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144 CHAPTER 2 • Modeling One-Variable Quantitative Data

LESSON 2.4 2.6 QUIZ LESSON APP 2 . 6
You can find a prepared quiz for Lessons Avoiding high cholesterol
2.4–2.6 by clicking on the link in the
TE-book or by logging into the teachers’ High levels of cholesterol in the blood increase the
resources on our digital platform. risk of heart disease. For teenage boys, the distribu-
tion of blood cholesterol is approximately normal
with mean µ = 151.6 milligrams of cholesterol per

LESSON APP 2.6 Answers deciliter of blood (mg/dl) and standard deviation

22
σ = 25 (mg/dl) .
1. (i) Table A: z = –0.84 gives 1. Find the 20th percentile of the distribution of
an area to the left of about. blood cholesterol for teen boys.
−
=
=
−0.84 (x 151.6)/25; x 130.6. For teenage girls, the distribution of blood Stockbroker/MBI/Alamy
(C) 2021 BFW Publishers -- for review purposes only.
cholesterol is approximately normal with mean
(ii) Tech: invNorm(area: 0.20,mean:151.6, µ = 157.5 milligrams of cholesterol per deciliter

=
SD:25) 130.6. A teen boy who has a of blood (mg/dl) . About 8.9% of teen girls have 2. Find the standard deviation of the distribution of ibution of

ind the standar
d devia
tion of the distr

F

cholesterol level of 130.6 would be at the high cholesterol—that is, levels of 200 mg/dl or blood cholesterol for teen girls.
20th percentile of the distribution. greater.
2. The area to the left of 200 is
1–0.089 = 0.911. A z-score of 1.35 Lesson 2.6
gives the closest value (0.9115).
=
=
1.35(200 −157.5)/SD; SD 31.48 mg/dl. WHA T DID Y OU LEARN ?
The standard deviation of the LEARNING TARGET EXAMPLES EXERCISES
distribution of blood cholesterol for Find the value that corresponds to a given percentile in a normal p. 142 5–8
teen girls is 31.48 mg/dl. distribution.
Find the mean or standard deviation of a normal distribution p. 143 9–12
given the value of a percentile.
FULL SOLUTIONS TO LESSON 2.6
EXERCISES
You can find the full solutions for this Exercises
lesson by clicking on the link in the Building Concepts and Skills a mean of $4200 and a standard deviation of $250.
TE-book or by logging into the teachers’ 1. What is the first step when finding values from Find the 30th percentile of this distribution.

resources on our digital platform. areas in a normal distribution? 6. Mail the letter A local post office weighs outgoing
2. What formula do you use to solve for the standard mail and finds that the weights of first-class letters

are approximately normally distributed with a mean
Answers to Lesson 2.6 Exercises deviation when given a percentile and the mean of of 0.69 ounce and a standard deviation of 0.16
a normal distribution? ounce. Find the 60th percentile of this distribution.
1. draw a normal distribution 3. Find the 40th percentile of a standard normal 7. Fire! A fire department in a rural county reports

−
valuemean distribution. that its response time to fires is approximately nor-
2. z = 4. Find the value in a standard normal distribution mally distributed with a mean of 22 minutes and a

SD with area 0.34 to its right. standard deviation of 6.9 minutes. Assume that this
3. z = –0.25 Mastering Concepts and Skills claim is true. One percent of response times take at

least how many minutes?
4. z = 0.41 5. Expensive mountain bike! The average sale price 8. Helmet sizes The army reports that the distribution

pg 142 (online) for a certain brand of professional moun- of head circumference among soldiers is approxi-
5. (i) z = –0.52 gives an tain bike is approximately normally distributed with mately normal with mean 22.8 inches and standard
area to the left of about 0.3.
=
−0.52 (x − 4200)/250; x = $4070
(ii) invNorm(area: 0.30,mean: 4200,
=
SD:250)$4,070. About 30% of this
brand of professional mountain bikes (ii) invNorm(area: 0.99,mean: 22, (ii) invNorm(area: 0.95,mean: 22.8, SD:1.1) = 07/09/20 1:58 PM
03_StarnesSPA4e_24432_ch02_088_153.indd 144
have sales prices less than $4,068.90, SD:6.9) = 38.05minutes. One percent of 24.61inches. Soldiers with a head
so a bike with a sales price of $4,068.90 response times take at least 38.05 minutes. circumference less than 20.99 inches or greater
would be at the 30th percentile of the 8. Smallest 5%: than 24.61 inches get custom-made helmets.
distribution. (i) A z-score of –1.64 and –1.65 are equally 9. z = 0.84 gives an area to the left of about 0.8.
6. (i) z = 0.25 gives an area to the left of close to a left area of 0.05. Use z = –1.645. 0.84 = 290 276/SD; SD 16.67 yards
=
−
x 0.69/0.16;
about 0.6: 0.25 =− x = 0.73 −1.645 =− x = 20.99 The standard deviation of the distribution of
x 22.8/1.1;
(ii) invNorm(area: 0.60,mean: 0.69, (ii) invNorm(area: 0.05,mean: 22.8, SD:1.1) = Lexi’s driving distance is 16.67 yards.
SD:0.16) = 0.73 ounce. About 60% of 20.99inches. 10. z =1.88 gives an area to the left of
first-class letters weigh less than 0.73 Largest 5%: about 0.97: 1.882 1.25/SD; SD 0.4minute
=
=−
ounce, so a first-class letter with a weight (i) A z-score of 1.64 and 1.65 are equally
of 0.73 ounce would be at the 60th close to a left area of 0.95. Use z =1.645. The standard deviation of the distribution of
percentile of the distribution. 1.645 =− x = 24.61 the amount of time spent on the smartphone
x 22.8/1.1;
is 0.4 minute.
7. (i) z = 2.33 gives an area to the left of
about 0.99: 2.33 =− x = 38.08
x 22/6.9;
144 CHAPTER 2 • Modeling One-Variable Quantitative Data
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