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CHAPTER 5 ANSWERS
Section 5.1
Statistical Literacy and Critical Thinking
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The word “normal” has a special meaning in statistics. It refers to a
specific type of distributions which are characterized by their bell-shaped
curve.
The bell shape starts low, rises to a maximum, and then decreases, being
symmetric around the point at which the maximum occurs. There is a formula
that produces the shape known as the normal distribution. This
distinguishes the normal curve from others which may be similar, but are not
exactly normal.
No. The ten different possible digits are all equally likely, so the graph
of the distribution will be flat, not bell-shaped.
Since the total area under a normal distribution curve is 1.0, if the area
under the curve to the left a particular value is 0.2, then the area to the
right of that value is 0.8.
This statement is not sensible. A normal distribution is symmetric, so the
mean and the median are equal.
This statement is not sensible. A normal distribution has only one peak, so
there can be only one mode.
This statement is not sensible. Since the mean is 100 and the distribution
is symmetric, 50% of the scores are greater than 100. The number of scores
greater than 105 must be something less than 50%, not more.
This statement makes sense. The actual volumes will vary a little bit, and
it is common for such actual measurements to have a normal distribution.
Her initial measurements of 100 cans could easily give her information about
the nature of the distribution, so it makes sense that the distribution is
normal.
Concepts and Applications
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Distribution (b) is not normal since it is not symmetric. Distribution (c)
has the larger standard deviation since it is more spread out than
distribution (a).
Distribution (a) is not normal since it is not symmetric. Distribution (b)
has the larger standard deviation since it is more spread out than
distribution (c).
Normal. It is common for manufactured products, such as CDs, to have a
distribution that is normal. The weights typically vary above and below the
mean weight by the same amounts, so the distribution has one peak and is
symmetric.
Not normal. Many people have relatively small incomes, while a smaller
number have quite large incomes. You may have heard of expressions like the
90-10 rule, which states that 90 percent of the people have 10% of the
income and 10% of the people have 90% of the income. The exact values are
not important, but the idea behind them conveys the impression that incomes
are not normally distributed or even symmetrically distributed.
Not normal. The numbers 1 through 6 should occur with approximately equal
frequency, so the distribution is uniform, not normal. A graph of the
distribution of the distribution will be flat, not bell-shaped.
Normal. There is a small number of scores that are very high and a small
number that are very low, and the distribution is likely to peak at the
value of the mean, which is around 1518. There is a possibility that the
distribution might not be exactly symmetric (and not exactly normal) since
the SAT is taken by students intending to go to college. Many of those
students who do not take the SAT would score low on the test if they were to
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SECTION 5.2, PROPERTIES OF THE NORMAL DISTRIBUTION
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take it, so the lower end of the otherwise normal distribution may be
“chopped off” or truncated.
Normal. Such physical measurements generally tend to be normally
distributed. A small number of males will have extremely high grip strength
measurements, and a very small number will have low measurements. The
distribution will tend to peak around the value of the mean.
Not normal. Many flights experience little or no delay, and only a few
flights experience lengthy delays, so the distribution of flight delay times
will be quite right-skewed.
Not normal. The waiting times will tend to be evenly distributed over a 10minute interval and will therefore have a uniform distribution.
Not normal. Most cars will have no tickets. Those that do have tickets
will most likely be subject to a small number of different amounts according
to the types of parking violations (expired meter, too close to fire hydrant
or crosswalk, loading zones, etc.). It is not possible to determine the
exact shape of the distribution, but it will not be normal due to the number
with no tickets.
The movie lengths are not normally distributed because movies generally have
a minimum length, but no maximum length. There are very few movies that are
a lot shorter than a typical length, but there are some very long movies.
Heart rates are nearly normal because data values are distributed
symmetrically about the mean. Many physiological variables have normal
distributions.
The quarter weights should be nearly normal because the deviations from the
mean are symmetrical above and below the mean.
While the data values are distributed fairly symmetrically about the mean,
the distribution does not have the left and right tails for the more extreme
weights. In other samples, this variable could be normally distributed, but
in this particular sample it is not. If the aspirin were taken from a
bottle that had been purchased, it is possible that tablets that were either
too light or too heavy were rejected in the packing process. This would
explain the lack of tails.
a)
The total area under the curve is 1.
b)
0.20
c)
0.80
d)
0.40
e)
0.45
a)
The total area under the curve is 1.
b)
0.50
c)
0.30
d)
0.70
e)
0.20
a)
The mean is 115.
b)
15%
c)
50%
d)
15%
a)
The mean is 155.
b)
15%
c)
15%
d)
45%
Section 5.2
Statistical Literacy and Critical Thinking
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z = standard score =
value - mean
mean - mean
=
= 0
standard deviation
standard deviation
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CHAPTER 5, A NORMAL WORLD
Since the total percentage under the curve is 100%, the percentage of values
to the right is 100% - 75% = 25%.
No. The distribution of the outcomes from rolling a die is a uniform
distribution, not normal. The rule applies only to normal distributions.
