≤ a, then this expression simplifies to

The case for which there is no such a results in an improper integral (i.e., an integral over an infinite range) which is discussed in Section 7.2.

There are several nice properties of CDFs (opposed to PDFs). For example, if X is a random variable with a CDF F, then

Thus, when you are given a CDF, computing probabilities is much easier as no integration needs to be performed. If you are only given the PDF, you are stuck with doing the integration one way or the other.

A CDF F(x) for a random variable X is characterized by the following properties.

The first property requires that the fraction data with values less than x are nondecreasing with x. Intuitively, as x increases, one is only including more data in the interval (−∞, x]; hence, the fraction cannot get smaller. The second property ensures that all the data lie somewhere on the real line. The third property ensures that there can be no jump discontinuities in F(x) from the right. For discrete random variables, there always are jump discontinuities from the left. For example, if X = 0 with probability and X = 1 with probability , then the CDF for X is

Clearly, F(x) has jump discontinuities at x = 0, 1. However, as expected, F(x) = and = 1. It is worth noting, as we saw in Example 9, that the CDFs for continuous random variables are continuous and, therefore, easily satisfy the third property.

CDFs can arise quite naturally from differential equation models, as the following example illustrates.

Example 10 is a particular instance of the exponential distribution that arises in many applications. The general exponential distribution and additional applications are discussed in this chapter's problem sets, but to emphasize the importance of exponential distribution, we define it here.

Since CDFs are nondecreasing continuous functions, it is often easier to fit functions to empirically derived CDFs than to empirically derived PDFs, which might be less smooth. When fitting CDFs, however, we need to be careful—as the following example illustrates.

Example 11 suggests that for some populations, we could have a problem constructing F(x) if we do not have an estimate of the maximum life span of individuals. At the beginning of this millennium, for example, the Guinness Book of Records reported that the oldest fully authenticated age to which any human has ever lived is attributed to a French woman, Jeanne-Louise Calment, who was born on February 21, 1875 and died at age 122 years and 164 days. Individuals who appear to be older than this are alive today, but authentication of their birth date is required for them to be listed in the Guinness book of records.

Percentiles

Using the CDF, we can define quantities called percentiles, which play a special role in statistics and probability.

Level 1 DRILL PROBLEMS

In Problems 1 to 4 construct a histogram for the given data sets.

5. If X denotes a score in Problem 1, find

1. P(50 ≤ X ≤ 59)
2. P(50 ≤ X < 69)
3. P(70 ≤ X ≤ 89)
4. P(90 ≤ X < 100)

6. If X denotes a score in Problem 2, find

1. P(1 ≤ X ≤ 10)
2. P(1 ≤ X < 21)
3. P(31 ≤ X < 41)
4. P(51 ≤ X ≤ 60)

7. If X denotes a score in Problem 3, find

1. P(X < 71)
2. P(1 ≤ X < 141)

8. If X denotes a score in Problem 4, find

1. P(X < 500)
2. P(X ≥ 500)

In Problems 9 to 12, find a constant a so that the given function is a PDF and find the values of x that correspond to the median, the first quartile, and the third quartile.

9. f(x) = 2ax, 0 ≤ x ≤ 2

10. f(x) = 5ax, 1 ≤ x ≤ 5

11. f(x) = ax², 0 ≤ x ≤ 1

12. f(x) = 3ax², 1 ≤ x ≤ 4

In Problems 13 to 16, use the CDC charts in Figure 7.8 to estimate the length for age and weight for age percentiles for the following boys of age a months, w kg, and l cm.

13. a = 21, w = 12.5, l = 85 cm

14. a = 20, w = 12.2, l = 87 cm

15. a = 15, w = 13.4, l = 87 cm

16. a = 18, w = 10.9, l = 83 cm

20. If = −0.15y, y(0) = y₀

1. Find y(t).
2. If we define F(t) = 0 for t ≤ 0 and F(t) = 1 − y(t)/y₀, verify that F(t) is a CDF.
3. If X is a random variable whose CDF is given by F(t), find P(0 < X ≤ 1).

21. Consider the function g(x) whose graph is shown below:

1. For what value of c is f(x) = cg(x) a PDF?
2. For a continuous random variable with PDF f(x), find P(2 ≤ X ≤ 3).

22. Consider the function g(x) whose graph is shown below:

1. For what value of c is f(x) = cg(x) a PDF?
2. For a continuous random variable with PDF f(x), find P(3 ≤ X ≤ 12).

23. For Problem 21, find an expression for the CDF and plot it.

24. For Problem 22, find an expression for the CDF and plot it.

25. Consider

1. Verify that F(x) is a CDF.
2. Assume X is a continuous random variable with CDF F(x). Find P(0 ≤ X ≤ 1), P(2 ≤ X ≤ 10).

26. Consider

1. Verify that F(x) is a CDF.
2. Assume X is a continuous random variable with CDF F(x). Find P(0 ≤ X ≤ 1) and P(2 ≤ X ≤ 10).

27. The distribution of fathers' heights from Example 5 is approximated by the PDF

Use numerical integration to approximate:

1. Fraction of fathers whose heights are between 5 and 6 feet
2. Fraction of fathers whose heights are greater than 7 feet or less than 5 feet

28. The distribution of sons' heights from Example 5 is approximated by the PDF

Use numerical integration to approximate:

1. Fraction of sons whose heights are between 5 and 6 feet
2. Fraction of sons whose heights are greater than 7 feet or less than 5 feet

Level 2 APPLIED AND THEORY PROBLEMS

29. The following distribution table gives the distribution of cholesterol level for 6,000 children, 4 to 19 years old. Cholesterol level is measured in milligrams per 100 milliliters of blood. The class intervals include the left endpoint but exclude the right endpoint.

1. Sketch the histogram for the given intervals.
2. Find the probability that a randomly selected child in this group has a cholesterol level of ≥ 140.
3. Find the probability that a randomly selected child in this group has a cholesterol level between 100 and 220.

30. A study of grand juries compared the demographic characteristics of jurors with those of the general population to see if the jury panels were representative. Here are the results for age. Only persons 21 and older are considered; countywide age distribution is known from public health department data.

Sketch the histogram for countywide percentage and the number of jurors. What do you notice? For simplicity, assume that the last bin is [60, 70).

31. According to the US Census Bureau's International Data Base, these were the life expectancies in 2000 for the following countries:

1. Sketch a histogram with the following bins: less than 50 (plot as if on the interval [45, 50)), [50, 55), [55, 60), [60, 65), [65, 70), [70, 75), [75, 80), and greater than 80 (plot as if on the interval [80, 85)).
2. Say you are selecting one of the countries at random (i.e., each country is equally likely to be selected). Give the probability of getting a country with a life expectancy of

   

32. Let f(x) represent the PDF for the weight of a field mouse in Williamsburg, Virginia, where x is measured in grams. Express the following probabilities as integrals:

1. A randomly chosen field mouse weighs between 20 and 30 grams.
2. A randomly chosen field mouse weighs less than 40 grams.

33. Let f(x) represent the PDF for the weight of a pigeon in New York City where x is measured in ounces. Express the following probabilities as integrals:

1. A randomly chosen pigeon does not weigh between 13 and 14 ounces.
2. A randomly chosen pigeon is in the weight class 12–15 ounces, but does not weigh between 13 and 14 ounces.

34. If you are really bad at darts, then the PDF for the distance x (in inches) that your dart is from the center of a 12- inch dartboard may be given by

1. Verify that f(x) is a PDF.
2. Compute the probability that your dart is more than 9 inches from the center.
3. Compute the probability that you dart is less than 3 inches from the center.

Note: This PDF assumes that you are equally likely to hit any point on the dartboard (a fact that you are asked to verify in Problem 25 of Problem Set 7.3).

35. Suppose you are a champion dart player with a PDF for the distance x (in inches) that your dart is from the center of a 12-inch dartboard given by

1. Verify that f(x) is a PDF.
2. Compute the probability that your dart is more than 1 inch from the center.
3. Compute the probability that your dart is between 1/4 and 1/2 inch from the center.

