## Section4.6Chapter exercises

### Exercises4.6.1Exercises

In the game of roulette, a wheel is spun and you place bets on where it will stop. One popular bet is that it will stop on a red slot; such a bet has an $18/38$ chance of winning. If it stops on red, you double the money you bet. If not, you lose the money you bet. Suppose you play 3 times, each time with a $1 bet. Let $Y$ represent the total amount won or lost. Write a probability model for $Y\text{.}$ Solution 0 wins (-$3): 0.1458. 1 win (-1$): 0.3936. 2 wins (+1$): 0.3543. 3 wins (+$3): 0.1063. ###### 2.Speeding on the I-5, Part I. The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. 1 1. What percent of passenger vehicles travel slower than 80 miles/hour? 2. What percent of passenger vehicles travel between 60 and 80 miles/hour? 3. How fast do the fastest 5% of passenger vehicles travel? 4. The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5. S. Johnson and D. Murray. “Empirical Analysis of Truck and Automobile Speeds on Rural Interstates: Impact of Posted Speed Limits”. In: Transportation Research Board 89th Annual Meeting. 2010. ###### 3.University admissions. Suppose a university announced that it admitted 2,500 students for the following year's freshman class. However, the university has dorm room spots for only 1,786 freshman students. If there is a 70% chance that an admitted student will decide to accept the offer and attend this university, what is the approximate probability that the university will not have enough dormitory room spots for the freshman class? Solution Want to find the probability that there will be 1,786 or more enrollees. Using the normal approximation, with $\mu= np = 2,500 \times 0.7 = 1750$ and $\sigma = \sqrt{np(1-p} = \sqrt{2,500 \times 0.7 \times 0.3} \approx 23\text{,}$ $Z = 1.61\text{,}$ and $P(Z > 1.61) = 0.0537\text{.}$ With a 0.5 correction: 0.0559. ###### 4.Speeding on the I-5, Part II. Exercise 4.6.1.2 states that the distribution of speeds of cars traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. The speed limit on this stretch of the I-5 is 70 miles/hour. 1. A highway patrol officer is hidden on the side of the freeway. What is the probability that 5 cars pass and none are speeding? Assume that the speeds of the cars are independent of each other. 2. On average, how many cars would the highway patrol officer expect to watch until the first car that is speeding? What is the standard deviation of the number of cars he would expect to watch? ###### 5.Auto insurance premiums. Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California is approximately normal with a mean of$1,650. The article also states that 25% of California residents pay more than $1,800. 1. What is the $Z$-score that corresponds to the top 25% (or the $75^{\text{th}}$ percentile) of the standard normal distribution? 2. What is the mean insurance cost? What is the cutoff for the $75^{\text{th}}$ percentile? 3. Identify the standard deviation of insurance premiums in California. Solution (a) $Z=0.67$ (b) $\mu=1650\text{,}$ $x=1800\text{.}$ $0.67=\frac{1800-1650}{\sigma} \rightarrow \sigma = 233.88\text{.}$ ###### 6.SAT scores. SAT scores (out of 1600) are distributed normally with a mean of 1100 and a standard deviation of 200. Suppose a school council awards a certificate of excellence to all students who score at least 1350 on the SAT, and suppose we pick one of the recognized students at random. What is the probability this student's score will be at least 1500? (The material covered in Section 3.2 would be useful for this question.) ###### 7.Married women. The American Community Survey estimates that 47.1% of women ages 15 years and over are married. 2 1. We randomly select three women between these ages. What is the probability that the third woman selected is the only one who is married? 2. What is the probability that all three randomly selected women are married? 3. On average, how many women would you expect to sample before selecting a married woman? What is the standard deviation? 4. If the proportion of married women was actually 30%, how many women would you expect to sample before selecting a married woman? What is the standard deviation? 5. Based on your answers to parts (c) and (d), how does decreasing the probability of an event affect the mean and standard deviation of the wait time until success? U.S. Census Bureau, 2010 American Community Survey, Marital Status. Solution (a) $(1-0.471)^2 \times 0.471 = 0.1318\text{.}$ (b) $0.471^3 = 0.1045\text{.}$ (c) $\mu = 1/0.471 = 2.12\text{,}$ $\sigma = \sqrt{2.38} = 1.54\text{.}$ (d) $\mu = 1/.30 = 3.33\text{,}$ $\sigma = 2.79\text{.}$ (e) When p is smaller, the event is rarer, meaning the expected number of trials before a success and the standard deviation of the waiting time are higher. ###### 8.Survey response rate. Pew Research reported that the typical response rate to their surveys is only 9%. If for a particular survey 15,000 households are contacted, what is the probability that at least 1,500 will agree to respond? 3 ###### 9.Overweight bags. Suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee. Solution $Z=1.56\text{,}$ $P(Z > 1.56)= 0.0594\text{,}$ i.e. 6%. ###### 10.Heights of 10 year olds, Part I. Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches. 1. What is the probability that a randomly chosen 10 year old is shorter than 48 inches? 2. What is the probability that a randomly chosen 10 year old is between 60 and 65 inches? 3. If the tallest 10% of the class is considered “very tall”, what is the height cutoff for “very tall”? ###### 11.Buying books on Ebay. The textbook you need to buy for your chemistry class is expensive at the college bookstore, so you consider buying it on Ebay instead. A look at past auctions suggest that the prices of that chemistry textbook have an approximately normal distribution with mean$89 and standard deviation $15. 1. What is the probability that a randomly selected auction for this book closes at more than$100?