A z core can be negative if the value is less than the mean. A percentile
cannot be negative since percentiles represent the cumulative percentage of
scores from the left. The smallest possible such percentage is 0%.
This statement is not sensible. If we assume that the minimum test score is
zero, then the standard deviation cannot be greater than the mean. If it
were greater than the mean, at least 16% of the scores would be negative,
those that are more than one standard deviation below the mean.
The statement makes sense. For those who can’t visualize a 3420 g baby,
this is about 7 lbs, 8 oz. The standard deviation of 495 g is about 1.09
lbs
The statement makes sense. The given values are realistic and a bell-shaped
distribution is plausible.
This statement does not make sense. Marc’s score is less than one standard
deviation above the mean. An unusually high score would be one that is at
least two standard deviations above the mean.
Concepts and Applications
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a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
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a)
b)
c)
d)
e)
f)
50%. Half of all values are less than the mean.
0.84. 120 is 1 standard deviation to the right of the mean. There is
0.50 to the left of 100 and another 0.34 between 100 and 120.
97.5%. 140 is 2 standard deviations to the right of the mean. There
is 0.50 to the left of 100 and another 0.475 between 100 and 140.
16%. 80 is 1 standard deviation to the left of the mean. The percent
of the scores between 80 and the mean is 34%, so there is 16% to the
left of 80.
0.025. 60 is two standard deviations to the left of the mean. The
relative frequency of scores between 60 and the mean is 0.475, so
there is 0.025 to the left of 60.
16%.
120 is 1 standard deviation to the right of the mean. The
percent of the scores between 120 and the mean is 34%, so there is 16%
to the right of 120.
2.5%. 140 is two standard deviations to the right of the mean. The
percentage of scores between 140 and the mean is 47.5%, so there is
2.5% to the right of 140.
0.84. From part (d), 16% of the scores are less than 80, so 84% must
be greater than 80.
68%. 80 and 120 are both 1 standard deviation from the mean.
81.5%. 80 is 1 standard deviation below the mean and 140 is 2
standard deviations above the mean. There is 34% between 80 and 100
and 47.5% between 100 and 140, a total of 81.5%.
50%. Half of all values are less than the mean.
16%. 55 is 1 standard deviation to the left of the mean. The percent
of the scores between 55 and the mean is 34%, so there is 16% to the
left of 55.
0.025. 40 is two standard deviations to the left of the mean. The
relative frequency of scores between 40 and the mean is .475, so there
is 0.025 to the left of 40.
0.84. 85 is 1 standard deviation to the right of the mean. There is
0.50 to the left of 70 and another 0.34 between 70 and 85.
0.975. 100 is 2 standard deviations to the right of the mean. There
is 0.50 to the left of 70 and another 0.475 between 70 and 100, making
a total of 0.975 to the left of 100.
16%.
85 is 1 standard deviation to the right of the mean. The
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SECTION 5.2, PROPERTIES OF THE NORMAL DISTRIBUTION
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percent of the scores between 85 and the mean is 34%, so there is 16%
to the right of 85.
g)
84%. From part (b), 16% of the scores are less than 55, so 84% must
be greater than 55.
h)
0.975. 40 is 2 standard deviations to the left of the mean. The
relative frequency of scores between 40 and the mean is 0.475, so
there is 0.025 to the left of 40 and 0.975 to the right of 40.
i)
68%. 55 and 80 are both 1 standard deviation from the mean.
j)
97.5%. 40 and 100 are both 2 standard deviations from the mean.
a)
Since 55 seconds is the standard deviation, 68% of the times will be
within 55 seconds of the mean of 184.0.
b)
Since 55 seconds is the standard deviation, 95% of the times will be
within 110 seconds (2 standard deviations) of the mean of 184.0.
c)
Since 55 seconds is the standard deviation, 99.7% of the times will be
within 165 seconds (3 standard deviations) of the mean of 184.0.
a)
From 24.4 in. to 29.6 in. is 27.0 + 2(1.3). Within two standard
deviations of the mean, there are 95% of the women.
b)
30.9 in. is three standard deviations (3.9 in.) above the mean. Since
99.7% of the women will have forward grip reaches within three
standard deviations of the mean, only 0.3% will fall outside this
range and half of those will be greater than 3 standard deviations
above the mean. Thus, 0.15% of women will have forward grip reaches
greater than 30.9 in.
c)
68% of the women will have forward grip reaches within one standard
deviation of the mean (27.0). Half of those (34%) will fall between
the mean and 28.3 in. (one standard deviation above the mean).