36. According to Thomson and colleagues,* the elimination constant for lidocaine for patients with hepatic impairment is 0.12 per hour. Hence, for a patient who has received an initial dose of y₀ mg, the lidocaine level y(t) in the body can be modeled by the differential equation

1. Solve for y(t).
2. Write an expression, call it F(t), that represents the fraction of drug that has left the body by time t ≥ 0.
3. If F(t) = 0 for t ≤ 0, verify that F(t) is a CDF.
4. What is the probability that a randomly chosen molecule of drug leaves the body in the first two hours?
5. What is the probability that a randomly chosen molecule of drug leaves the body between the second hour and fourth hour?

37. Consider a drug that has an elimination rate constant of c. If y is the amount of drug in the body and there is no further input of drug into the body, we can model the drug dynamics by

where t denotes times in hours and y₀ is the initial amount of drug in the body.

1. Solve for y(t).
2. Write an expression, call it F(t), that represents the fraction of drug that has left the body by time t ≥ 0.
3. If F(t) = 0 for t ≤ 0, verify that F(t) is a CDF.
4. Find an expression that allows one to calculate for any value c > 0 and times 0 < r < s what proportion of the drug is removed on the interval [r, s].

38. In the early 1960s, Robert MacArthur of Princeton University and Edward O. Wilson of Harvard University developed a theory to explain why big islands generally have more species than smaller islands, and why the numbers of species on islands of similar sizes are inversely related to island distance from continental landmasses. They argued that the number of species on an island represents a dynamic balance between the rate at which new species arrive at that island and the rate at which species on the island go extinct. The simplest model of island biodiversity assumes that the rate of change of the number N of species is given by a constant rate I of immigration of new species from the mainland and that species on the island go extinct at a rate proportional to N.

If the proportionality constant is c, then we obtain

where t denotes time in years. To know what the species immigration rate I might be for a particular island, we need to know the number of species on the mainland that serve as a source for the colonization process. On the other hand, the extinction rate c on each island is a characteristic of the island alone, rather than of the surrounding mainlands and the distance of the island to these mainlands. To understand the likelihood a species already on the island has gone extinct by time t, we can ignore the immigration process (i.e., keep track only of the species currently on the island) and consider the model

1. Solve for N(t).
2. Write an expression, F(t), for the fraction of species that has gone extinct by year t.
3. Donald Levin, a botany professor at the University of Texas, Austin, was quoted in Science Daily: “Roughly 20 of the 297 known mussel and clam species and 40 of about 950 fishes have perished in North America in the last century.”* Use these data to approximate the extinction constants c for mussel and clam species and for fish species.
4. Using your estimates from part b, estimate the probability that a specific clam or mussel species goes extinct in the next decade.
5. Using your estimates from part b, estimate the probability that a specific fish species goes extinct in the next decade.

7.2 Improper Integrals

As we saw in the previous section in defining cumulative distribution functions, we often encounter integrals in which the limits of integration are not finite. These improper integrals come in three varieties:

where a is a real number. In this section, we discuss when these integrals are well defined.

One-sided improper integrals

What is the area under the graph of y = e^−x for x ≥ 0? At first, one might reason this way: Since the region under the curve goes on forever, the area is infinite. To evaluate this statement, define A(t) to be the area under e^−x from x = 0 to x = t, as illustrated in Figure 7.9. In other words,

Figure 7.9 Area under e^−x from x = 0 to x = t

Computing A(t) yields

A(t) is always less than 1 for any t > 0. Therefore, the area under e^−x for x ≥ 0 cannot be infinite. In fact, it is natural to define the area under e^−x for x ≥ 0 to be

Thus, even though the curve is of infinite length, the area under this curve is finite. Our first guess was wrong!

Inspired by this example, we propose the following definition.

Example 1 shows that while the curves and are similar (i.e., both decreasing to zero as x goes to ∞), the areas under these curves are infinitely different: encloses an infinite area for x ≥ 2, while encloses a finite area for x ≥ 2. Figure 7.10 shows that decreases to zero much slower than . This observation suggests the following question: How fast does the function have to approach zero to ensure convergence? The following example formulates a precise answer to this question for p-integrals.

Figure 7.10 Area under curves that go to zero at different rates

Example 2 illustrates that convergence depends subtly on the speed at which f(x) approaches zero as x approaches ∞. For instance, while seems to go to zero only slightly faster than , the integral is convergent (i.e., p = 1.0001 > 1) while the integral is divergent. Although this might appear shocking at first, notice that the former integral converges to a very large value: = 10,000. More generally, as p > 1 approaches 1 from above, the area under approaches ∞ because .

The p-integrals are related to the Pareto distribution, named after the Italian economist Vilfredo Pareto (1848–1923). Pareto originally used this power distribution to describe the allocation of wealth among individuals; it also has been used to describe social, scientific, geophysical, and many other types of observable phenomena. In the next example, we examine the Pareto distribution and its use to describe the frequency of individuals visiting websites.

Example 3 leads us to the following definition.

Example 3 also illustrates how we go from PDFs to CDFs by integrating over the interval (−∞, x). Conversely, suppose that you are given a CDF for a continuous random variable. How do you find the associated PDF? The following theorem says all you have to do is differentiate. Hence, integrate to go from PDF to CDF and differentiate to go from CDF to PDF.

Convergence tests

As we have seen, the integral of a function cannot be always expressed in terms of elementary functions (e.g., f(x) = e^−x2). One way to get around this issue is to numerically estimate f(x)dx. However, if the upper limit is +∞, then numerical estimates only make sense if the integral converges because, informally,

if the left-hand side is convergent and b is sufficiently large. Consequently, it is important to have methods that determine whether an improper integral is convergent or not. A powerful yet simple test for convergence is the comparison test. The basic idea of this test is to compare the integral in question (the one for which convergence is not understood) to an integral for which convergence is understood.

Improper integrals can lead to maddening paradoxes, as the following example illustrates.

Two-sided improper integrals

We conclude this section by defining

A first attempt at this definition might be

Unfortunately this definition is flawed, as the following example illustrates.

To avoid the faulty path taken in Example 7, we need to avoid simultaneously confounding two computations involving ∞ (one in the positive direction and one in the negative direction). We avoid this by separating out the two computations, as presented in the following property box:

In Problem Set 7.2, you will be asked to show that for convergent integrals

for any a. Hence, for convergent integrals, we cannot make the infinite from nothing as in Example 7.

Many PDFs have bi-infinite tails (i.e., f(x) > 0 for all x (−∞, ∞)). One such example is the Laplace distribution.

Given the importance of the Laplace distribution, we conclude by providing a general definition for this distribution.

The derivation of the Laplace CDF is left to Problem 32 in Problem Set 7.2.

Level 1 DRILL PROBLEMS

Determine whether the integrals in Problems 1 to 10 are convergent or divergent. If convergent, determine their value.

For Problems 11 to 14, use the comparison test to determine whether the integrals are convergent or divergent.

For Problems 15 to 18, find the CDF of the given PDF.

For Problems 19 to 22, find the PDF of the given CDF.

In many species of birds, the age at which individuals die is assumed to follow an exponential distribution. In Problems 23 to 28, estimate the age of death corresponding to the first, second (median), and third quartiles of the distribution for the named species and given rate-of-death parameter b (units are per year rates: data from D. B. Botkin and B. S. Miller, “Mortality Rates and Survival of Birds,” American Naturalist, 108(1974): 181–192).

23. Blue tit, b = 0.72

24. Starling, b = 0.52

25. Grey heron, b = 0.31

26. Alpine swift, b = 0.18

27. Yellow-eyed penguin, b = 0.10

28. Royal albatross, b = 0.06 (an average of two different estimates)

Level 2 APPLIED AND THEORY PROBLEMS

29. Estimate the numerical value of e^−x2 by writing it as the sum of e^−x2 dx and e^−x2 dx. Approximate the first integral using Simpson's rule with n = 8. Show that the second integral is smaller than 0.0000001. Hint: Compare to e^−4x dx.

