EWU – CSBS 320 – Introductory Statistics – Briand
Handout. Chapter 4: The Normal Curve. Practice Problems. ANSWERS.
1.
a..0668 b.The test scores are normally distributed
2.
a. .0831 b..2912 c..4778
3.
x
z
a. z score = -3.00 b. z score = .67 c. z score = .20 d. z score = .00

4.
A quick check of your answer can be made by comparing it with the proportion of students who have
IQs of 120 or higher, .0918. The proportion with IQs of 70 or lower will be less than the proportion
with IQs of 120 or higher. Is your calculated proportion less? For IQ = 70, z = (70-100)/15 = -2.00.
The proportion beyond z = 2.00 is .0228. Thus, .0228 or 2.28 percent of the population is expected to
have IQs of 70 or lower.
5.
(.0228)(4000) = 91.2 or 91 students
6.
z = (110 – 100)/15 = 10/15 = .67. The proportion beyond z = .67 is .2514, which is the expected
proportion of people with IQs of 110 or higher.
7.
a. .2514 x 250 = 62.85 or 63 students b. 250 – 62.85 = 187.15 or 187 students c.½ x 250 – 125.
8.
The z score that includes a proportion of .02 is 2.06. If z = 2.06, than X= 100  (2.06)(15) = 100 + 30.9
= 130.9. In fact, Mensa requires an IQ of 130 on tests that have a standard deviation of 15.
9.
a.The z score you need is 1.65. (1.64 will include more than the tallest percent.) 1.65 = (X - 64.6)/2.5; X
= 64.6 + 4.1 = 68.7 inches. b. z = (58 – 64.6)/2.5 = -6.6/2.5 = -2.64. The proportion excluded is .0041.
10.
a. z = (60 – 69.8)/3.0 = -9.8/3.0 = -3.27. The proportion excluded is .0006. b. z = (62 – 69.8)/3.0 =
7.8/3.0 = -2.60. The proportion taller than Napoleon is .9953 (.5000 + .4953).
11.
a. z = (3.20 – 3.11)/.05 = 1.80, proportion = .0359. b. The z score corresponding to a proportion of .1000
is 1.28. Thus X = 3.11 + (1.28)(.05) = 3.11 + .064 = 3.17 grams, X = 3.11 - .064 = 3.02. Thus, the
middle 80 percent of the pennies weigh between 3.05 and 3.17 grams.
12.
a. z = (30 – 35)/6 = -.83, proportion = .2967; z = (40 – 35)/6 = .83, proportion = .2967; (2)(.2967) =
.5934 = the proportion of students with scores between 30 and 40. b. The probability is also .5934.
13.
No, because the distance between 30 and 40 straddles the mean, where scores that occur frequently are
found. The distance between 20 and 30 is all in one tail of the curve, where scores occur less frequently.
14.
z = (20 – 35)/6 = -2.5, proportion = .4938; z = (30 – 35)/6 = -.83, proportion = .2967; 4938 - .2967 =
.1971. The proportion of students with scores between 35 and 30 is subtracted from the proportion with
scores between 35 and 20. There are other ways to set up this problem that yield the same proportion.
15.
800 people x .1971 = 157.7 = 158 students.
16.
The z score that is conventionally used for a proportion of .0050 is 2.58. Thus X =   Z = 100 +
(2.58)(15) = 61.3 or 61. IQ scores of 139 or higher or 61 or lower are achieved by 1 percent of the
population.
17.
Because .01 of the curve lies outside the limits of the scores 61 and 139, the probability is .01. The
probability is .005 that the randomly selected person has an IQ of 139 or higher and .005 that is 61 or
lower.
18.
z = (.85 - .99)/.17 = -.14/.17 = -.82. The proportion associated with z = -.82 is .2061, and
(.2061)(185,822) = 38,298 truck drivers. Because the normal curve corresponds only approximately to
the empirical curve, the usual procedure is to report this result as “About 38,000 truck drivers would get
raises.”
19.
This problem has two parts: determining the number of trees over 8 inches DBH on 1 acre and
multiplying by 100:z = (8 - 13.68)/4.83 = -1.18, proportion = .5000 + .3810 = .8810; .8810(199) =
175.319 trees on 1 acre; 175.319(100) = 17,532 trees on 100 acres (about 17500 trees).