Since 100 is the mean, z = 0, and 50% of the scores are less than 100.
z = (84 - 100)/16 = -1.00; from Table 5.1, 15.87% of scores are less than
84.
z = (116 – 100)/16 = 1.00; from Table 5.1, 84.13% of scores are less than
116, so 100% - 84.13 = 15.87% of scores are greater than 116.
z = (76 – 100)/16 = -1.50; from Table 5.1, 6.68% of scores are less than 76.
z = (132 – 100)/16 = 2.00; from Table 5.1, 97.72% of scores are less than
132, so 100% - 97.72% = 2.28% of scores are greater than 132.
z = (80 – 100)/16 = -1.25, which we’ll round to –1.30; from Table 5.1, 9.68%
of z-scores are less than –1.3.
z = (65 – 100)/16 = -2.2; from Table 5.1, 1.39% of scores are less than 65.
z = (129 – 100)/16 = 1.80; from Table 5.1, 96.41% of scores are less than
129.
Since 84 and 116 are each 1 standard deviation away from the mean, the ascores are –1 and +1. From Table 5.1, the area under the curve between
these two values is 84.13 – 15.87 = 68.26%.
For 76, the z-score is (76 – 100)/16 = -1.50. For 124, the z-score is (124
– 100)/16 = 1.50. From Table 5.1, the area between z = -1.5 and z = 1.5 is
93.32% - 6.68% = 86.64%.
See Exercise 18. The z-score for 80 is –1.30. The z-score for 120 is
+1.30. The area under the curve to the left of –1.30 is about 9.68%. The
area to the left of +1.25 is 90.32%. Then the area between –1.30 and +1.30
is 90.32% - 9.68% = 80.64%.
For 92, z = (92 – 100)/16 = -0.5; for 115, z = (115 – 100)/16 = 0.94. The
closest table value of z is 0.95, so we’ll use that. The area between z = 0.50 and z = 0.95 is 82.89% - 30.85% = 52.04%.
Since the mean is 162 cm, the percentage of heights greater than 162 cm is
50%.
z = (168 – 162)/6 = 1.0; from Table 5.1, the percentage less than 168 is
84.13%.
z = (156 – 162)/6 = -1.0; from Table 5.1, the percentage less than 156 is
15.87%. The percentage greater than 156 is 100% - 15.87% = 84.13%.
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CHAPTER 5, A NORMAL WORLD
z = (171 – 162)/6 = 1.5; from Table 5.1, the percentage less than 171 is
93.32%. Therefore the percentage greater than 171 is 100% - 93.32% = 6.68%.
z = (177 – 162)/6 = 2.5; from Table 5.1, the percentage less than 177 is
99.38%.
z = (147 – 162)/6 = -2.5; from Table 5.1, the percentage less than 147 is
0.62%.
z = (144 – 162)/6 = -3.0; from Table 5.1, the percentage less than 144 is
0.13%.
z = (179 – 162)/6 = 2.80; from Table 5.1, the percentage less than 179 is
99.74%. Therefore the percentage greater than 179 is 100% - 99.74% = 0.26%.
The two values of z are z = (156 – 162)/6 = -1.0 and z = (168 – 162)/6 =
1.0. From table 5.1, the percentage between 156 and 168 is 84.13% – 15.87%
= 68.26%.
The two values of z are z = (159 – 162)/6 = -0.50 and z = (165 – 162)/6 =
0.50. From table 5.1, the percentage between 159 and 165 is 69.15% – 30.85%
= 38.30%.
The two values of z are z = (148 – 162)/6 = -2.3 and z = (170 – 162)/6 =
1.3. From table 5.1, the percentage between 148 and 170 is 90.32% – 1.07% =
89.25%.
The two values of z are z = (146 – 162)/6 = -2.7 and z = (156 – 162)/6 = 1.0. From table 5.1, the percentage between 156 and 168 is 15.87% – 0.35% =
15.52%.
In all cases, 5% of coins are rejected. The ranges of weights that are
acceptable to the vending machine are
Cent
2.44-2.56 grams
Nickel
4.88-5.12 grams
Dime
2.208-2.328 grams
Quarter
5.53-5.81 grams
Half dollar 11.06-11.62 grams
a)
z
value mean
s tan dard deviation
250 268
15
1.20
From Table 5.1, the percentage lasting less than 250 days is 11.51%.
b)
z
value mean
s tan dard deviation
300 268
15
2.10
From Table 5.1, the percentage lasting less than 300 days is 98.21%.
Therefore, the percentage lasting more than 300 days is 100% - 98.21%
= 1.79%.
c)
z
value mean
s tan dard deviation
238 268
15
2.00
From Table 5.1, the percentage lasting less than 238 days is 2.28%.
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a)
z
value mean
s tan dard deviation
2000 1518
325
1.50
From Table 5.1, the percentage less than 2000 is 93.32%. Therefore,
the percentage higher than 2000 is 100% - 93.32% = 6.68%.
b)
z
value mean
s tan dard deviation
1500 1518
325
0.05
From Table 5.1, the percentage less than 1500 is 48.01%.
c)
The two standard scores are
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SECTION 5.2, PROPERTIES OF THE NORMAL DISTRIBUTION
z
value mean
s tan dard deviation
1600 1518
325
0.25
z
value mean
s tan dard deviation
2100 1518
325
1.80
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From Table 5.1, the percentage between 1600 and 2100 is 96.41% 59.87% = 36.54%.