30. Determine how large a needs to be to ensure that

Hint: Compare to .

31. If is convergent, show that

for all a.

32. Show that

is the CDF to PDF

33. Consider a marine species whose larvae disperse northward and southward according to the Laplace distribution f(x) = e^−2|x|. For all the individuals that travel in both directions:

1. Determine the fraction of individuals that travel north at least 2 km.
2. Determine the fraction of individuals that travel south at least 2 km.
3. Determine the fraction of individuals that travel at most 2 km north.

34. Consider a large pine tree situated in a long, narrow valley. Suppose the seeds of this pine tree disperse up and down the valley according to the Laplace distribution f(x) = e^−|x|/2.

1. What proportion of all seeds disperse up the valley at least 1 km.
2. What proportion of seeds disperse down valley at least 3 km.
3. Determine the fraction of seeds that disperse at most 2 km in either direction of the tree.

35. The following graph shows a plot on a log-log scale of the frequency of trips of different durations (in hours) made by albatrosses (data from A. M. Edwards et al., “Revisiting Levy Flight Search Patterns of Wandering Albatrosses, Bumblebees and Dear,” Nature 449 (2007): 1044–1048).

From this plot deduce the proportion of trips that lie between 5 and 20 hours.

36. Journal Problem College Mathematics Journal 24 (September 1993): 343. Peter Lindstrom reported that a student handled an ∞/∞form as follows:

What is wrong, if anything, with this student's solution?

37.

Evangelista Torricelli studied in Galileo's home near Florence. When Galileo died, Torricelli succeeded his teacher as mathematician and philosopher for the Grand Duke of Tuscany, their friend and patron.

Torricelli described his amazement at discovering an infinitely long solid with a surface that calculates to have an infinite area, but a finite volume: “It may seem incredible that although this solid has an infinite length, nevertheless none of the cylindrical surfaces we considered has an infinite length but all of them are finite.” (See http://curvebank.calstatela.edu/torricelli/torricelli.htm.)

In Example 6, we introduced Torricelli's trumpet and paraphrased Thomas Hobbes, that is, to solve Torricelli's conundrum one needs to be mad rather than a geometrician or a logician. Present an argument that resolves this paradox.

38.

Newton and Leibniz have been credited with the discovery of calculus, but much of its development was due to the mathematicians Pierre-Simon Laplace, Joseph-Louis Lagrange, and Karl Gauss.

These three great mathematicians of calculus were contrasted by W. W. Rouse Ball*:

“The great masters of modern analysis are Lagrange, Laplace, and Gauss, who were contemporaries. It is interesting to note the marked contrast in their styles. Lagrange is perfect both in form and matter, he is careful to explain his procedure, and though his arguments are general they are easy to follow. Laplace, on the other hand, explains nothing, is indifferent to style, and, if satisfied that his results are correct, is content to leave them either with no proof or with a faulty one. Gauss is exact and as elegant as Lagrange, but even more difficult to follow than Laplace, for he removes every trace of the analysis by which he reached his results, and strives to give a proof which while rigorous will be as concise and synthetical as possible.”

Pierre-Simon Laplace taught Napoleon Bonaparte, who appointed him for a short time as France's Minister of Interior. Today, Laplace is best known as a major contributor to probability, taking it from gambling to a true branch of mathematics. He was one of the earliest to evaluate the improper integral

which plays an important role in the theory of probability. Use the web to find the value of this improper integral and its applications in mathematics, particularly probability theory.

7.3 Mean and Variance

As we have seen in Section 7.1, histograms provide visual summaries of large data sets. Sometimes these histograms are nicely approximated by the graph of a continuous function, the probability density function (PDF). When this occurs, a scientist can describe concisely his or her data set to another scientist by describing the PDF. Many important PDFs can be expressed in terms of families of functions whose parameters provide some basic information about the shape of the PDF. These parameters are often related to the mean and variance of the PDF. The mean is a measurement of the centrality of a data set. The variance, on the other hand, describes the spread of the data set around the mean; the greater the variance, the greater the spread in the data.

Means

There are numerous ways to characterize the central tendency in a set of data, including several ways of defining the concept of a mean—the most important being the arithmetic mean. For a data set {x₁,..., x_n} of real numbers, the arithmetic mean, or average, is given by the sum of the data values divided by the number of data values; namely, . It is useful to express this well-known expression in a slightly different manner, as illustrated in the following example.

Example 1 motivates the following more general definition.

One inspiration for this definition is Blaise Pascal's statement that the excitement felt by a gambler is equal to the amount he might win times the probability of winning it. From a gambling perspective, if x₁,..., x_k are the amounts you can win and p₁,..., p_k are the likelihoods of winning these amounts, then the mean is what you expect to win. Each term in the sum corresponds to the “amount you might win times the probability of winning it.”

Now suppose X is a continuous random variable with PDF f(x). To find the mean of X, we may consider approximating X by a discrete random variable by dividing the real line into intervals of length Δx with endpoints:

The probability of X taking on a value in the interval [x, x + Δx) is approximately f(x)Δx. Using the definition of the mean for a discrete random variable as motivation, the mean of X should be approximately given by the sum of the values weighted by their probabilities, that is,

Taking the limit as Δx goes to zero yields the integral xf(x) dx, which suggests the following definition.

As discussed in Problem Set 7.1, exponential distribution can be used to model extinction times for species.

Examples 3 and 4 illustrate that the fraction of data to the left of the mean can be significantly greater than 50%. This raises two questions. First, what is the geometrical interpretation of the mean? To answer this question, imagine that we take a (infinitely) long board and cut out the area lying under the PDF. If we placed this wooden PDF as shown in Figure 7.15 on a fulcrum at the mean, then the PDF would balance perfectly. Second, for what type of PDFs is 50% of the area to the left (and to the right) of the mean? A partial answer to this question is provided in the following example, using the concept of a symmetrical function and an odd function. Recall from Chapter 1 that an odd function g(x) satisfies g(x) = −g(−x) for all x.

Figure 7.15 PDF with fulcrum at the mean

In summary, the preceding example proves that for symmetric PDFs with a convergent mean, the mean corresponds to the point of symmetry of the PDF. Furthermore, by symmetry, the area under the PDF below the point of symmetry is equal to the area under the PDF above the point symmetry. Each contains half of the total area, so that, by definition, the point of symmetry is the 50th percentile, or median. Thus, we have the following result:

Sometimes the mean of a random variable can be infinite or not well defined, as the following example illustrates.

Variance and standard deviation

The importance of going beyond the mean was captured by the English mathematician Sir Francis Galton (1822–1911). He famously queried why statisticians of his day typically limited their enquiries to computing averages; he commented (pp. 62–63 in Galton, F., 1889, Natural Inheritance. Macmillan & Co., London), that souls of these statisticians “seem as dull to the charm of variety as that of the native of one of our flat English counties, whose retrospect of Switzerland was that, if its mountains could be thrown into its lakes, two nuisances would be got rid of at once.” One method to calculate the spread of data around the mean is to compute variance.

For a data set taking on the distinct values x₁, x₂, x₃,..., x_k with relative frequencies p₁, p₂, p₃,..., p_k and mean μ, the variance and standard deviation for the data set are defined exactly as for the discrete random variable. Some technologies employ a correction factor for calculating the variance associated with a sample of size n by multiplying the variance, as defined above, by the factor n/(n − 1) to get what statisticians call an unbiased estimate of the variance. If the value you obtain in solving a problem does not agree with the value we provide as solution, it may be that your technology employed this correction factor.

In describing the spread of data around the mean, scientists typically use the standard deviation rather than the variance. Unlike the variance, the standard deviation and the mean have the same units as the data. We will show, at the end of this section, that three-quarters of a data set always lies within two standard deviations of its mean. Equivalently, the probability that a random variable takes on a value within two standard deviations of its mean is at least three-quarters. In the next section, we will show that for certain bell-shaped distributions, approximately “two-thirds” of a data set lies within one standard deviation of its mean.

