Chapter 5: Probability and Random Variables

In Exercises 5.16-5.26, express your probability answers as a decimal rounded to three places.

Cardiovascular Hospitalizations. From the Florida State Center for Health Statistics report Women and Cardiovascular Disease Hospitalization, we obtained the following table showing the number of female hospitalizations for cardiovascular disease, by age group, during one year.

One of these case records is selected at random. Find the probability that the woman was

(a) in her 50s.

(b) less than 50 years old.

(c) between 40 and 69 years old, inclusive.

(d) 70 years old or older.

Multiple-Choice Exams. A student takes a multiple-choice exam with 10 questions, each with four possible selections for the answer. A passing grade is 60% or better. Suppose that the student was unable to find time to study for the exam and just guesses at each question. Find the probability that the student

(a) gets at least one question correct.

(b) passes the exam.

(c) receives an "A" on the exam (90% or better).

(d) How many questions would you expect the student to get correct?

(e) Obtain the standard deviation of the number of questions that the student gets correct.

Love Stinks? J. Fetto, in the article "Love Stinks" (American Demographics, Vol. 25. No. 1. pp. 10-11), reports that Americans split with their significant other for many reasons-including indiscretion, infidelity, and simply "growing apart." According to the article, 35% of American adults have experienced a breakup at least once during the last 10 years. Of nine randomly selected American adults, find the probability that the number, X, who have experienced a breakup at least once during the last 10 years is

(a) exactly five; at most five; at least five.

(b) at least one; at most one.

(c) between six and eight, inclusive.

(d) Determine the probability distribution of the random variable X.

(e) Strictly speaking, why is the probability distribution that you obtained in part (d) only approximately correct? What is the exact distribution called?

Carbon Tax. A poll commissioned by Friends of the Earth and conducted by the Mellman Group found that 72% of American voters are in favor of a carbon tax. Suppose that six voters in the United States are randomly sampled and asked whether they favor a carbon tax. Determine the probability that the number answering in the affirmative is

(a) exactly two. (b) exactly four. (c) at least two.

(d) Determine the probability distribution of the number of American voters in a sample of six who favor a carbon tax.

(e) Strictly speaking, why is the probability distribution that you obtained in part (d) only approximately correct? What is the exact distribution called?

Video Games. A pathological video game user (PVGU) is a video game user that averages 31 or more hours a week of gameplay.

According to the article "Pathological Video Game Use among Youths: A Two-Year Longitudinal Study" (Pediatrics, Vol. 127. No. 2, pp. 319-329) by D. Gentile et al., in 2011, about 9% of children in grades 3-8 were PVGUs. Suppose that, today, seven youths in grades 3-8 are randomly selected.

(a) Assuming that the percentage of PVGUS in grades 3-8 is the same today as it was in 2011, determine the probability distribution for the number, X, who are PVGUs.

(b) Determine and interpret the mean of X.

(c) If, in fact, exactly three of the seven youths selected are PVGUs, would you be inclined to conclude that the percentage of PVGUs in grades 3-8 has increased from the 2011 percentage? Explain your reasoning. Hint: First consider the probability $P(X\ge 3)$.

(d) If, in fact, exactly two of the seven youths selected are PVGUs, would you be inclined to conclude the percentage of PVGUs in grades 3-8 has increased from the 2011 percentage? Explain your reasoning.

Recidivism. In the Scientific American article "Reducing Crime: Rehabilitation is Making a Comeback," R. Doyle examined rehabilitation of felons. One aspect of the article discussed recidivism of juvenile prisoners between 14 and 17 years old, indicating that 82% of those released in 1994 were rearrested within 3 years. Suppose that, today. six newly released juvenile prisoners between 14 and 17 years old are selected at random.

(a) Assuming that the recidivism rate is the same today as it was in 1994, determine the probability distribution for the number, Y, who are rearrested within 3 years.

(b) Determine and interpret the mean of Y.

(c) If, in fact, exactly two of the six newly released juvenile prisoners are rearrested within 3 years, would you be inclined to conclude that the recidivism rate today has decreased from the 82% rate in 1994? Explain your reasoning. Hint: First consider the probability $P(Y<2)$.

(d) If, in fact, exactly four of the six newly released juvenile prisoners are rearrested within 3 years, would you be inclined to conclude that the recidivism rate today has decreased from the 82% rate in 1994? Explain your reasoning.

Roulette. A success, s, in Bernoulli trials is often đerived from a collection of outcomes. For example, an American roulette wheel consists of 38 numbers, of which 18 are red, 18 are black, and 2 are green. When the roulette wheel is spun, the ball is equally likely to land on any one of the 38 numbers. If you are interested in which number the ball lands on, each play at the roulette wheel has 38 possible outcomes. Suppose, however, that you are betting on red. Then you are interested only in whether the ball lands on a red number. From this point of view, each play at the wheel has only two possible outcomes-either the ball lands on a red number or it doesn't. Hence, successive bets on red constitute a sequence of Bernoulli trials with success probability $\frac{18}{38}$. In four plays at a roulette wheel, what is the probability that the ball lands on red

(a) exactly twice? (b) at least once?

Sampling and the Binomial Distribution. Refer to the discussion on the binomial approximation to the hypergeometric distribution that begins on page 240.

(a) If sampling is with replacement, explain why the trials are independent and the success probability remains the same from trial to trial-always the proportion of the population that has the specified attribute.

(b) If sampling is without replacement, explain why the trials are not independent and the success probability varies from trial to trial.

Sampling and the Binomial Distribution. Following is a gender frequency distribution for students in Professor Weiss's introductory statistics class.

Two students are selected at random. Find the probability that both students are male if the selection is done

(a) with replacement.

(b) without replacement.

(c) Compare the answers obtained in parts ( a ) and ( (b).

Suppose that Professor Weiss's class had 10 times the students, but in the same proportions, that is, 170 males and 230 females.

(d) Repeat parts (a)-(c), using this hypothetical distribution of students.

(e). In which case is there less difference between sampling without and with replacement? Explain why this is so.

The Hypergeometric Distribution. In this exercise, we discuss the hypergeometric distribution in more detail. When sampling is done without replacement from a finite population, the hypergeometric distribution is the exact probability distribution for the number of members sampled that have a specified attribute. The hypergeometric probability formula is

$P\left(X=x\right)=\frac{\left(\begin{array}{c}Np\\ x\end{array}\right)\left(\begin{array}{c}N\left(1-p\right)\\ n-x\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)}$,

where Xdenotes the number of members sampled that have the specified attribute, Nis the population size, nis the sample size, and pis the population proportion.

To illustrate, suppose that a customer purchases 4 fuses from a shipment of 250, of which 94 % are not defective. Let a success correspond to a fuse that is not defective.

(a) Determine N, n, and p.

(b) Apply the hypergeometric probability formula to determine the probability distribution of the number of nondefective fuses that the customer gets.

Key Fact 5.6 shows that a hypergeometric distribution can be approximated by a binomial distribution, provided the sample size does not exceed 5% of the population size. In particular, you can use the binomial probability formula

$P(X=x)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x}$

with $n=4\mathrm{and}p=0.94$, to approximate the probability distribution of the number of nondefective fuses that the customer gets.

(c) Obtain the binomial distribution with parameters $n=4\mathrm{and}p=0.94$.

(d) Compare the hypergeometric distribution that you obtained in part (b) with the binomial distribution that you obtained in part (c).