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Lesson 2.6
and it looks like those who don’t play is
Normal Distributions: about 11 times as many as gamers, when
in reality there are 4.6 times as many
Finding Values from Areas that play some as gamers and 6.5 times
as many who don’t play as there are Lesson 2.6
gamers.
L E AR N I N G TAR G E T S
(b)

• Find the value that corresponds to a given percentile in a normal
distribution. 1100

• Find the mean or standard deviation of a normal distribution given the value 1000
of a percentile. 900
800
(C) 2021 BFW Publishers -- for review purposes only.
When the distribution of a quantitative variable can be modeled with a normal curve, 700
600
we can use the methods of Lesson 2.5 to find the percentile for a given value. We do Frequency 500

this by finding the appropriate area in a normal distribution. What if we want to 400
find the value in a normal distribution corresponding to a given percentile? That will 300
require us to reverse the process to go from an area to a value. 200
100
0
Finding Percentiles in a Normal Distribution Gamer Play some Don’t play
Video game player status
Let’s return to the distribution of ITBS vocabulary scores among all Gary, Indiana,
seventh-graders. Recall that this distribution is approximately normal with mean

.

µ = 6.84 and standard deviation σ = 1.55 What score would a student have to earn

to be at the 90th percentile of the distribution?
Figure 2.18(a) shows what we are trying to find: the ITBS score x with 90% of TEACHING TIP

the area to its left under the normal curve. Figure 2.18(b) shows the standard normal

distribution with the corresponding area shaded. The learning targets in this lesson
are easier if students have just a little
algebraic skill. Consider doing a short
review of solving linear equations and
Area = 0.90 Area = 0.90 systems of two linear equations with two
unknowns sometime during this lesson.
Most of the examples and exercises
can be completed without algebra, but
2.19 3.74 5.29 6.84 8.39 x 9.94 11.49 –3 –2 –1 0 1 z 2 3 a little algebra will go a long way to
ITBS vocabulary score z-score
making the job easier.
FIGURE 2.18 (a) Normal distribution showing the 90th percentile of ITBS vocabulary scores for Gary,
Indiana, seventh-graders. (b) The 90th percentile in the standard normal distribution.
We can use Table A to find the z -score with an area of 0.90 to its left in a standard z .07 .08 .09 LEARNING T AR GET KEY

normal distribution. Because Table A gives the area to the left of a specified z -score, 1.1 .8790 .8810 .8830
all we have to do is find the value closest to 0.90 in the middle of the table. From the The problems in the test bank are

.
table excerpt in the margin, you see that the desired value is =z 1.28 1.2 .8980 .8997 .9015 keyed to the learning targets using
1.3 .9147 .9162 .9177
these numbers:
• 2.6.1
139
• 2.6.2
03_StarnesSPA4e_24432_ch02_088_153.indd 139 07/09/20 1:57 PM
(b) (i) z = (2.1 2.04)/0.032; the 25. The z-score that is closest to having BELL RINGER
=
−
proportion of z-scores greater than z = 2 an area of 0.8 to the left is z = 0.84. We
is 0.0228. also know that the mean = 50 and SD = Mr. Tyson’s dog Zeus is a Vizsla (a breed of
(ii) Applet/normalcdf(lower:2.1, 6. Substituting into the z-score formula dog). Vizslas are famously energetic, so
upper:1000,mean: 2.04,SD: 0.03) = gives us 0.84(value 50)/6. The unknown Mr. Tyson takes Zeus for a walk following
=
−
0.0228. About 2.28% of bottles have a value, which is the 80th percentile of this the same route just about every day.
volume greater than 2.10 liters. distribution, is (0.84)(6)50 55.04. The duration of these walks follows an
+
=
approximately normal distribution with
(c) (i) z = (1.952.04)/0.03 =−3 and 26. The distribution of weights of women a mean of 20 minutes and a standard
−
=
−
z = (2.052.04)/0.030.33; the aged 20 to 29 is not approximately normal. deviation of 2 minutes. If the walk takes
proportion of z-scores between z = –3 In a normal distribution, Q 1 and Q 3 should longer than 25 minutes, Mr. Tyson will be
=
and z = 0.33 is 0.6293–0.00130.6280. be about the same distance from the late for breakfast. About what percent of
(ii) Applet/normalcdf(lower:1.95, median. However, the distance from Q 1 to these walks result in Mr. Tyson being late
=
=
upper: 2.05,mean: 2.04,SD: 0.03)0.6292. the median (133.2–118.3 14.9) is smaller for breakfast? Show your work.
About 62.92% of bottles are within than the distance from the median to Q 3
0.05 liter of the target volume. (157.3–133.2 = 24.1).
(d) Yes. Too many bottles have less than 27. ( a) By starting the vertical scale at 100
the advertised amount of 2 liters (9.12%). rather than 0, it looks like those who play
some is about 8 times as many as gamers,
LESSON 2.6 • Normal Distributions: Finding Values from Areas 139
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