The following example illustrates that standard deviations measure the spread of the data around its mean.

Consider a continuous random variable X with PDF f(x). Assume the mean μ = xf(x)dx is well defined. To define the variance of X, we can approximate X by a discrete random variable by dividing the real line into intervals of length Δx with endpoints:

Since the probability X takes on a value between x and x + Δx is approximately f(x)Δx, the variance should approximately equal

Equivalently,

Taking the limit as Δx goes to zero yields (x − μ)² f(x)dx.

The following example studies the effect of the standard deviation on the shape of the distribution.

The following example provides an easier way of computing variances.

Chebyshev's inequality

As we saw in the previous example, through its square root (i.e., standard deviation), the variance provides both a measurement and a sense of the spread around the mean. Larger standard deviations suggest greater spread around the mean. A basic inequality from probability theory provides a general method of estimating what fraction of the data is within a certain number of standard deviations of the mean. This inequality is Chebyshev's inequality, named after the mathematician Pafnuty Chebyshev (1821–1894), who first proved it.

Since this theorem holds for all PDFs, no matter how peaked or how flat, the bounds in some cases can be rather weak.

Example 15 illustrates that Chebyshev's inequality states that at least of the data (k = 2) are within two standard deviations of the mean, at least of the data (k = 3) are with three standard deviations of the mean, and so on.

Level 1 DRILL PROBLEMS

Compute the mean, variance, and standard deviation of the data set given in Problems 1 to 8.

1. 1, 1, 0, 1, 1

2. 2, 0, 2

3. 1, 1, 1, 1, 1

4. 1, 2, 3, 4, 5, 6 (a die)

5. 1, 5, 7

6. −1, −2, 1, 4

7. The set of numbers that contains 2 zeros, 6 ones, 17 twos, and 8 threes.

8. The set of numbers that contains 7 negative twos, 5 negative ones, 3 zeros, 8 ones, and 12 twos.

Compute the mean of the random variable with the indicated PDF in Problems 9 to 16.

9. f(x) = for 0 ≤ x ≤ 2 and f(x) = 0 elsewhere

10. f(x) = for x ≥ 1 and f(x) = 0 elsewhere

11. f(x) = for |x| ≥ 1 and f(x) = 0 elsewhere

12. f(x) = e^−x for x ≥ 0 and f(x) = 0 elsewhere

13. f(x) =

14. f(x) =

15. f(x) = xe^−x for x ≥ 0 and f(x) = 0 elsewhere

16. f(x) = for x ≥ 0 and f(x) = 0 elsewhere (Hint: See Example 5 of Section 5.6)

Compute the variance of the PDFs in Problems 17 to 20.

17. f(x) = for 0 ≤ x ≤ 2 and f(x) = 0 elsewhere

18. f(x) = for x ≥ 1 and f(x) = 0 elsewhere

19. f(x) = for |x| ≥ 1 and f(x) = 0 elsewhere

20. f(x) = xe^−x for x ≥ 0 and f(x) = 0 elsewhere

21. Consider the following data set:

1. Find the mean and standard deviation.
2. According to Chebyshev's inequality, what fraction of data (at the bare minimum) has to lie in the interval [μ −2σ, μ + 2σ]? What fraction of the data does lie in this interval?

22. Consider the following data set:

1. Find the mean and standard deviation.
2. According to Chebyshev's inequality, what fraction of data (at the bare minimum) has to lie in the interval [μ −2σ, μ + 2σ]? What fraction of the data does lie in this interval?

23. Suppose that a random variable X has a PDF of

1. Find P(0.2 ≤ X ≤ 0.6). Sketch this probability on a graph of the PDF.
2. Find and graph the CDF.
3. Find the mean value.

24. Suppose that a random variable X has a PDF of

1. Find P(1.2 ≤ X ≤ 2.4). Sketch this probability on a graph of the PDF.
2. Find and graph the CDF.
3. Find the mean value.

Level 2 APPLIED AND THEORY PROBLEMS

25. Let X be the distance from the center that a dart lands on a dartboard with a radius of 2 feet.

1. Show that the PDF for x is given by

   

2. Find the mean and variance of this PDF.

26. In spring 1994, the number of bird species in forty different oak woodland sites in California were collected. Each site was around 5 hectares in size—the equivalent of about 12 acres, or 0.019 square miles— and the sites were situated in a relatively homogeneous habitat. The numbers of bird species found in these sites are listed below:

Let X be the number of species in a randomly chosen site. Using technology, compute the mean and standard deviation of X.

27. The following numbers are length in seconds of scenes showing tobacco use in six animation movies from a film studio:

Let X be the length of a scene from a randomly chosen movie.

1. Compute the mean for X.
2. Compute the standard deviation of X.

28. The following numbers are belly bristles per fruit fly in a sample size of six:

1. Find the relative frequency of 30.
2. Compute the mean number of belly bristles in this sample.
3. Compute the standard deviation of this data set.
4. According to Chebyshev's inequality, what is the minimum fraction of the data taking on values between 26.5 and 34.5? What is the actual fraction of data taking on values between 26.5 and 34.5?

29. Consider the exponential random variable X with PDF given by

Find the variance and standard deviation of X. Compare these numbers to the mean of the exponential distribution. What do you notice?

30. According to Thomson et al. (see Problem 36 in Problem Set 7.1), the elimination constant for lidocaine for patients with hepatic impairment is 0.12 per hour.

1. Determine the mean time μ for a lidocaine molecule to be eliminated.
2. Determine the fraction of lidocaine eliminated by time t = μ.

31. Donald Levin was quoted in Science Daily (see Problem 38 in Problem Set 7.1) as stating: “Roughly 20 of the 297 known mussel and clam species and 40 of about 950 fishes have perished in North America in the last century.”

1. Use these data to approximate the mean time to extinction constants for mussel and clam species and for fish species.
2. Determine the fraction of mussel and clam species and fish species that will be lost in the next century.

32. In Example 3 of Section 7.2, the following Pareto PDF was used to describe how many AOL users visited certain websites on one day in 1997:

1. Find the mean of this PDF.
2. Compute the variance for this PDF. What does the variance suggest about the variability in the number of hits that a website can experience?

33. Let X denote the number of years a patient lives after receiving treatment for an acute disease such as cancer. Under appropriate conditions, X is exponentially distributed. Suppose that the probability that a patient will live at least five years after treatment is 0.85.

1. Find the mean value of X.
2. Find the probability a patient will live at least ten years.

34. Based on data from 1974 to 2000 in Humboldt and Del Norte counties in California, the mean time to the next earthquake of magnitude ≥ 4 is approximately 2.5 weeks. Assuming that the time to the next earthquake is exponentially distributed, find the probability there will be an earthquake of magnitude ≥ 4 in the next week.

35. According to a newspaper article titled “Babies by the Dozen for Christmas: 24-Hour Baby Boom,” a record forty-four babies were born in one 24-hour period at the Mater Mothers' Hospital, Brisbane, Australia, on December 18, 1997. The article listed the times of birth for all of the babies.* The histogram of the times between births is as follows:

1. If this histogram has proportions {0.35, 0.25, 0.16, 0.12, 0.08, 0.02, 0.0, 0.0, 0.0, 0.02} centered on the values 7.5, 22.5, and so on every 15 minutes up to 142.5, then what is the mean time between births?
2. If this histogram is approximately exponentially distributed with a mean of 33.26 minutes between births, then what fraction of the times between births were less than 30 minutes? According to the histogram, what fraction of times between births were less than 30 minutes?
3. According to the exponential distribution, what fraction of the times between births were more than 75 minutes? Compare this with the actual fraction of times between births that are more than 80 minutes, as depicted in the histogram.

36. Here is a fun but challenging exercise: Construct a data set so that only 75.1% of the data lie in the interval [μ −2σ, μ + 2σ].