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a)
z
value mean
s tan dard deviation
650 497
115
1.30
The percentage of scores less than 650 is 90.32, so a score of 650 is
in the 90th percentile.
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b)
The 95th percentile corresponds to a standard score (from Table 5.1)
of 1.70. Thus (value – 497)/115 = 1.70. Multiplying both sides of
this equation by 115, we get value – 497 = 195.5. Now adding 497 to
each side of the equation, we get value = 497 + 195.5 = 692.5 or 693.
a)
z
value mean
s tan dard deviation
31 30.4
0.23
2.60
From Table 5.1, the percentage less than 31 is 99.53%. Therefore, the
percentage higher than 31 is 100% - 99.53% = 0.47%. [Note that of 50
barometers, this means that only 0.23 barometers read higher than 31,
so most of the time, there will be no barometers reading over 31.]
b)
z
value mean
s tan dard deviation
30 30.4
0.23
1.70
From Table 5.1, the percentage less than 30 is 4.46%.
c)
For the low side, we have (value – 30.4)/0.23 = -1.50. Multiplying
both sides of this equation by 0.23, we get value – 30.4 = -0.345.
Now adding 30.4 to each side of the equation, we get value = 30.4 –
0.345 = 30.055. For the high side, we have (value – 30.4)/0.23 =
1.50. Multiplying both sides of this equation by 0.23, we get value –
30.4 = 0.345. Now adding 30.4 to each side of the equation, we get
value = 30.4 + 0.345 = 30.745. Barometers will be rejected if they
read below 30.055 or above 30.745/
d)
The mean of the 50 barometers is the best measure of the actual
atmospheric pressure at the time the barometers were read. The
distribution of readings could be expected to be approximately normal,
so the median could also be used if there were any outliers in the
values. Outliers would not affect the median, but may have an effect
of the mean.
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z
For the girls,
value mean
s tan dard deviation
75 71
6
0.70
From Table 5.1, the percentage less than 75 is 75.80%. Therefore 24.2% of
the girls’ scores are above 75. Of the 60 girls, 24.2%, or 16, will go to
the state bee.
For the boys,
z
value mean
s tan dard deviation
75 66
5
1.80
From Table 5.1, the percentage less than 75 is 96.41%. Therefore 3.59% of
the boys’ scores are above 75. Of the 50 boys, 3.59%, or 2, will go to the
state bee.
Thus, 16 out of the 18, or 89%, of those going to the state bee are girls.
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The two standard scores are
z
value mean
s tan dard deviation
64 69
2.8
1.80
z
value mean
s tan dard deviation
78 69
2.8
3.20
From Table 5.1, the percentage between 64 and 78 is 99.87% - 3.59% =
96.28%.
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a)
z
value mean
s tan dard deviation
120 110.5
22.4
0.40
From Table 5.1, the percentage less than 2 hours (120 minutes) is
65.54%. Therefore, the percentage greater than 2 hours is 100% 65.54% = 34.46%.
b)
z
value mean
s tan dard deviation
90 110.5
22.4
0.90
From Table 5.1, the percentage less than 1 1/2 hours (90 minutes) is
18.41%.
c)
z
value mean
s tan dard deviation
150 110.5
22.4
1.80
From Table 5.1, the percentage less than 2 1/2 hours (150 minutes) is
96.41%, so the probability that a randomly selected movie will be less
than 2.5 hours long is 0.9641.
Section 5.3
Statistical Literacy and Critical Thinking
1
No. The sample is a convenience sample and is subject to a bia that would
not be expected with a random sample. The cannot assume that her
convenience sample has the same characteristics as a sample that was
randomly selected.
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SECTION 5.3, THE CENTRAL LIMIT THEOREM
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The symbol  represents the standard deviation of the population, and n
represents the size of the sample.
No. The Central Limit Theorem indicates that the sample means will be
approximately normally distributed when sample sizes are large, but samples
of size 2 cannot be considered large unless the population itself if
normally distributed.
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The standard deviation of the sample mean is
V/ n
V/ 4
V /2 .
Concepts and Applications
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a)
The mean of the distribution of sample means is the population mean,
100. With n = 64, the standard deviation of the distribution of the
b)
V / n 16/ 64
2.0 .
With a sample size of n = 100, the mean of the distribution of sample
mean is
16/8
means is the population mean, 100.
c)
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a)
distribution of the mean is V / n 16/ 100 16/10 1.6 .
As the sample size n increases, the means tend to be closer together
(less variation) and the value of the standard deviation of the mean
decreases due to the presence of the square root of n in the
denominator.
The mean of the distribution of sample means is the population mean,
1518. With n = 100, the standard deviation of the distribution of the
V/ n
325/ 100 325/10 3.25 .
With a sample size of n = 2500, the mean of the distribution of sample
sample means is
b)
means is the population mean, 1518.
c)
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a)
b)
c)
8
a)
b)
9
The standard deviation of the
The standard deviation of the
distribution of the sample means is V / n 325/ 2500 325/50
6.50 .
