Answers: The Normal Curve
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1. Suppose SAT scores are normally distributed with a
mean of 1000 and a standard deviation of 100.
 a) What percentage will be between 900 and 1100?
b) What percentage of scores will be between 1100 and
1200?
c) What percentage of scores will be below 850?
d) How high must the score be to be in the top 2% of all
scores?
a) 68.27; b) 13.59; c)
6.68%
d) z = 2.054 = (x1000)/100; x = 1000 + 205 =
1205.
2. Suppose that the length of time people can hold their
breath is normally distributed with a mean of 80 seconds and
a standard deviation of 12 seconds.
 a) What percentage of people will be able to hold
their breath more than 80 seconds?
b) What percentage of people will be able to hold their
breath between 60 and 80 seconds?
d) What percentage of people will be able to hold their
breath more than 110 seconds?
e) What percentage of people will be able to hold their
breath between 90 and 100 seconds?
f. How long would you have to be able to hold your breath
to be in the top 5 percent of this population?
a) 50%; b) 45.22%; c) 0.6%;
d) 15.45%
f) z = 1.545 = (x80)/12; x = 80 + 18.5 =
98.5
3. Suppose a large orchard produces apples that have a
mean weight of 112 grams with a standard deviation of 8
grams. (Please draw a picture showing what you are looking
for.)
 a) What percentage of apples will weigh less than 120
grams?
P(x < 120)
 b) What percentage of apples will weigh between 112
and 120 grams?
P(112 < x < 120) =
 c) What is the probability that an apple chosen at
random will weigh between 120 and 132 grams?
P(120 < x < 132) =
 d) What percentage will weigh between 100 and 120
grams?
P(100 < x < 120) =
 e) How small must an apple be to be in the smallest
10% of apples?
 f) What range of apple sizes will make up the middle
50% of the population?
a) 84.13%; b) 43.13%; c)
15.24%; d) 77.45%
e) z = 1.28 = (x112)/8; x = 112 + 10.24 = 122.24.
f) z = .6745 = (x112)/8; x = 112  5.4 = 106.6 for the
lower number; z = .6745 = (x112)/8; x = 112 + 5.4 = 117.4
for the upper number.
4. Assume that a machine can fill cereal boxes with a
mean of 36 ounces and a standard deviation of 1 ounce.
 a) What percentage of boxes will have between 37 and
38 ounces?
b) What percentage of boxes will have less than 35
ounces?
c) How light must a box be to be in the lowest five
percent of all boxes?
a) 13.59%; b) 15.87%
c) z = 1.64 = (x  36)/1; x = 36 1.64 =
34.36.
5. Suppose a variety of bananas produces bunches that
have a mean weight of 70 pounds with a standard deviation of
3 pounds. (Please draw a picture showing what you are
looking for.)
 a) What is the probability that a bunch will weigh
between 70 and 74 pounds?
P(70 < x < 74) =
 b) What is the probability that a bunch chosen at
random will weigh between 71 and 75 pounds?
P(71 < x < 75) =
 c) What is the probability that a bunch will weigh
between 68 and 76 pounds?
P(68 < x < 76) =
 d) What is the probability that a bunch will weigh
less than 63 pounds?
P(x < 63) =
 e) How small must a bunch be to be in the smallest
12% of bunches?
a) .4087; b) .3217; c) .7248;
d) .0098
e) z = 1.175 = (x  70)/3; x = 70  3.5 =
76.5
6. Suppose that the life of a particular brand of watch
battery is 1000 days on average with a standard deviation of
50. Suppose further that the lifetimes of these batteries
are normally distributed.
 a) What percentage of these batteries will last more
than 920 days?
b) How long must a battery last so that it is in the top
10% of all batteries?
a) 94.52
b) z = 1.28 = (x  1000)/50; x = 1000 + 64 =
1064.
7. IQ scores are supposed to be normally distributed with
a mean of 100 and a standard deviation of 15.
 a) If a person scores 120 on an IQ test, what
percentage of the population is above him or her?
 b) How low would a person have to score to be in the
lowest 1% of the population.
a) 9.12%
b) z = 2.33 = (x  100)/15; x = 100 + 15*(2.33) =
65.
8. Suppose that the amount of time (in minutes) need for
students to complete a test is N(40,10)
 a) What percentage of students will finish in less
than 50 minutes?
 b) What percentage will finish in less than 25
minutes?
 c) What is the probability that a student selected at
random will take between 35 and 45 minutes?
 d) How many minutes will elapse before 90% of the
students are finished?
a) 84.13; b) 6.68% c)
38.92%
d) z = 1.28 = (x  40)/10; x = 40 + 12.8 =
52.8
9. A bakery sells and average of 1200 donuts per day with
a standard deviation of 200 donuts.
 a) If it bakes 1300 donuts, what is the probability
that it will sell all of them before the end of the day.
(Assume a Normal distribution.)
 b) Assuming a Normal distribution, how many donuts
must it bake to have a probability of .80 that it will
not run out?
a) .3085
b) z = .8416 = (x  1200)/200; x = 1200 + 168 =
1368
10. Suppose a variable is distributed normally with a
mean of 15 and a standard deviation of 3.
 a) What is the probability that x<13?
 b) What is the probability that 14<x<17?
 c) What is the probability that 13<x<14?
 d) What values include the middle 40% of the
distribution?
 e) If two items are drawn from this distribution,
what is the probability that at least one will have a
value of 12 or less?
a) .2525; b) .3781; c)
11.69;
d) lower z = .5244 = (x  15)/3; x = 15  1.6 = 13.4; upper
z = .5244 = (x  15)/3; x = 15 + 1.6 = 16.6;
e) The probability that at least one will have a value of 12
or less is 1  probability that neither will have a value of
12 or less. Assuming that they are independent, the
probability of both being 12 or greater can be found by
finding the probability that one draw will be twelve or
greater and multiplying it by itself. Twelve is one standard
error below the mean, the probability of being one standard
error below the mean or more is .1587 and the probability of
being 12 or above is .8414. Therefore the probability of
both draws being above 12 is .84.14*.8414 = .7079. The
probability of at least on draw being below 12 is 1  .7079
= .2921.
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