37. Let X be a continuous random variable with PDF f(x). Show that g(x) = f(x + a) is a PDF for the continuous random variable Y = X − a.

7.4 Bell-Shaped Distributions

An important class of PDFs has graphs that are bell-shaped (see center panel in Figure 7.18). In this section, we investigate two PDFs with this property: the logistic PDF and the normal PDF. Both are symmetric about their means and nonzero on the interval (−∞, ∞). Although logistic and normal PDFs have similar shapes, each is used to represent biological data arising from different types of processes. In this section, we also study random variables that are normal on a log scale. This lognormal distribution is right skewed or right tailed (also referred to as positively skewed—see Figure 7.18) because it is zero on (−∞, 0].

Figure 7.18 Symmetric and skewed PDFs

The logistic distribution

The logistic growth equation studied in Chapter 6 describes how populations change over time. In the next example, we show that solutions to the logistic equation can lead to a CDF for the logistic distribution.

In the previous example, we selected y(0) = 0.5, resulting in a symmetric PDF around zero. Thus, the mean is zero, provided that the associated improper integral is convergent. More generally, we can derive a logistic PDF for any initial condition y(0), as well as for any arbitrary r > 0 in which case the PDF is symmetric around a value other than zero. In particular, in Problem Set 7.4 asks you to show the following:

Note that a > 0 if y(0) (0, 0.5) and a < 0 if y(0) (0.5, 1).

Example 2 illustrated how the mean and variance of the logistic distribution are affected by the parameters r and y(0). The following example determines the mean of the logistic distribution.

In addition to determining the mean, it is possible—but more challenging—to compute the variance of the logistic PDF.

The logistic distribution can also describe the spread of an organism across a landscape.

An important type of regression analysis is associated with the logistic PDF. In Chapter 1, we demonstrated fitting linear models y = ax + b to data and inferring values for y from values of x. Suppose we want to infer the probability p of a certain event occurring associated with some measurement t. For example, t could be the age of a healthy cow and p could be the probability that this cow will die during the next year. If we have data on the proportion p(t) of cows that have died at or before t, then we may consider fitting the logistic CDF to data. A method for fitting the logistic CDF to data is illustrated in the following two examples.

Example 5 implies that if we have a set of n data points, (t₁, p₁),...,(t_n, p_n), whose distribution we want to describe with the logistic CDF, it suffices to find the best-fitting line through the transformed data, . Finding the best-fitting parameters r and a in this manner is called logistic regression.

Normal distribution

The most ubiquitous probability distribution in the natural and behavioral sciences is the normal distribution. For example, the normal distribution describes the distributions of heights (see Example 5 of Section 7.1) and weights of many organisms, crop yields (see Example 7), human IQs (see Example 9), and much more. Its ubiquity stems from this fact: if each data point is under the influence of many, independent, additive effects, then one can prove that the distribution of the data is well approximated by the normal distribution. The normal distribution is also known as the Gaussian distribution, after the German mathematician Karl Friedrich Gauss (1777–1855), who is shown in Figure 7.22 with the normal distribution in the background.

Figure 7.22 Deutsche mark showing Karl Gauss and the normal distribution in the background

The effect of increasing μ on this distribution is to move the graph to the right. The standard deviation σ, on the other hand, controls the spread of the distribution about its center. For small σ, the distribution is more peaked or concentrated around the mean; for larger σ, the distribution is fatter with a lower, broader peak, as illustrated in Figure 7.23. As we discussed in Chapter 5, there is no elementary representation of the antiderivative of f(x). Hence, we use numerical estimates.

Figure 7.23 Normal distributions, with mean 0 and standard deviations of 1 (blue) and 2 (red)

Aside from using numerical integrators, we can use tables to estimate areas under normal densities. At first, you might think that we need an infinite number of tables to deal with all possible values of μ and σ. However, this is not the case. Using a simple substitution, we can reduce everything to a question about one normal distribution, the standard normal distribution.

The following example illustrates how all questions about normal distributions can be reformulated as a question about the standard normal distribution.

z Scores

Since questions about normally distributed random variables or data can be converted to questions about the standard normal distribution, we can use z scores to determine what fraction of normally distributed data lies between the data's mean and z standard deviations above the mean. Table 7.3 reports z scores where rows determine the first two digits of z and columns the third digit of z. For example, in the row labeled (at the left) 1.0 and in the column headed 0.00, the entry is 0.3413. Therefore, for data with a standard normal distribution, the fraction of data lying in the interval [0, 1] is 34.13%. Alternatively, suppose the data have a standard normal distribution and we want to know what fraction of the data lies in the interval (−∞, 1.68]. The z table tells us the fraction of data lying in the interval [0, 1.68] is 0.4535. Since 50% of the data lies in (−∞, 0], the fraction in the interval (−∞, 1.68] is 0.4535 + 0.5 = 0.9535, as shown in the top panel of Figure 7.25. By contrast for z < 0, the symmetry of the standard normal distribution around 0 implies that subtraction, rather than addition, is required. For example, if z = −1.2, as shown in the bottom panel of Figure 7.25, 50%−34.13% = 15.87% of the area under the standard normal PDF lies on the interval (−∞, −1.2].

Figure 7.25 Using z scores to calculate area under a standard normal PDF

Table 7.3 Standard normal distribution

Lognormal distribution

One potential problem with a normally distributed random variable X is that it can be negative; typically, however, biological data only assume positive values (e.g., height or weight). This issue may not be a problem if only the extreme tail of the distribution is associated with a negative value of X, as in the wheat yield example illustrated in Figure 7.24. Sometimes, however, the “normality” of a data set is not apparent until it has been appropriately transformed to take values on (−∞, ∞). For example, a common transformation for data sets of positive values is taking the natural logarithm of all the data values. Data that exhibit a normal distribution after such a transformation are said to be lognormally distributed.

The following example determines the PDF for the lognormal distribution and explores the effects of the parameters μ and σ on the shape of the distribution.

In Problem 27 in Problem Set 7.4, you will be asked to show that the mean and variance of the lognormal distribution satisfy the relationships given below.

Level 1 DRILL PROBLEMS

Assume that a data set is normally distributed with a mean of 0 and a standard deviation of 1. A value X is randomly selected. Find the probability requested in Problems 1 to 4.

1. P(0 ≤ X < 0.85)

2. P(X ≤ 0)

3. P(X ≥ 0.55)

4. P(−1.00 < X < 0.75)

5. In the PDF of the standard normal distribution, find the area under the PDF bounded by the lines z = 1.20 and z = 1.90 and compare this with the value of z = 1.90 − 1.20 = 0.70 in Table 7.3.

6. For X normally distributed with mean μ = −1 and standard deviation σ = 1, calculate P(X ≥ 0).

7. For X normally distributed with mean μ = 1 and standard deviation σ = 2, calculate P(X > 0).

8. For X normally distributed with mean μ = −2 and standard deviation σ = 2, calculate P(−3.00 < X < −1.00).

9. For X lognormally distributed with log mean μ = −2 and log standard deviation σ = 2, calculate P(e⁻³ < X < e⁻¹).

10. For X lognormally distributed with log mean μ = 0, calculate P(0 < X < 1).

11. For X lognormally distributed with log mean μ = 0 and log standard deviation σ = 1, calculate P(0 < X < 0.5).

12. For X lognormally distributed with log mean μ = 1 and log standard deviation σ = 2, calculate P(1 < X < 4).

13. Example 4 discussed the spatial spread of the large tunicate Pyura praeputialis on the Chilean coast. In this example, the fraction of habitat occupied by this tunicate species at time t equals

where t is measured in hundreds of days.

1. What fraction of habitat was occupied on day t = 0?
2. What fraction of habitat was occupied at 100 days?
3. At what point in time will 95% of the habitat be covered?

14. Suppose the fraction of habitat occupied by a tunicate species at time t equals

where t is measured in hundreds of days. Let T be the time a randomly chosen location becomes occupied. Find the mean, median, and variance of T.