2. Ebay allows you to set your maximum bid price so that if someone outbids you on an auction you can automatically outbid them, up to the maximum bid price you set. If you are only bidding on one auction, what are the advantages and disadvantages of setting a bid price too high or too low? What if you are bidding on multiple auctions?

3. If you watched 10 auctions, roughly what percentile might you use for a maximum bid cutoff to be somewhat sure that you will win one of these ten auctions? Is it possible to find a cutoff point that will ensure that you win an auction?

4. If you are willing to track up to ten auctions closely, about what price might you use as your maximum bid price if you want to be somewhat sure that you will buy one of these ten books?

Solution

(a) $Z = 0.73\text{,}$ $P(Z > 0.73) = 0.2327\text{.}$

(b) If you are bidding on only one auction and set a low maximum bid price, someone will probably outbid you. If you set a high maximum bid price, you may win the auction but pay more than is necessary. If bidding on more than one auction, and you set your maximum bid price very low, you probably won't win any of the auctions. However, if the maximum bid price is even modestly high, you are likely to win multiple auctions.

(c) An answer roughly equal to the 10th percentile would be reasonable. Regrettably, no percentile cutoff point guarantees beyond any possible event that you win at least one auction. However, you may pick a higher percentile if you want to be more sure of winning an auction.

(d) Answers will vary a little but should correspond to the answer in part (c). We use the $10^{th}$ percentile: $Z = -1.28 \rightarrow 69.80\text{.}$

###### 12.Heights of 10 year olds, Part II.

Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches.

1. The height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches. What percent of 10 year olds cannot go on this ride?

2. Suppose there are four 10 year olds. What is the chance that at least two of them will be able to ride Batman the Ride?

3. Suppose you work at the park to help them better understand their customers' demographics, and you are counting people as they enter the park. What is the chance that the first 10 year old you see who can ride Batman the Ride is the 3rd 10 year old who enters the park?

4. What is the chance that the fifth 10 year old you see who can ride Batman the Ride is the 12th 10 year old who enters the park?

###### 13.Heights of 10 year olds, Part III.

Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches.

1. What fraction of 10 year olds are taller than 76 inches?

2. If there are 2,000 10 year olds entering Six Flags Magic Mountain in a single day, then compute the expected number of 10 year olds who are at least 76 inches tall. (You may assume the heights of the 10-year olds are independent.)