As the sample size n increases, the means tend to be closer together
(less variation) and the value of the standard deviation of the sample
mean decreases due to the presence of the square root of n in the
denominator.
The mean of the distribution of sample means is the population mean,
6.5. With n = 81, the standard deviation of the distribution of the
sample means is V / n 3.452/ 81 3.452/9
0.3836 .
The mean of the distribution of sample means is the population mean,
6.5. With n = 100, the standard deviation of the distribution of the
sample means is V / n 3.452/ 100 3.452/10
0.3452 .
As the sample size n increases, the means tend to be closer together
(less variation) and the value of the standard deviation of the sample
mean decreases due to the presence of the square root of n in the
denominator.
The mean of the distribution of sample means is the population mean,
5.5. With n = 49, the standard deviation of the distribution of the
sample means is V / n 2.872/ 49 2.872/7
0.4103 .
The mean of the distribution of sample means is the population mean,
5.5. With n = 400, the standard deviation of the distribution of the
sample means is. V / n 2.872/ 400 2.872/20
0.1436
c)
As the sample size n increases, the means tend to be closer together
(less variation) and the value of the standard deviation of the sample
mean decreases due to the presence of the square root of n in the
denominator.
For individual aircraft, we use the population standard deviation, 7.9, to
find the standard score.
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CHAPTER 5, A NORMAL WORLD
z
value mean
s tan dard deviation
15 13.0
7.9
0.25 .
From Table 5.1, the percentage with
ages less than 15 years is 59.87%. Therefore, the percentage with ages
greater than 15 years is 100% - 59.87% = 40.13%.
For a sample of size 49, we use the standard deviation of the sample mean,
V/ n
z
10
7.9/ 49 7.9/7
value mean
s tan dard deviation
1.1286 , to find the standard score.
15 13.0
1.80 . From Table 5.1, the percentage of
1.1286
sample means less than 15 years is 96.41%. Therefore, the percentage of
sample means greater than 15 years is 100% - 96.41% = 3.59%.
For individual aircraft, we use the population standard deviation, 7.9, to
find the standard score.
z
value mean
s tan dard deviation
10 13.0
7.9
0.40 .
From Table 5.1, the percentage with
ages less than 10 years is 34.46%.
For a sample of size 84, we use the standard deviation of the sample mean,
V/ n
z
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7.9/ 84 7.9/9.165 0.8620 , to find the standard score.
value mean
10 13.0
3.48 . From Table 5.1, the percentage of
s tan dard deviation
0.8620
sample means less than 10 years is 0.02%.
For individual aircraft, we use the population standard deviation, 7.9, to
find the standard scores.
For 10,
value mean
s tan dard deviation
z
10 13.0
7.9
0.40 .
From Table 5.1, the
percentage with ages less than 10 years is 34.46%.
For 16,
value mean
s tan dard deviation
z
16 13.0
7.9
0.40 .
From Table 5.1, the
percentage with ages less than 16 years is 65.54%. Thus the percentage with
ages between 10 and 16 years is 65.54% – 34.46% = 31.08%.
For a sample of size 81, we use the standard deviation of the sample mean,
V/ n
z
z
12
7.9/ 81 7.9/9
value mean
s tan dard deviation
0.8778 , to find the standard scores.
10 13.0
3.40
0.8778
value mean
s tan dard deviation
16 13.0
0.8778
.
3.40
From Table 5.1, the percentage of sample means less than 10 years is 0.02%
and the percentage less than 16 years is 99.98% Thus, the percentage of
sample means between 10 and 16 is 99.98% - 0.02% = 99.96%.
For individual aircraft, we use the population standard deviation, 7.9, to
find the standard scores.
For 12.5,
z
value mean
s tan dard deviation
12.5 13.0
7.9
0.0633 .
From Table 5.1, the
percentage with ages less than 10 years is 48.01%.
For 13.5,
z
value mean
s tan dard deviation
13.5 13.0
7.9
0.0633 .
percentage with ages less than 13.5 years is 51.99%.
From Table 5.1, the
Thus the percentage
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SECTION 5.3, THE CENTRAL LIMIT THEOREM
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with ages between 12.5 and 13.5 years is 51.99% – 48.01% = 3.98%.
For a sample of size 400, we use the standard deviation of the sample mean,
V/ n
z
z
13
7.9/ 400 7.9/20 0.395 , to find the standard scores.
value mean
12.5 13.0
1.30
s tan dard deviation
0.395
.
value mean
13.5 13.0
1.30
s tan dard deviation
0.395
From Table 5.1, the percentage of sample means less than 12.5 years is 9.68%
and the percentage less than 13.5 years is 90.32% Thus, the percentage of
sample means between 10 and 16 is
90.32% - 9.68% = 80.64%.
a)
For a sample of size 36, we use the standard deviation of the sample
V / n 0.11/ 36 0.11/6 0.0183 , to find the standard score.