15. Consider Example 1 with r = 0.1 (units per month) and y(0) = 0.5 (a relatively slow-spreading disease).

1. Solve the differential equation for y(t).
2. Verify that y(t) is a CDF.
3. Find the probability that a randomly chosen individual is infected with the disease in the next two months.

16. Consider Example 1 with r = 3 (units per month) and y(0) = 0.5 (a relatively fast-spreading disease).

1. Solve the differential equation for y(t).
2. Verify that y(t) is a CDF.
3. Find the probability that a randomly chosen individual is infected with the disease in fifteen days (i.e., 0.5 months).

17. Consider Example 1 with r = 1 (units per month) and y(0) = 0.1 (i.e., 10% of the population have the disease).

1. Solve the differential equation for y(t).
2. Verify that y(t) is a CDF.
3. Find the probability that a randomly chosen individual is infected with the disease in one month.

18. Consider Example 1 with r = 0.5 (units per month) and y(0) = 0.3.

1. Solve the differential equation for y(t).
2. Verify that y(t) is a CDF.
3. Find the probability that a randomly chosen individual is infected with the disease in the 1.5 months.

Use logistic regression to find the best-fitting functions p(t) to the data in the sets D = {(t₁, p₁),...,(t_n, p_n)} given in Problems 19 to 22.

19. D = {(1, 0.10), (2, 0.15), (3, 0.30), (4, 0.49), (5, 0.58), (6, 0.76), (7, 0.87), (8, 0.95), (9, 0.93), (10, 0.98)}

20. D = {(1, 0.03), (2, 0.02), (3, 0.08), (4, 0.09), (5, 0.21), (6, 0.30), (7, 0.52), (8, 0.61), (9, 0.84), (10, 0.88)}

21. D = {(1, 0.01), (3, 0.01), (5, 0.03), (7, 0.03), (9, 0.10), (11, 0.18), (13, 0.29), (15, 0.48), (17, 0.73), (19, 0.85), (21, 0.87)}

22. D = {(1, 0.16), (3, 0.17), (5, 0.27), (7, 0.34), (9, 0.44), (11, 0.58), (13, 0.63), (15, 0.77), (17, 0.78), (19, 0.85), (21, 0.92)}

Level 2 APPLIED AND THEORY PROBLEMS

23. In a large study, human birth weights were found to be approximately normally distributed with mean of 120 ounces and standard deviation of 18 ounces (1 pound = 16 ounces; 1 ounce = 28.35 grams).

1. Find the probability that a randomly chosen baby has a birth weight of 8 pounds or less.
2. Find the probability that a randomly chosen baby weighs between 6 and 8 pounds at birth.
3. Find the probability that a randomly chosen baby weighs more than 9 pounds at birth.

24. A patient is said to be hyperkalemic (high levels of potassium in the blood) if the measured level of potassium is 5.0 milliequivalents per liter (meq/L) or more. In a population of students at Ozark University, the distribution of potassium levels is normally distributed with mean 4.5 meq/L and standard deviation 0.4 meq/L. Estimate the proportion of students who are hyperkalemic.

25. The gestation period of a pregnant woman is normally distributed with mean of 279 days and standard deviation of 16 days.

1. Find the probability that the gestation period is between 263 and 295 days.
2. Find the probability that the gestation period is greater than 303 days.

26. Answer the following questions for the data in Example 9.

1. What is the IQ value that corresponds to the 95th percentile for each of the two six-year-old low-birth-weight groups in Table 7.4?
2. In the normal-birth-weight urban and suburban communities, what is the change from age 6 to age 11 in the estimated proportion of individuals who have an IQ of 140 and above?
3. In the two eleven-year-old low-birth-weight communities, the 50th percentile of the suburban community corresponds to which percentile in the urban community?

In Problems 27 to 30, we emphasize that we are dealing with the lognormal distribution and note that the value e^μ is not the mean but the median (see Problem 35) and that the dispersion parameter σ is not the square root of the variance v of the distribution.

27. The latent period of disease is the time from a person initially getting infected to the moment the person exhibits first symptoms. In a paper by Sartwell, as referenced in Figure 7.26, Sartwell reported that the latency period (measured in days) of salmonellosis was approximately lognormally distributed. Taking the natural logs of the latency periods that are measured in days, he estimated that μ = ln(2.4) and σ = ln(1.47). Using these estimates, find the following quantities:

1. The fraction of individuals who start exhibiting symptoms within the first three days
2. The fraction of individuals who start exhibiting symptoms after four days
3. The fraction of individuals who start exhibiting symptoms between the start of the second day and end of the third day

28. The latent period of disease is the time from a person initially getting infected to the moment the person exhibits first symptoms. Sartwell found that the latency period (measured in days) of poliomyelitis was approximately lognormally distributed. Taking the natural logs of the latency periods that are measured in days, he estimated that μ = ln(12.6) and σ = ln(1.5). Using these estimates, find the following quantities:

1. The fraction of individuals who start exhibiting symptoms within the first two weeks
2. The fraction of individuals who start exhibiting symptoms after ten days
3. The fraction of individuals who start exhibiting symptoms between the start of the twelfth day and the end of the fifteenth day.

29. The survival time after cancer diagnosis is the number of days a patient lives after being diagnosed with cancer. In an article titled “Variation in the Duration of Survival of Patients with the Chronic Leukemias,” Blood. 1960 Mar; 15: 332–349, M. Feinleib and B. McMahon reported that the survival time for female patients diagnosed with lymphatic leukemia (measured in months) was approximately lognormally distributed. Taking the natural logs of the survival times, they estimated that μ = ln(17.2) and σ = ln(3.21). Using these estimates, find the following quantities:

1. The fraction of individuals who survived less than one year
2. The fraction of individuals who survived at least two years
3. The fraction of individuals who survived between 1 and 1.5 years

30. The survival time after cancer diagnosis is the number of days a patient lives after being diagnosed with cancer. Feinleib and McMahon, in a study cited in the previous problem, found that the survival time for female patients diagnosed with myelocytic leukemia (measured in months) was approximately lognormally distributed. Taking the natural logs of the survival times, they estimated that μ = ln(15.9) and that σ = ln(2.80). Using these estimates, find the following quantities:

1. The fraction of individuals who survived less than one year
2. The fraction of individuals who survived at least two years
3. The fraction of individuals who survived between the start of thirteen months and end of eighteen months (i.e., between 1 and 1.5 years)

31. In looking over her data, the entomologist mentioned in Example 12 found that she had transposed the number of individuals dying in weeks 7 and 8. After fixing this mistake, redo all the calculations covered in Example 12. Note differences to the estimates of the mean and variance associated with the actual data and the log mean μ and log variance σ² for the associated log normal PDF.

32. Linguist G. Herdan (see “The Relation Between the Dictionary Distribution and the Occurrence Distribution of Word Length and Its Importance for the Study of Quantitative Linguistics,” Biometrika 45 (A58): 222–228) found that length of spoken words in n = 738 phone conversations was lognormally distributed, with mean m = 5.05 letters and variance v = 2.16 letters. Find the probability that a randomly chosen word had six or more letters. Hint: See Problem 38.

33. The Gompertz growth equation normalized so that the variable y has the interpretation of a proportion (i.e., y = 1 is an equilibrium and upper bound) is given by

This equation can be used to model a variety of population processes, including tumor growth (y is proportion of maximum size), population growth (y is proportion of environmental carrying capacity), and acquisition of new technologies, as illustrated in the following example.

The Gompertz equation has been used to model mobile phone uptake, where y(t) is the fraction of individuals who have a mobile phone by time t (say, in years) and r is a parameter that can be fitted to the actual data. Using this model, we can derive a probability density function that represents the time at which an individual acquired her first mobile phone. To illustrate this idea, assume that y(0) = 1/e (i.e., currently 36.79% of people have mobile phones) and r = 1.