3. Using the binomial distribution, compute the probability that 0 of the 2,000 10 year olds will be at least 76 inches tall.

4. The number of 10 year olds who enter Six Flags Magic Mountain and are at least 76 inches tall in a given day follows a Poisson distribution with mean equal to the value found in part (b). Use the Poisson distribution to identify the probability no 10 year old will enter the park who is 76 inches or taller.

Solution

(a) $Z = 3.5\text{,}$ upper tail is 0.0002. (More precise value: 0.000233, but we'll use 0.0002 for the calculations here.)

(b) $0.0002 \times 2000 = 0.4\text{.}$ We would expect about 0.4 10 year olds who are 76 inches or taller to show up.

(c) ${2000 \choose 0}(0.0002)^0(1-0.0002)^2000 = 0.67029\text{.}$

(d) $\frac{0.4^0 \times e^{-0.4}}{0!} = \frac{1\times e^{-0.4}}{1} = 0.67032\text{.}$

###### 14.Multiple choice quiz.

In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. What is the probability that

1. the first question she gets right is the $3^{\text{rd}}$ question?

2. she gets exactly 3 or exactly 4 questions right?

3. she gets the majority of the questions right?

###### 15.Overweight baggage, Part II.

Suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. What is the probability that the total weight of 10 bags is greater than 460 lbs?

Solution

This is the same as checking that the average bag weight of the 10 bags is greater than 46 lbs. $SD_{\bar{x}}= \frac{3.2}{\sqrt{10}} = 1.012\text{;}$ $z=\frac{46-45}{1.012} = 0.988\text{;}$ $P(z > 0.988) = 0.162 = 16.2%\text{.}$

Students are asked to count the number of chocolate chips in 22 cookies for a class activity. The packaging for these cookies claims that there are an average of 20 chocolate chips per cookie with a standard deviation of 4.37 chocolate chips.

1. Based on this information, about how much variability should they expect to see in the mean number of chocolate chips in random samples of 22 chocolate chip cookies?

2. What is the probability that a random sample of 22 cookies will have an average less than 14.77 chocolate chips if the company's claim on the packaging is true? Assume that the distribution of chocolate chips in these cookies is approximately normal.

3. Assume the students got 14.77 as the average in their sample of 22 cookies. Do you have confidence or not in the company's claim that the true average is 20? Explain your reasoning.

###### 17.Young Hispanics in the US.

The 2012 Current Population Survey (CPS) estimates that 38.9% of the people of Hispanic origin in the Unites States are under 21 years old. 4  Calculate the probability that at least 35 people among a random sample of 100 Hispanic people living in the United States are under 21 years old.

United States Census Bureau. 2012 Current Population Survey. The Hispanic Population in the United States: 2012. Web.
Solution

First we need to check that the necessary conditions are met. There are $100 \times 0.389 = 38.9$ expected successes and $100 \times (1 -0.389) = 61.1$ expected failures, therefore the success-failure condition is met. Calculate using either (1) the normal approximation to the binomial distribution or (2) the sampling distribution of $\hat{p}\text{.}$ (1) The binomial distribution can be approximated by $N(\mu = 0.389, \sigma = 4.88)\text{.}$ $P( \gt 35) = P(Z > -0.80 = 1-0.2119 = 0.7881\text{.}$ (2) The sampling distribution of $\hat{p} ~ N(\mu = 0.389, \sigma = 0.0488\text{.}$ $P(\hat{p}>0.35)=P(Z>-0.8) = 0.7881\text{.}$

###### 18.Poverty in the US.

The 2013 Current Population Survey (CPS) estimates that 22.5% of Mississippians live in poverty, which makes Mississippi the state with the highest poverty rate in the United States. 5  We are interested in finding out the probability that at least 250 people among a random sample of 1,000 Mississippians live in poverty.

1. Estimate this probability using the normal approximation to the binomial distribution.

2. Estimate this probability using the distribution of the sample proportion.