12.05 12.00
value mean
2.70
.
s tan dard deviation
0.0183
mean,
z
b)
14
a)
From Table 5.1, the percentage of sample means less than 12.05 oz is
99.65%. Thus, the percentage of sample means greater than 12.05 oz is
100% - 99.65% = 0.35%.
If the true population mean is 12.00 oz, it is very unlikely that we
would get a mean amount as high as 12.05 oz for a sample of 36 cans.
More than likely, the cans are being filled with more than 12.00 oz.
In this case, since the consumers are getting more than they thought
they were getting, no one is likely to feel cheated.
For individual times, we use the population standard deviation, 0.4,
to find the standard scores.
For 4.0,
b)
z
value mean
s tan dard deviation
4.0 3.0
0.4
2.5 .
From Table 5.1, the
percentage of times less than 4 seconds is 99.38%. Therefore, the
percentage of times greater than 4.9 is
100% - 99.38% = 0.62%.
For a sample of size 36, we use the standard deviation of the sample
V / n 0.4/ 60 0.0516 , to find the standard score.
value mean
4.0 3.0
19.4
.
s tan dard deviation
0.0516
mean,
z
c)
15
a)
From Table 5.1, the percentage of sample means less than 4.0 seconds
is greater than 99.98%. Thus, the percentage of sample means greater
than 4.0 seconds is less than 0.02%.
The behavior of individual times is more important than the behavior
of batches of 60 times since a pilot is only concerned about other
pilots seeing one plane – his.
For individual sizes, we use the population standard deviation, 1.0
inches, to find the standard scores.
For 6.2,
b)
z
value mean
s tan dard deviation
6.2 6.0
1.0
0.2 .
From Table 5.1, the
percentage of male head breadths less than 6.2 inches is 57.93%%.
For a sample of size 100, we use the standard deviation of the sample
mean,
V / n 1.0/ 100
0.1 , to find the standard score.
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96
CHAPTER 5, A NORMAL WORLD
z
c)
16
a)
value mean
s tan dard deviation
z
value mean
s tan dard deviation
z
a)
.
12 11.4
2.7
0.2222 .
From Table 5.1, the
percentage of those guests who shower less than 12 minutes is 57.93%.
Therefore, the percentage of guests showering more than 12 minutes is
42.07%.
There will not be enough hot water if the mean shower time is greater
than 12 minutes. For a sample of size 84, we use the standard
deviation of the sample mean,
standard score.
17
2.0
From Table 5.1, the percentage of sample means less than 6.2 inches is
97.72%.
Although the mean head breadth of 100 men is very likely to be less
than 6.2 inches, from part (a), we see that only about 58% of men have
head breadths less than 6.2 inches. Therefore, about 42% of men have
head breadths larger than 6.2 inches and they could not use the
helmets.
For individual sizes, we use the population standard deviation, 2.7
minutes, to find the standard scores.
For 12,
b)
6.2 6.0
0.1
value mean
s tan dard deviation
12 11.4
0.2946
V/ n
2.04
2.7/ 84
0.2946 , to find the
.
From Table 5.1, the percentage of sample means less than 12 minutes is
97.72%. Therefore, the percentage of means greater than 12 minutes is
only 2.28%. The system appears to be effective because it will be
overloaded only rarely. Considering that it is also highly unlikely
that all 84 guests will shower at the same time, it is even less
likely that the system will ever be overloaded.
For individual women, we use the population standard deviation, 29
pounds, to find the standard scores.
For 140 lbs,
value mean
s tan dard deviation
z
140 143
29
0.10 .
From Table 5.1,
the percentage with weights less than 140 pounds is 46.02%.
For 211 lbs,.
b)
z
value mean
s tan dard deviation
211 143
29
2.30
From Table 5.1,
the percentage with weights less than 211 pounds is 98.93%. Thus the
percentage with weights between 143 and 211 pounds is 98.93% – 46.02%
= 52.91%.
For a sample of size 36, we use the standard deviation of the sample
V/ n
29/ 36 4.833 , to find the standard scores.
140 143
value mean
0.60 .
For 140 lbs, z
s tan dard deviation
4.833
value mean
211 143
14.0 .
For 211 lbs, z
s tan dard deviation
4.833
mean,
From Table 5.1, the percentage of sample means less than 140 pounds is
27.43%. For 211 pounds, the standard score is off the chart, so at
least 99.98% of the weights are less than 211 pounds. Therefore,
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SECTION 5.3, THE CENTRAL LIMIT THEOREM
c)
18
a)
there is about a 99.98% - 27.43% = 72.55% chance that the mean weight
is between 140 and 211 pounds.
The result from part (a) is more important in redesigning the ejection
seats since the seats are occupied by only one person, not a group of
36.
For individual pieces of candy, we use the population standard
deviation, 0.0518 g, to find the standard scores.
z
For 0.8535 g,
b)
97
value mean
s tan dard deviation
0.8535 0.8565
0.0518
0.05 .