1. Solve the differential equation for r = 1 and y(0) = 1/e.
2. Verify that F(t) = 1 − y(t), where y(t) is the solution found in part a, is a CDF.
3. Find the PDF for your CDF.
4. Compute the probability that a randomly chosen individual acquires a mobile phone two years from now.

34. Consider Example 1 with r > 0 and y(0) = y₀ (0, 1).

1. Solve the differential equation for y(t).
2. Verify that F(t) = 1 − y(t) is a CDF.

35. Show that the PDF of a normal curve has its maximum (i.e., median) at x = μ and points of inflection at x = μ + σ and x = μ − σ.

36. Consider Example 1 with r > 0 and y(0) = y₀ (0, 1).

1. Verify that y(t) can be written as y(t) = where a = ln(1/y₀ − 1).
2. Find the PDF for this CDF.

37. For the lognormal distribution defined by

show that the mean m and variance v are given by

and

38. If ln X is a normally distributed random variable with mean μ and variance σ², and X is a lognormally distributed random variable with mean m and variance v, then show that

and

7.5 Life Tables

In Section 6.1 we introduced the simplest differential equation model of population growth:

This model implicitly assumes that all individuals, whether young or old, have the same mortality and fecundity rates. Although this assumption is a useful first approximation, mortality and fecundity are often age dependent. For instance, many animals become sexually mature only after they have reached a particular age. Additionally, the risk of mortality is often higher at younger and older ages. In this section, we consider models that account for age-specific mortality and reproduction.

Survivorship functions

Biology professor Gregory Erickson and colleagues studied fossils of four North American tyrannosaurs—Albertosaurus, Tyrannosaurus, Gorgosaurus, and Daspletosaurus. Using the femur bones of these fossils, the scientists estimated that the life spans of the dinosaurs ranged from birth to 28 years. Based on these estimates, the scientists created a life table for each of the dinosaurs. These life tables keep track of what fraction l(t) of individuals survived to age t. For example, the life table for Albertosaurus sarcophagus (see Figure 7.28) is reported in Table 7.6.

Figure 7.28 Albertosaurus sarcophagus

Table 7.6 Life table for A. sarcophagus

Survivorship functions have a natural relationship to CDFs of an appropriate random variable, as the following example shows.

Using Table 7.6, we can determine how the mortality rates of A. sarcophagus vary with age. In particular, imagine (as did a famous movie!) that on a remote island scientists were able to create 100 A. sarcophagus babies. The life table implies that of these 100, 60 survive to age 2 and 56 survive to age 4. Hence, 4 of 60 individuals die from age 2 to 4. Thus, the mortality rate over this two-year period is 4/60 = 6.7% and the annual mortality rate is approximately 3.3% per year. Equivalently, we estimate the mortality rate as

In the following example, we compute and interpret the mortality rates for the remaining age classes.

In Example 3, we computed mortality rates using the relationship

where Δt is the step size between measurements. Multiplying both sides of this equation by −l(t) yields

Taking the limit as Δt approaches 0 provides the following result:

Life expectancy

Given a survival function l(t) for a population, we can ask: What is the life expectancy of an individual? To answer this question, let X be the age at which a randomly chosen individual dies. The mean of X is the mean life span of an individual in the population. To compute this mean, recall that the CDF for X is given by F(t) = 1 − l(t) for t ≥ 0 and 0 otherwise. Hence, the PDF for X (assuming l is differentiable!) is −l′(t) for t ≥ 0 and 0 otherwise. The mean of X is given by

provided the improper integral is convergent. Let's assume it is convergent. We can simplify the integral by integrating by parts. Define u = t and dv = −l′(t)dt, so that du = dt and v = −l(t). This yields

Evaluating this integral from 0 to b and taking the limit as b → ∞ yields

If we assume bl(b) = 0, then

Hence, we proved the result shown in Theorem 7.4.

Reproductive success

So far we have only considered the likelihood of an individual surviving until a certain age. To better understand the dynamics of a population, we also need to know how the reproductive success of individuals depends on their ages. In other words, how many progeny do individuals of a particular age produce on average? In developing the models, we let b(t) denote the average number of progeny produced by an individual of age t. The likelihood l(t) of an individual surviving to age t in conjunction with b(t) provides a considerable amount of information about the demography of a population, as the following example illustrates.

Figure 7.31 The vole Microtus agrestis

Example 9 illustrates how to use a life table to determine the average number of daughters produced by a female during her lifetime. To generalize the computations in Example 9 to an arbitrary survival function l(t) and an arbitrary birth function b(t) ≥ 0, assume that initially there are N females (e.g., N = 100 in Example 9) and that Δt is the width of the time intervals for life table (e.g., Δt = 8 in Example 9). The number of females that survive to age t₁ = Δt is Nl(t₁). Each of these females produces b(t₁)Δt daughters. Hence, by time t₁, there are Nl(t₁)b(t₁)Δt daughters. The number of females that survive to age t₂ = 2Δt is Nl(t₂). Each of these females produces approximately b(t₂)Δt daughters in the time interval [t₁, t₂]. Hence, by time t₂, there are approximately

daughters produced. Continuing in this manner, there are approximately

daughters produced. Taking the limit as Δt → 0 yields the expected number of daughters D to be

If we now define the reproductive number R₀ to be the number of daughters that each individual female is expected to produce in her lifetime—that is R₀ = D/N—then we obtain the following relationship:

Ignoring the role of males, if R₀ > 1, then each female more than replaces herself in each generation and the population grows. On the other hand, if R₀ < 1, then each female fails to fully replace herself in each generation and the population declines.

In addition to applications in demography, life tables can be used to understand the spread of disease in a population. As a striking parallel to the demographic process of survivorship and reproduction, consider the following: An individual who contracts a disease will be subject to a maturation process known as a latent period and then will become infective, which is akin to reaching sexual maturity. Then, in each period, the infected individual may or may not infect another individual, which is akin to reproduction. Along the way, of course, the infected individual may either recover from the disease or die, which is akin to mortality.

Level 1 DRILL PROBLEMS

Use Life Table 7.6 for Albertosaurus sarcophagus to compute the quantities in Problems 1 to 4.

1. The fraction of A. sarcophagus that died between 14 and 20 years

2. The fraction of A. sarcophagus that died between 20 and 28 years

3. The fraction of A. sarcophagus that lived at least 6 years

4. The fraction of A. sarcophagus that lived at least 8 years

Use Life Table 7.7 for Microtus agrestis to compute the quantities in Problems 5 to 8.

5. The fraction of female voles that lived fewer than 24 weeks

6. The fraction of female voles that lived fewer than 40 weeks

7. The fraction of female voles that lived between 24 and 48 weeks

8. The fraction of female voles that lived between 40 and 64 weeks

Use the survivorship curves for men and women in Example 7 to compute the quantities in Problems 9 to 12.

9. The fraction of women who lived at least 75 years

10. The fraction of men who lived at least 75 years

11. The fraction of women who lived between 25 and 75 years

12. The fraction of men who lived between 25 and 75 years

13. Find the survivorship function l(t) when m(t) = a + bt with a > 0 and b > 0.

14. Find the survivorship function l(t) when m(t) = with a > 0 and b > 0. Discuss how a and b influence the shape of the survivorship function.

15. Use Life Table 7.7 for Microtus agrestis to approximate the mortality rates for all age classes of the female vole. Discuss any pattern in the mortality rates that you observe.

16. Use Life Table 7.7 for Microtus agrestis to compute the life expectancy of the female vole.

Compute the life expectancy for populations with the (hypothetical) survivorship functions in Problems 17 to 22. Assume t is measured in years. Don't be surprised if one of them turns out to be infinite.

Compute R₀ for populations with the (hypothetical) survivorship and reproduction functions in Problems 23 to 28. Assume t is measured in years.

