Skip to main content

Advanced High School Statistics: Third Edition

David M Diez, Mine Çetinkaya-Rundel, Leah Dorazio, Christopher D Barr

Section 4.4 Chapter exercises

Exercises Exercises

1. 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.

2. 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.)

3. Overweight baggage.

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.
\(Z=1.56\text{,}\) \(P(Z \gt 1.56)= 0.0594\text{,}\) i.e. 6%.

4. 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?
 1 
Pew Research Center, Assessing the Representativeness of Public Opinion Surveys, May 15, 2012.

5. 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. \(SD_{\bar{x}}= \frac{3.2}{\sqrt{10}} = 1.012\text{;}\) \(z=\frac{46-45}{1.012} = 0.988\text{;}\) \(P(z \gt 0.988) = 0.162 = 16.2\%\text{.}\)

6. Chocolate chip cookies.

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.

7. Young Hispanics in the US.

The 2019 Current Population Survey (CPS) estimates that 36.0% of the people of Hispanic origin in the Unites States are under 21 years old.
 2 
United States Census Bureau. https://www.census.gov/data/tables/2019/demo/hispanic-origin/2019-cps.html The Hispanic Population in the United States: 2019. Web.
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.
Solution.
First we need to check that the necessary conditions are met. There are \(100 \times 0.360 = 36.0\) expected successes and \(100 \times (1 -0.360) = 64.0\) 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 \(N(\mu = 0.360, \sigma = 4.8)\text{.}\) \(P(X \ge 35) = P(Z \gt -0.208) = 0.5823\text{.}\) (2) The sampling distribution of \(\hat{p} ~ N(\mu = 0.360, \sigma = 0.048\text{.}\) \(P(\hat{p} \gt 0.35)=P(Z \gt -0.208) = 0.5823\text{.}\)

8. 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.
 3 
United States Census Bureau. 2013 Current Population Survey. Historical Poverty Tables - People. Web.
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.
  3. How do your answers from parts (a) and (b) compare?
You have attempted of activities on this page.
PrevTopNext