An Introduction To Statistical Inference And Data Analysis

70 CHAPTER 3. DISCRETE RANDOM VARIABLES

Theorem 3.5 If X is a discrete random variable, then

VarX = E(X ¡ ¹)2

=E(X2 ¡ 2¹X + ¹2)

=EX2 + E(¡2¹X) + E¹2

=EX2 ¡ 2¹EX + ¹2

=EX2 ¡ 2¹2 + ¹2

=EX2 ¡ (EX)2:

A straightforward way to calculate the variance of a discrete random variable that assumes a fairly small number of values is to exploit Theorem 3.5 and organize one's calculations in the form of a table.

Example 8 Suppose that X is a random variable whose possible values are X(S) = f2; 3; 5; 10g. Suppose that the probability of each of these values is given by the formula f(x) = P (X = x) = x=20.

(a)Calculate the expected value of X.

(b)Calculate the variance of X.

(c)Calculate the standard deviation of X.

(a) ¹ = EX = :2 + :45 + 1:25 + 5 = 6:9.

(b) ¾2 = Var X = EX2 ¡ (EX)2 = (:4 + 1:35 + 6:25 + 50) ¡ 6:92 = 58 ¡ 47:61 = 10:39.

p:

(c)¾ = 10:39 = 3:2234.

Now suppose that X : S ! < is a discrete random variable and Á : < ! < is a function. Let Y = Á(X). Then Y is a discrete random variable and

We conclude this section by stating (and sometimes proving) some useful consequences of De¯nition 3.5 and Equation 3.4.

Theorem 3.6 Let X denote a discrete random variable and suppose that c 2 < is constant. Then

Var(X + c) = VarX:

Although possibly startling at ¯rst glance, this result is actually quite intuitive. The variance depends on the squared deviations of the values of X from the expected value of X. If we add a constant to each value of X, then we shift both the individual values of X and the expected value of X by the same amount, preserving the squared deviations. The variability of a population is not a®ected by shifting each of the values in the population by the same amount.

Theorem 3.7 Let X denote a discrete random variable and suppose that c 2 < is constant. Then

Var(cX) = E [cX ¡ E(cX)]2

=E [cX ¡ cEX]2

=E [hc(X ¡ EX)]2 i

=E c2(X ¡ EX)2

=c2E(X ¡ EX)2

=c2VarX:

To understand this result, recall that the variance is measured in the original units of measurement squared. If we take the square root of each expression in Theorem 3.7, then we see that one can interchange multiplying a random variable by a nonnegative constant with computing its standard deviation.

Theorem 3.8 If the discrete random variables X1 and X2 are independent, then

Var(X1 + X2) = VarX1 + VarX2:

Theorem 3.8 is analogous to Theorem 3.4. However, in order to ensure that the variance of a sum equals the sum of the variances, the random variables must be independent.

3.4Binomial Distributions

Suppose that a fair coin is tossed twice and the number of Heads is counted. Let Y denote the total number of Heads. Because the sample space has four equally likely outcomes, viz.

and f(y) = 0 if y 62Y (S) = f0; 1; 2g.

Referring to representation (c) of Example 2 in Section 2.5, the above experiment has the following characteristics:

²Let X1 denote the number of Heads observed on the ¯rst toss and let X2 denote the number of Heads observed on the second toss. Then the random variable of interest is Y = X1 + X2.

²The random variables X1 and X2 are independent.

²The random variables X1 and X2 have the same distribution, viz.

X1; X2 » Bernoulli(:5):

We proceed to generalize this example in two ways:

1.We allow any ¯nite number of trials.

2.We allow any success probability p 2 [0; 1].

De¯nition 3.7 Let X1; : : : ; Xn be mutually independent Bernoulli trials, each with success probability p. Then

Xn

Y = Xi

i=1

74 CHAPTER 3. DISCRETE RANDOM VARIABLES

for the speci¯c example in question and the general formula for the binomial

Except for very small numbers of trials, direct calculation of (3.5) is rather tedious. Fortunately, tables of the binomial cdf for selected values of n and p are widely available, as is computer software for evaluating (3.5). In the examples that follow, we will evaluate (3.5) using the S-Plus function pbinom.

As the following examples should make clear, the trick to evaluating binomial probabilities is to write them in expressions that only involve probabilities of the form P (Y · k).

Example 1 In 10 trials with success probability :5, what is the probability that no more than 4 successes will be observed?

Here, n = 10, p = :5, and we want to calculate

P (Y · 4) = F (4):

We do so in S-Plus as follows:

> pbinom(4,size=10,prob=.5) [1] 0.3769531

Example 2 In 12 trials with success probability :3, what is the probability that more than 6 successes will be observed?

