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LESSON 2.5 • Normal Distributions: Finding Areas from Values 135

TEACHING TIP
In these figures and in the previous
examples in this lesson, every normal
distribution has been drawn with a Lesson 2.5
horizontal axis marked with the mean
2.19 3.74 5.29 6 6.84 8.39 9 9.94 11.49 –3 –2 –1 –0.54 0 1 1.39 2 3
(a) ITBS vocabulary score (b) z-score and the points 1, 2, and 3 standard
FIGURE 2.17 (a) Normal distribution approximating the proportion of seventh-graders in Gary, deviations from the mean. Insist that
Indiana, with ITBS vocabulary scores between 6 and 9. (b) The corresponding area in the standard normal your students do the same every time
distribution.
they draw a normal distribution. Also
insist that they mark the appropriate
Using Table A: The table makes this process a bit trickier because it only shows areas
(C) 2021 BFW Publishers -- for review purposes only.
to the left of a given z-score. The visual shows one way to think about the calculation. boundary value(s) and shade the area
of interest.
5 –
AL TERNA TE EX AMPLE
–3 –2 –1 0 1 2 3 –3 –2 –1 0 1 2 3 –3 –2 –1 0 1 2 3
–0.54 1.39 –0.54 More sweet sugary goodness?
z-score 1.39 z-score z-score
Area between z =−0.54 and z = 1.39 Finding area between two values in a
−
= (Areatothe left of z = 1.39)(Area to theleftof z =−0.54) normal distribution
−
= 0.9177 0.2946
= 0.6231 PROBLEM: Machines that fill bags
=
−
Using technology: Applet/normalcdf(lower:0.54, upper: 1.39,mean: 0, SD:1)0.6231 with powdered sugar, also called
confectioner’s sugar, are supposed to
Option (ii): Applet/normalcdf(lower:6, upper: 9, mean:6.84, SD:1.55) = 0.6243. dispense exactly 32 ounces of powdered
sugar into each bag. The amount of
About 62% of Gary, Indiana, seventh-graders earned grade-equivalent scores
between 6 and 9. sugar dispensed in the bags from one
Notice that the answer obtained using standardized scores (0.6231) is differ - manufacturer follows an approximately
ent from the one obtained directly from the unstandardized distribution of ITBS normal distribution with mean 32 ounces
vocabulary scores (0.6243). This difference is due to rounding the z-scores to
2 decimal places before finding the area under the standard normal curve with and standard deviation 0.6 ounces.
Option (i). About what proportion of bags contain
31 to 33 ounces of sugar?
EXAMPLE
SOLUTION:
Can Rory reach the green?
Finding area between two values in a normal distribution Area = 0.9047
PROBLEM: When professional golfer Rory McIlroy hits his driver, the distance the ball travels can be
modeled by a normal distribution with mean 304 yards and standard deviation 8 yards. On another
golf hole, McIlroy has the opportunity to drive the ball onto the green if he hits the ball between 305
and 325 yards. What proportion of Rory’s drives travel a distance that falls in this interval?
30.2 30.8 31.4 32.0 32.6 33.2 33.8
31 33
Amount of powdered sugar
dispensed (oz)
33 −32
z = =1.67
0.6
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31 −32
z = =−1.67
0.6
=
(i) Using Table A: 0.9525–0.04750.9050
Using technology:
Applet/normalcdf(lower:–1000,
upper: –1.67, mean:0,SD:1) = 0.9051
(ii) Applet/normalcdf(lower:31,
=
upper: 33,mean: 32,SD: 0.6) 0.9044
About 90.5% of bags from this
manufacturer will contain less than
31 ounces of powdered sugar.

LESSON 2.5 • Normal Distributions: Finding Areas from Values 135

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