From Table
5.1, the percentage with weights less than 0.8535 g is 48.01%. It
follows that 51.99% have weights greater than 0.8535 g.
For a sample of size 465, we use the standard deviation of the sample
V/ n
0.0518/ 465 0.0024 , to find the standard scores.
value mean
0.8535 0.8565
1.25 .
For 0.8535 g, z
s tan dard deviation
0.0024
mean,
c)
19
a)
From Table 5.1, the percentage of sample means less than 0.8535 g is
11.51%. It follows that 88.49% of sample means are greater than
0.8535 g.
If the company actually puts exactly 465 M&Ms in each bag, it appears
that it is doing a good job of filling the bags since most
(approximately 8 out of every 9) will exceed the weight printed on the
bag. More likely, the bags will be filled by weight, not by count.
Furthermore, the quality inspection process will probably result in
underweight bags not reaching the market.
For individual quarters, we use the population standard deviation,
0.062 g, to find the standard scores.
For 5.550 g,
z
value mean
s tan dard deviation
5.550 5.670
0.062
1.90 .
From Table 5.1, the percentage with weights less than 5.550 g is
2.87%.
For 5.790 g,
b)
z
value mean
s tan dard deviation
5.790 5.670
0.062
1.90 .
From Table 5.1, the percentage with weights less than 5.790 g is
97.13%. Therefore 2.87% of the quarters will weigh more than 5.790 g.
Altogether 5.74% of the quarters will be rejected. If 280 quarters
are inserted in the vending machine, 5.74% of them, or 16, can be
expected to be rejected.
For a sample of size 280, we use the standard deviation of the sample
V/ n
0.062/ 280 0.0037 , to find the standard scores.
5.550 5.670
value mean
32.43 .
For 5.550 g, z
s tan dard deviation
0.0037
mean,
c)
20
a)
From Table 5.1, the percentage of sample means less than 5.550 g is
less than 0.02%.
Similarly, for 5.790 g, z = 32.43, and the percentage of means less
than 5.790 is greater than 99.98%. There is at least a 99.96% chance
that the mean of 280 quarters is between the limits.
The result from part (a) is more important. No one puts 280 quarters
in a vending machine. Individual legitimate quarters rejected may
mean the loss of a sale. However, the configuration is designed to
prevent losing merchandise when someone puts in a slug or foreign coin
about the size of a quarter.
If his total weight is greater than 195 pounds, then he weighs more
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley
98
CHAPTER 5, A NORMAL WORLD
than 175 pounds.
z
b)
value mean
s tan dard deviation
V/ n
From Table 5.1, 53.98% of men
29/ 213
For 175 pounds,
z
1.9870 , to find the standard scores.
value mean
175 172
1.50 .
s tan dard deviation
1.9870
a)
From Table 5.1, the percentage of sample means less than 175 is
93.32%, so there is a 6.68% chance that the mean weight of the men is
over 175 pounds, making the mean total weight over 195 pounds. A
pilot should be concerned because a 6.68% chance that such an
overweight condition could occur is too high. However, there are
other factors that contribute to the total weight being carried,
including the weight of the checked baggage (which is normally weighed
when it is checked in, and the amount of fuel carried, which may
depend on the length of the flight and expected weather conditions.
These factors are all taken into account in preparation for the flight
to ensure that the plane is not overloaded.
The standard deviation of the distribution of sample means will be
b)
0.1 since it is found by V / n V / 100 V /10 . Sample means tend to
cluster more closely around the population mean than do individual
values. The smaller spread of the means is reflected in the smaller
standard deviation.
The standard deviation of the distribution of sample means for n =
c)
22
0.10 .
will weigh less than 175 pounds, so 46.02% will weigh more than 175
pounds, making his total weight greater than 195 pounds.
For a sample of size 213, we use the standard deviation of the sample
mean,
21
175 172
29
a)
1000 will be V / n V / 1000 V /31.62 0.0316V . Sample means tend to
cluster even more closely around the population mean for samples of
size 1000 than they do for samples of size 100, the standard deviation
being about 1/3 the size it was for samples of size 100. The smaller
spread of the means is reflected in the smaller standard deviation,
which is now only about 3/100 of the size of the population standard
deviation.
As the sample size increases, the standard deviation of the sample
means decreases due to the presence of n in the denominator of the
formula for the standard deviation of the sample mean.
For an individual, we use the population standard deviation, 13.1 mm,
to find the standard scores.
For 125 mm,
b)
z
value mean
s tan dard deviation
125 114.8
13.1
0.80 .
From Table 5.1, the percentage with pressures less than 125 mm g is
78.81%. It follows that there is a 21.19% chance that the woman has a
blood pressure over 125 mm.
For a sample of size 300, we use the standard deviation of the sample
V/ n
13.1/ 300 0.7563 , to find the standard scores.
value mean
125 114.8
13.5 .
For 125 mm, z
s tan dard deviation
0.7563
mean,
From Table 5.1, the percentage of sample means less than 125 is
greater than 99.98%, so there is less than a 0.02% chance that the
mean blood pressure of the women is over 125 mm.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley
CHAPTER 5 REVIEW EXERCISES
c)
z
For 114 mm,
value mean
s tan dard deviation
114 114.8
0.7563
99
1.057 | 1.10 .
From Table 5.1, the percentage of sample means less than 114 is
13.57%.
Chapter 5 Review Exercises
1
a)
b)
c)
2
a)
b)
c)
3
The numbers will occur with equal likelihood. Therefore, they have a
uniform distribution, not a normal distribution.
Weights of a homogeneous population, such as Golden Retriever dogs,
typically have a normal distribution.
Reaction times of 18-year-old drivers may have a nearly normal
distribution since the population is quite homogeneous. If they are
not normally distributed, they are probably slightly right-skewed
since the reaction times cannot go below zero, but there may be a few
people who react slowly and have long reaction times.
Since 9.6 is two standard deviations, about 95% of the act scores are
within 9.6 of the mean of 21.1.
Since 14.4 is three standard deviations, about 99.7% of the act scores
are within 14.4 of the mean of 21.1.
Yes. It is more than three standard deviations above the mean and
therefore would happen less 0.3% of the time. [It would also be
unusual because ACT scores are reported as whole numbers.]
value mean
s tan dard deviation
99.00 98.20
0.62
a)
z
b)
c)
d)
percentile.
From part (a), the standard score is 1.29.
No, the data value lies less than 2 standard deviations from the mean.
For a sample of size n = 50, the standard deviation of the
1.29 ; this corresponds to the 90th
V / n 0.062/ 50 0.0877 . For 97.98,
97.98 98.20
2.50 . This corresponds to the
0.0877
distribution of means is
z
value mean
s tan dard deviation
0.62 percentile, so the likelihood that the mean body temperature is
97.98 degrees is 0.0062.
e)
f)
g)
h)
z
For 101.00,
value mean
s tan dard deviation
101.00 98.20
0.62
4.52 .
This is an
unusual temperature since it lies more than two standard deviations
above the mean. We conclude that the person has a fever.
The 95th percentile is associated with a standard score of 1.65. Thus
the temperature must be 1.65 standard deviations above the mean, or
1.65 x 0.62 = 1.02 degrees above 98.2 degrees, or 99.22. [If you
choose to use the closest table value, use z = 1.6. Then the
temperature is 98.20 + 0.99 = 99.19, not very different.]
The 5th percentile is associated with a standard score of -1.65.
Thus the temperature must be 1.65 standard deviations below the mean,
or 1.02 degrees below 98.2 degrees, or 97.18. [Again, if you choose
to use -1.60, the temperature will be 0.99 degrees below 98.20, or
97.21.]
For 100.6,
z
value mean
s tan dard deviation
100.60 98.20
0.62
3.87 .
This
corresponds to a percentile higher than 99.98. Thus fewer than 0.02%
of healthy adults would be expected to have a temperature above 100.6.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley
100
CHAPTER 5, A NORMAL WORLD
i)
It should be very safe to conclude that someone with a temperature of
100.6 or higher has a fever. On the other hand, using such a high
cutoff point will certainly result in concluding that a number of
people do not have fevers when, in fact, they do have a fever.
For a sample of n = 106, the standard deviation of the distribution of
V / n 0.062/ 106 0.0602 For a temperature of 98.2,
value mean
98.20 98.60
6.64 . Thus the sample mean is
s tan dard deviation
0.0602
sample means is
z
about 6.6 standard deviations below the mean; the chance of selecting
such a sample is extremely small if the assumed mean is correct. The
assumed mean (98.60) may be incorrect.
Chapter 5 Quiz
1
2
3
4
Statements (b) and (e) are correct. Statement (a) is not correct because
there are many distributions that are not unusual that are not normal (e.g.,
uniform). Statement (c) is not correct because the mean and median of a
normal distribution are equal. Statement (d) is not correct because if the
standard deviation were zero, all of the values would be the same. The
distribution would not be bell-shaped.
Twenty points is two standard deviations, so 95% of the scores fall within
20 points of the mean of 50.
The mean of the sample means is always the same as the population mean, so
in this case, it is 50.
The standard deviation of the sample means is
value mean
s tan dard deviation
value mean
s tan dard deviation
70 50
10
40 50
10
V/ n
5/ 100
0.5 .
5
z
6
z
7
8
9
Since 50 is the mean, 50% of the scores are greater than 50.
The percentage of scores greater than 70 is 100% - 97.72% = 2.28%.
Since 70 is 2 standard deviations above the mean of 50 and 30 is 2 standard
deviation below the mean of 50, the percentage of scores below 30 is the
same as the percentage above 70, namely 2.28%.
Since the sample size is largest in part (c), those means will have a
distribution that is closest to normal. The outcomes in part (b) have a
uniform distribution. The outcomes in part (b) have a distribution that
looks like a triangle with its peak at 3.5.
10
2
1
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