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142 CHAPTER 2 • Modeling One-Variable Quantitative Data

AL TERNA TE EX AMPLE EXAMPLE
How fast is that serve?
Stop the car!
Finding a value from an area in a normal Finding a value from an area in a normal distribution
distribution
PROBLEM: Many studies on automobile safety suggest that when drivers must make emergency stops, the

PROBLEM: Recall from a previous stopping distances follow an approximately normal distribution. Suppose that for one model of car traveling

alternate example, the distribution of at 60 mph under typical conditions on dry pavement, the mean stopping distance is µ =165 feet with a stan-
dard deviation of σ = 4 feet. One percent of all such emergency stops take more than what distance?

Roger Federer’s first serve speeds at one
Wimbledon Championship averaged SOLUTION:
115 mph. Suppose that the distribution 1. Draw a normal distribution. If 1% of
of first serve speeds is approximately emergency stops take more than a certain
(C) 2021 BFW Publishers -- for review purposes only.

normal with a standard deviation of 4 mph. Area = 0.99 distance x, then 99% of emergency stops
take less than or equal to that distance. So we
How fast was a serve if it was one of the just need to find the 99th percentile of the
fastest 10% of Federer’s serves? Area = 0.01 distribution of stopping distance.
SOLUTION: 153 157 161 165 169 173 177
Stopping distance (ft) X
z
(i) Using Table A : 0.99 areato theleft →=2.33 2. Perform calculations—show your work!
Area = 0.90 Using technology: Applet/invNorm(area:0.99,mean:0,SD:1)2.33 (i) Use Table A or technology to find the
=

Area = 0.10 value of z with the appropriate area under the
x −165
2.33 = standard normal curve, then “unstandardize”; or

4 (ii) Use technology to find the desired value
2.33(4) +165 =x without standardizing. Label the inputs you
103 107 111 115 119 123 127 174.32 =x used for the applet or calculator.
First serve speed (mph) (ii) Applet/invNorm(area:0.99,mean:165,SD:4) =174.31
One percent of all such emergency stops take more than about Be sure to answer the question that was asked.
Note that an area of 0.10 to the right of a 174.3 feet.
boundary value implies that there is an FOR PRACTICE TRY EXERCISE 5.
area of 0.90 to the left of that value.
(i) Using Table A: 0.10areatotheright S
z
0.90areatotheleft S = 1.28 Another approach to finding the 99th percentile in the example is to use the inter-
pretation of the z -score. A standardized score of =z 2.33 means we are looking for the

Using technology: Applet/invNorm(area: value that is 2.33 standard deviations above the mean:
=
0.90, mean:0,SD:1) 1.28 mean 2.33(SD) = 165 2.33(4) = 174.32
+
+
x − 115 So 1 percent of all such emergency stops take more than 174.32 feet.
1.28 =
4 Finding the Mean or Standard Deviation from Areas
x
1.28(4) =− 115 in a Normal Distribution
120.12 = x
You have seen how to find the value corresponding to a given percentile in a normal
(ii) Applet/invNorm(area: 0.90,mean:115, distribution with known mean and standard deviation. It is also possible to find the
SD:4) 120.13 mean or standard deviation of a normal distribution using the value of one or more
=
percentiles.
A serve must be 120.1 mph or greater
to be one of the fastest 10% of Federer’s
serves.
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142 CHAPTER 2 • Modeling One-Variable Quantitative Data

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