Level 2 APPLIED AND THEORY PROBLEMS

29. Show that m(t) = ln[l(t)] provided that l(t) is differentiable.

30. If l(t)dt is convergent and b(t) ≤ B for all t, show that R₀ = l(t)b(t)dt is convergent.

31. According to the work of Erikson and colleagues reported in Table 7.6, the survivorship function l(t) for the dinosaur species Albertosaurus is well approximated by

for t ≥ 2. This function plotted against the data is shown here:

Find and plot the mortality function m(t) for t ≥ 2.

32. Using l(t) from Example 7, find the mortality function m(t) for females in the United States in 2005.

33. According to the work of Erikson and colleagues, the mortality rate for the dinosaur species Gorgosaurus is given by

Find and plot the survivorship function l(t).

34. According to the work of Erikson and colleagues, the mortality rate for the dinosaur species Daspletosaurus is given by

Find and plot the survivorship function l(t).

35. According to the work of Erikson and colleagues, the survivorship function for the dinosaur species Tyrannosaurus is

Find and plot the mortality rate m(t).

36. According to a National Vital Statistics Report (volume 54, number 14), the life table for people in the United States in 2003 was as follows:

Using right endpoints, estimate the life expectancy of a human.

37. During an outbreak of a SARS-like coronavirus, data were collected that resulted in the construction of the following table:

Use this table to answer the following questions:

1. If several infectious individuals are introduced into another population, is the epidemic expected to spread?
2. If the proportion of individuals vaccinated in a population reduces the expected number of individuals infected per infectious individual by this same proportion, then what proportion of the population should be vaccinated to ensure that the disease will not spread?

38. Communicable diseases often have at least two stages: a latent stage in which the individual is infected but not infectious, and an infectious stage in which the individual can infect others. For a deadly disease where the time to death is exponentially distributed with mean 1/q days, the fraction of individuals surviving t days with the disease is l(t) = e^−qt. Using differential equations to model the infection with two stages, latent and infectious, the infectiousness of an average infected individual (i.e., the number of people infected per day) is given by

where 1/a is the mean duration of the latent period, 1/c is the mean duration of the infectious period, and k is the rate an infectious individual infects others. For this model find R₀.

39. The parameters of the HIV epidemic vary considerably from country to country. The following table shows survival (including both death and drop-out rates) for treated and untreated segments of the sexually promiscuous population. The numbers reflect the fact that we expect all individuals to die within ten years if they are infected, unless they are treated. In the latter case, we assume that the individuals leave or drop out of the sexually promiscuous population after being part of it for twenty years. Also their infectivity is less for some of the infectivity period because the levels of virus in their body fluids are reduced by treatment. Infectivity comes back later as the efficacy of treatment is reduced over time.

1. Compare the R₀ for the treated and untreated segments of the population. What do you conclude?
2. What levels of condom use in the two subpopulations are needed to control the epidemic, assuming that condom use reduces the probability of transmission by 95%?

40. Botswana is a midsized country in central Africa. With a population of just over two million people, it is one of the most sparsely populated countries in the world. Using 2006 data from Human Life-Table database (see Example 5), one can approximate the survival functions of males and females as

These approximations work fairly well as shown below, with males in blue and females in red.

Use numerical integration to estimate the life expectancy for males and females in Botswana.

1. Consider the following data set corresponding to scores on a test. Let X denote a randomly chosen test score.

   Score	Frequency
   50–59	6
   60–69	14
   70–79	26
   80–89	10
   90–99	4

   1. Construct a histogram.
   2. Find P(0 ≤ X ≤ 89).
   3. Find P(X > 79).
2. Let f(x) = ax³, 0 ≤ x ≤ 4.
   1. Find a constant a so that it is a PDF.
   2. Find the mean of this PDF.
3. Consider the hyperbolic function

   

   for any k > 0.

   1. Show that F(x) is a CDF.
   2. Let X be a random variable with a CDF F(x). Find P(1 ≤ X ≤ 2).
4. Use the comparison test to prove that the integral

   

   is convergent for any r > 0.

5. Determine for which p > 0 values the integral is convergent.
6. Use the convergence test to determine whether the given integrals converge or diverge.

   

7. Show that f(x) = is a PDF on [1, 2] and find its CDF.
8. Compute the mean, variance, and standard deviation for a pair of dice; that is, this data set:

   2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 11, 11, 12

9. Compute the mean and variance of the random variable with PDF f(x) = for x ≥ 1 and f(x) = 0 elsewhere.
10. According to Thomson et al. (see Problem 36 in Problem Set 7.1), the elimination constant for lidocaine for patients with congestive heart failure is 0.31 per hour. Hence, for a patient who has received an initial dosage of y₀ mg, the lidocaine level y(t) in the body can be modeled by the differential equation

    

    1. Solve for y(t).
    2. Write an expression, call it F(t), that represents the fraction of drug that has left the body by time t ≥ 0.
    3. If F(t) = 0 for t ≤ 0, verify that F(t) is a CDF.
    4. What is the probability that a randomly chosen molecule of drug leaves the body in the first two hours?
    5. What is the probability that a randomly chosen molecule of drug leaves the body between the start of the second and start of the fourth hour?
11. The 1999 American Academy of Physician Assistants (AAPA) Physician Assistant Census Survey found that the mean income for a clinically practicing physician's assistant (PA) working full-time was $68,164, with a standard deviation $17,408. Using Chebyshev's inequality, determine a lower bound for the fraction of PAs with an income between $42,052 and $94,276.
12. The time for a mosquito to mature from larva to pupa is approximately exponentially distributed with a mean of fourteen days. Find the probability that a mosquito has matured from larva to pupa in ten days or less. Find the probability a mosquito has taken at least fourteen days to mature from larva to pupa.
13. According to Alexei A. Sharov, Department of Entomology at Virginia Tech, mortality depends on many factors such as temperature, population, and density. Sharov also said that when life tables are built, the effect of these factors is averaged and only age is considered as a factor that determines mortality. Sharov developed a life table for a sheep population in which females are counted once a year, immediately after breeding season:

    

    Use this table to compute the life expectancy of a female sheep.

14. Consider a random variable X with the Pareto distribution with parameter p = 5.
    1. According to Chebyshev's inequality, what is a lower bound for the probability of X being within two standard deviations of its mean?
    2. Find the probability that X is within two standard deviations of its mean.
15. Using 2006 data from Human Life-Table Database (see Example 5 of Section 7.5) one can approximate the survival functions of females in Botswana as

    

    1. Find the fraction of women who live at least forty years.
    2. Find the mortality rate at age 40.
16. For loggerhead turtles, females become reproductively active around age 21 years. The fraction of individuals that live to age 21 is 0.0023, and the mortality rate for individuals older than 21 is approximately 0.2 per year. Reproductively active females produce, on average, 160 eggs per year of which half are daughters. Estimate the reproductive number R₀ for this population. What is the fate of the population? How much does one need to reduce the mortality rate of individuals ≥ 21 to reverse this fate?
17. Given a continuously differentiable survivorship function l(t), let X correspond to the lifetime of a randomly chosen individual. What is the PDF and CDF for X?
18. Find parameters a and r in the logistic PDF

    

    such that the mean of this PDF is 1 and the variance is π².

19. You are told that a set of data is normally distributed with mean and variance equal to 21 and 64, respectively. Estimate the proportion of these data that have a value greater than or equal to 29.
20. In an experiment involved in rearing cohorts of the human louse, Pediculus humanus, researchers Francis Evans and Fredrick Smith at the University of Michigan obtained data on the proportion of individuals that survive over time. From their data, one can calculate the proportion of adults surviving each week of their experiment as follows*: {0.124, 0.301, 0.273, 0.210, 0.083, 0.009}. Since these six values sum to one, they imply that all individuals are dead by the end of the sixth week. In the following figure, the cumulative proportion of adults dying each week is plotted along with the CDF of best-fitting lognormal distribution.

    

    If the equation of the best-fitting lognormal PDF is

    

    then what is the mean survival time for these lice? How does this value compare to the value obtained by calculating the mean survival directly from the data, if the six values correspond to survival midway between each week? What accounts for the difference in these two approaches to estimating the mean?