Here, n = 12, p = :3, and we want to calculate

P (Y > 6) = 1 ¡ P (Y · 6) = 1 ¡ F (6):

We do so in S-Plus as follows:

> 1-pbinom(6,12,.3) [1] 0.03860084

Example 3 In 15 trials with success probability :6, what is the probability that at least 5 but no more than 10 successes will be observed?

Here, n = 15, p = :6, and we want to calculate

P (5 · Y · 10) = P (Y · 10) ¡ P (Y · 4) = F (10) ¡ F (4):

We do so in S-Plus as follows:

> pbinom(10,15,.6)-pbinom(4,15,.6) [1] 0.7733746

Example 4 In 20 trials with success probability :9, what is the probability that exactly 16 successes will be observed?

Here, n = 20, p = :9, and we want to calculate

P (Y = 16) = P (Y · 16) ¡ P (Y · 15) = F (16) ¡ F (15):

We do so in S-Plus as follows:

> pbinom(16,20,.9)-pbinom(15,20,.9) [1] 0.08977883

Example 5 In 81 trials with success probability :64, what is the probability that the proportion of observed successes will be between 60 and 70 percent?

Here, n = 81, p = :64, and we want to calculate

P (:6 < Y=81 < :7) = P (:6 ¢ 81 < Y < :7 ¢ 81)

=P (48:6 < Y < 56:7)

=P (49 · Y · 56)

=P (Y · 56) ¡ P (Y · 48)

=F (56) ¡ F (48):

We do so in S-Plus as follows:

> pbinom(56,81,.64)-pbinom(48,81,.64) [1] 0.6416193

Many practical situations can be modelled using a binomial distribution. Doing so typically requires one to perform the following steps.

4. The necessary calculation can be performed in S-Plus as follows:

> 1-pbinom(58,64,.8) [1] 0.006730152

3.5Exercises

1.Suppose that a weighted die is tossed. Let X denote the number of dots that appear on the upper face of the die, and suppose that P (X = x) = (7 ¡ x)=20 for x = 1; 2; 3; 4; 5 and P (X = 6) = 0. Determine each of the following:

(a)The probability mass function of X.

(b)The cumulative distribution function of X.

(c)The expected value of X.

(d)The variance of X.

(e)The standard deviation of X.

2.Suppose that a jury of 12 persons is to be selected from a pool of 25 persons who were called for jury duty. The pool comprises 12 retired persons, 6 employed persons, 5 unemployed persons, and 2 students. Assuming that each person is equally likely to be selected, answer the following:

(a)What is the probability that both students will be selected?

(b)What is the probability that the jury will contain exactly twice as many retired persons as employed persons?

3.Suppose that 20 percent of the adult population is hypertensive. Suppose that an automated blood-pressure machine diagnoses 84 percent of hypertensive adults as hypertensive and 23 percent of nonhypertensive adults as hypertensive. A person is selected at random from the adult population.

(a)Construct a tree diagram that describes this experiment.

(b)What is the probability that the automated blood-pressure machine will diagnose the selected person as hypertensive?

(c)Suppose that the automated blood-pressure machine does diagnose the selected person as hypertensive. What then is the probability that this person actually is hypertensive?

(d)The following passage appeared in a recent article (Bruce Bower, Roots of reason, Science News, 145:72{75, January 29, 1994) about how human beings think. Please comment on it in whatever way seems appropriate to you.

And in a study slated to appear in COGNITION, Cosmides and Tooby confront a cognitive bias known as the \base-rate fallacy." As an illustration, they cite a 1978 study in which 60 sta® and students at Harvard Medical School attempted to solve this problem: \If a test to detect a disease whose prevalence is 1/1,000 has a false positive rate of 5%, what is the chance that a person found to have a positive result actually has the disease, assuming you know nothing about the person's symptoms or signs?"

Nearly half the sample estimated this probability as 95 percent; only 11 gave the correct response of 2 percent. Most participants neglected the base rate of the disease (it strikes 1 in 1,000 people) and formed a judgment solely from the characteristics of the test.

4.Koko (a cat) is trying to catch a mouse who lives under Susan's house. The mouse has two exits, one outside and one inside, and randomly selects the outside exit 60% of the time. Each midnight, the mouse emerges for a constitutional. If Koko waits outside and the mouse chooses the outside exit, then Koko has a 20% chance of catching the mouse. If Koko waits inside, then there is a 30% chance that he will fall asleep. However, if he stays awake and the mouse chooses the inside exit, then Koko has a 40% chance of catching the mouse.

(a)Is Koko more likely to catch the mouse if he waits inside or outside? Why?

(b)If Koko decides to wait outside each midnight, then what is the probability that he will catch the mouse within a week (no more than 7 nights)?

5.Three urns are ¯lled with colored balls: