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Chapter 3 Solutions
3.1
(a)
(b)
3.2
(a)
(b)
3.3
(a)
(b)
87% of the girls her daughter’s age weigh the same or less than she does
and 67% of girls her daughter’s age are her height or shorter.
According to the Los Angeles Times, the speed limits on California highways
are such that 85% of the vehicle speeds on those stretches of road are no
greater than the speed limit.
Peter’s time was better than 80% of his previous race times that season, but
it was better than only 50% of the racers at the league championship meet.
Although Molly’s league swim meet time was better than only 50% of her
meet times for the season, it was better than 80% of the other participants
in the league swim meet.
Francine’s bone density is 1.45 standard deviations below the mean hip bone
density for 25-year-old women of 956 grams/cm2.
−1.45 =
948 − 956
-8
⇒ standard devation =
= 5.517.
standard deviation
-1.45
The
standard
deviation of the reference population is 5.517 grams/cm2.
3.4
(a)
(b)
For their respective ages, Louise has the healthier bone hip density since her
bone density is 0.5 standard deviations above the mean bone hip density for
her age group while Francine’s was 1.45 standard deviations below the
average for her age group.
0.50 =
948 − 944
4
⇒ standard deviation =
= 8.
standard deviation
0.5
The result is logical since
we might have expected the variability in bone density values to increase as
women age.
3.5
(a)
Since 23 of the salaries are less than or equal to Lidge’s salary, his salary is
at the
23
= 79.3 percentile. Roughly 79% of the team had salaries less than
29
or equal to Lidge’s salary.
(b)
z=
6,350,000 − 3,388,617
= 0.79.
3,767,484
Lidge’s salary was 0.79 standard deviations
above the mean salary of $3,388,617.
3.6
Madson’s salary in 2008 was $1,400,000. Since 9 of the salaries were less
than or equal to Madson’s salary, his salary percentile is
9
= 31.0. Madson’s
29
Chapter 3
z-score is z =
33
1,400,000 − 3,388,617
= −0.53. Madson had a low salary compared
3,767,484
to the rest of the team since his percentile is less than 50% and his z-score
is negative, indicating a salary below the mean team salary.
3.7
(a)
(b)
(c)
(d)
The total area under the curve must equal 1. Since the base of the rectangle
is 1, the height must also be 1.
The balance point (mean) is 0.5.
The median is also 0.5 . It is the same as the mean since the density curve
is symmetric.
(0.4)(1) = 0.4. 40%.
3.8
(a)
The height of the curve is ½ so that the total area of the curve is 1
(2*1/2=1).
(b)
(c)
1.4 −1
= 0.2 . 20% of the observations lie between 1 and 1.4.
2
Q1 is the value such that 25% of the curve is less than this value:
Q1
⇒ Q1 = 0.5. The median and third quartile are found similarly:
2
Median
Q
0.5 =
⇒ Median = 1; 0.75 = 3 ⇒ Q3 = 1.5.
2
2
0.25 =
3.9
Answers will vary. One possibility:
The distribution is skewed to the right so that the median < mean.
median is approximately 3.5 and the mean is approximately 4.2.
The
Chapter 3
34
3.10
Answers will vary. One possibility:
The distribution is well approximated by a uniform distribution between 0 and
9 having mean=median=4.5.
3.11
(a)
(b)
The overall shape of the distribution is symmetric and mound-shaped.
Hence, the mean and median are equal and are at point A.
The overall shape of the distribution is skewed to the left. The mean is at A
and the median is at B.
3.12
3.13
(a) The distribution is skewed to the right, thus the mean is to the right of the
median. Point “A” clearly does not mark the 50th percentile, so the median
must be at “B” and the mean at “C”.
(b) The distribution is symmetric so the mean and the median both mark the
50th percentile of the distribution and are located at “A”.
(c) The distribution is skewed to the left, thus the mean is to the left of the
median. Point “C” clearly does not mark the 50th percentile so the median
must be at “B” and the mean at “A”.
3.14
It actually means that 90% of similar men have cholesterol levels that are
less than or equal to Martin’s level. Martin’s cholesterol is comparatively
high, not low.
Chapter 3
35
3.15
(a)
(b)
(c)
(d)
The mean of the 50 observations is calculated to be 5.598 percent. The
median is 5.45 percent. The standard deviation is 1.391 percent. The
distribution of unemployment rates is fairly symmetric, with a center of about
5.5 percent and a spread of 5.6 percent.
In the ordered data set, Illinois’s unemployment rate is 40th. Thus, the
percentile for Illinois is 40/50=.8. 80% of the unemployment rates are less
than or equal to Illinois’s unemployment rate. Illinois has a pretty high
unemployment rate compared to the remaining 49 states.
Since .4*50=20, the state whose unemployment rate is 20th among the
ordered observations represents the 40th percentile. Texas represents the
40th percentile with an unemployment rate of 5.1%. The z-score for this
state is z =
3.16
(a)
(b)
(c)
5.1− 5.598
= −0.36.
1.391
The unemployment rate for Illinois was 2.4% (6.9-4.5) higher in September,
2008 than in December, 2000.
The percentile for Illinois (its unemployment rate compared to the rest of the
country) did not change much from December of 2000 to September of
2008, it actually dropped slightly. It was at the 86th percentile in December
of 2000 (43/50=0.86) versus the 80th percentile in September of 2008.
The z-score for Illinois in December of 2000 was z =
4.5 − 3.47
= 1.03.
1
The
unemployment rate for Illinois was slightly higher in December of 2000 than
in September of 2008 relative to the rest of the country. It was 1.03
standard deviations above the mean in December of 2000 whereas it was
⎛
⎝
0.94 ⎜ z =
of 2008.
⎞
6.9 − 5.598
= 0.94 ⎟.standard deviations above the mean in September
⎠
1.391
Chapter 3
36
3.17
(a)
The only way to obtain a z-score of 0 is if the x value equals the mean.
Thus, the mean is 170 cm. To find the standard deviation, use the fact that
a z-score of 1 corresponds to a height of 177.5 cm. Then
1=
177.5 −170
⇒ standard deviation = 7.5 . Thus, the standard deviation of
standard deviation
the height distribution for 15-year-old-males is 7.5 cm.
(b)
Since 2.5 =
x −170
, x = 2.5 * 7.5 + 170 = 188.75 cm. A height of 188.75 cm has a
7.5
z-score of 188.75 cm.
3.18
(a)
Since a z-score of 0 corresponds to a height of 162.5, the mean of the
distribution of heights for 16-year-old females is 162.5. Since a z-score of 1
corresponds to a height of 169, we have
1=
169 −162.5
⇒ standard deviation = 6.5.
standard deviation
The standard deviation of the
height distribution for 16-year-old females is 6.5 cm.
(b)
Since −1.5 =
x −162.5
, x = −1.5 * 6.5 + 162.5 = 152.75 cm. A height of 152.75 cm
6.5
has a z-score of -1.5.
3.19
175 −170
= 0.67. Paul is slightly taller than average for his age. His height
7.5
(a)
z=
(b)
is 0.67 standard deviations above the average male height for his age.
75% of boys Paul’s age are shorter than or equal to Paul’s height.
3.20
170 −162.5
= 1.15 . Miranda is taller than average for her age. Her height is
6.5
(a)
z=
(b)
(c)
1.15 standard deviations above the average female height for her age.
Among 16-year-old girls, 88% are Miranda’s height, 170 cm, or shorter.
Relative to her peers, Miranda is taller than Paul since her height is at the
88th percentile and Paul’s is only at the 75th.
3.21
111 ± 2(11) = (89, 133).
3.22
3.23
100 −111
= −1. Since 68% of the scores are within one standard
11
deviation of the mean, 34% lie between the mean and z = −1. Thus,
50% + 34% = 84% are larger than 100. In our sample, 62 of the 74 scores
are greater than 100 so the sample percentage, 62/74=84%, is the same.
z100 =
144 has a z-score of 3 (111 + 3(11) = 144). We would expect half of 0.3%,
or 0.15% to be above 144 (since 99.7% are within three standard deviations
of the mean). It is not surprising that none of the students in the sample
Chapter 3
37
had an IQ score this large as we would only expect a value this large, on
average, about 1.5 times out of every thousand.
3.24
3.25
(a)
(b)
3.26
(a)
(b)
(c)
The mean is 10 and the standard deviation is about 2. The entire range of
scores is about 16 – 4 = 12. Since the range of values in a normal
distribution is about six standard deviations, we have 12/6 = 2 for σ.
336 ± 3(3) = (327, 345).
339 − 336
z339 =
= 1 ⇒ 16% of horse pregnancies last longer than 339 days.
3
This is because 68% are within one standard deviation of the mean, so 32%
are more than one standard deviation from the mean and half of those are
greater than 339.
Since 2.1 is 1 standard deviation above the mean, 50% + 34% = 84% of
the cartons weigh less than 2.1 pounds (50% below the mean of 2 pounds
and 34% between the mean and one standard deviation above the mean).
Since 1.8 is 2 standard deviations below the mean, 2.5% of the cartons
weigh less than 1.8 pounds (5% of the weights are more than 2 standard
deviations from the mean with 2.5% being at least 2 standard deviations
below and 2.5% being at least 2 standard deviations above).
Since 1.9 is 1 standard deviation below the mean, 84% of the cartons weigh
more than 1.9 pounds (50% above the mean of 2 pounds and 34% between
the mean and 1.9 pounds).
3.27
(a)
The area to the left of z=-0.37
is 0.3557.
(b)
The area to the right of z=-0.37
is 1-0.3557=0.6443.
38
Chapter 3
(c)
The area to the left of z=2.15 is
0.9842.
(d)
The area to the right of z=2.15
is 1-0.9842=0.0158.
3.28
(a)
The area to the left of z=-1.58
is 0.0571.
(b)
The area to the right of z=1.58
is 1-0.9429=0.0571.
(c)
The area to the right of z=-0.46
is 1-0.3228=0.6772.
Chapter 3
39
(d)
The area to the left of z=0.93 is
0.8238.
3.29
(a)
The area between z= -1.33 and
z=1.65 is 0.9505-0.0918
=0.8587.
(b)
The area between z =0.50 and
z=1.79 is 0.9633-0.6915
=0.2718.
3.30
(a)
The area between z=-2.05 and
z=0.78 is 0.7823-0.0202
=0.7621.
(b)
The area between z=-1.11 and
z=-0.32 is 0.3745-0.1335 =
0.241.
Chapter 3
40
3.31
(a)
The area to the
left of z is 0.20.
The value of z with 20% of the
area to the left is z=-0.84.
(b)
The value of z with 55% (1-.45)
of the observations less than z
is approximately z=0.13.
The area to the
right of z is 0.45.
3.32
(a)
The area to the left
of z is 0.63.
The value of z with 63% of the
observations less than z is 0.33.
(b)
The value of z with 25% (1-.75)
of the observations less than z is
-0.67.
3.33
The area to the
right of z is 0.75.
Step 1: State the Problem. Let x=the distance that Tiger’s ball travels. The
variable x has a Normal distribution with μ=304 and σ=8. Assuming Tiger is
safe hitting his driver as long as the ball lands before the creek, we want to
find the probability that x ≤ 317. Step 2: Standardize and draw a picture.
For x=317, we have
z=
x −μ
σ
=
317 − 304
= 1.63.
8
So, x ≤ 317 corresponds to z ≤ 1.63 under the standard Normal curve.
Chapter 3
41
Step 3: Use the table. From Table A, we see that the proportion of
observations less than 1.63 is 0.9484. The probability that Tiger is safe
hitting his driver is about 0.95. Step 4: Conclusion. Since the probability
that Tiger’s drive misses the creek is about 0.95, we can conclude that Tiger
is safe hitting his driver.
3.34
Step 1: State the problem and draw a picture. Let x=the score on the
Wechsler Adult Intelligence Scale of a randomly selected adult aged 20 to 34.
The variable x has a Normal distribution with μ=110 and σ=25. We want to
find the score that separates the top 25% from the bottom 75% of scores.
Area = 0.25
Step 2: Use the table. From Table A, we see that the value of z with an
area of 0.75 to its left is 0.67. Step 3: “Unstandardize”. We know that the
standardized value of the unknown x is z=0.67. We find x by solving the
following equation:
0.67 =
Step 4: Conclusion.
scores.
3.35
x −110
⇒ x = 0.67 * 25 + 110 = 126.75.
25
A score of 127 puts a person in the top 25% of all
Step 1: State the Problem. Let x=a randomly selected female’s score on the
SAT Math test. The variable x has a Normal distribution with μ=500 and
σ=111. We want to find the probability that x > 533, the male mean. Step
2: Standardize and draw a picture. For x=533, we have
z=
x −μ
σ
=
533 − 500
= 0.30.
111
So, x > 533 corresponds to z > 0.30 under the standard Normal curve.
Chapter 3
42
Step 3: Use the table. From Table A, we see that the proportion of
observations less than 0.30 is 0.6179.
Therefore, the proportion of
observations greater than 0.30 is 1-0.6179 = 0.3821. Step 4: Conclusion.
Approximately 38% of females scored higher than the male mean.
3.36
(a)
Step 1: State the problem and draw a picture. Let x=the score on the SAT
Math test of a randomly selected male. The variable x has a Normal
distribution with μ=533 and σ=118.
We want to find the score that
separates the top 15% from the bottom 85% of scores.
Area = 0.85
Step 2: Use the table. From Table A, we see that the value of z with an
area of 0.85 to its left is 1.04. Step 3: “Unstandardize”. We know that the
standardized value of the unknown x is z=1.04. We find x by solving the
following equation:
1.04 =
(b)
x − 533
⇒ x = 1.04 *118 + 533 = 655.72
118
Step 4: Conclusion. The 85th percentile of the SAT Math score distribution
for males is 655.7.
Step 1: State the Problem. Let x=a randomly selected female’s score on the
SAT Math test. The variable x has a Normal distribution with μ=500 and
σ=111. We want to determine the percentile corresponding to a value of
655.7. Step 2: Standardize and draw a picture. For x=655.7, we have
z=
x −μ
σ
=
655.7 − 500
= 1.40.
111
To find the percentile corresponding to x = 655.7, we find the area under the
standard Normal curve corresponding to z < 1.40.
Chapter 3
43
Step 3: Use the table. From Table A, we see that the proportion of
observations less than 1.40 is 0.9192. Step 4: Conclusion.
The 85th
percentile for the SAT Math score distribution for males corresponds to the
92nd percentile for the SAT Math score distribution for females.
3.37
(a)
Step 1: State the Problem. Let x=the weight of a randomly selected bag of
potatoes from the shipment. The variable x has a Normal distribution with
μ=10 and σ=0.5. We want to determine the percent of bags weighing less
than 10.25 pounds. Step 2: Standardize and draw a picture. For x=10.25,
we have
z=
x −μ
σ
=
10.25 −10
= 0.5.
0.5
To find the percent of bags weighing less than 10.25 pounds, we find the
area under the standard Normal curve corresponding to z < 0.5.
(b)
Step 3:
Use the table.
From Table A, we see that the percent of
observations less than 0.5 is 0.6915. Step 4: Conclusion. Approximately
69% of the bags in the shipment weighed less than 10.25 pounds.
Step 1: State the Problem. Let x=the weight of a randomly selected bag of
potatoes from the shipment. The variable x has a Normal distribution with
μ=10 and σ=0.5. We want to determine the percent of bags weighing
between 9.5 and 10.25 pounds. Step 2: Standardize and draw a picture.
From part (a), we know that the percent of bags weighing less than 10.25
pounds corresponds to P(z<0.5). For x=9.5, we have
z=
x −μ
σ
=
9.5 −10
= −1.
0.5
Thus, the area we would like to find is the area under the standard Normal
curve between -1 and 0.5.
To calculate this area, we need to find the percent of bags weighing less than
10.25 pounds and subtract from this the percent of bags weighing less than
Chapter 3
44
9.5 pounds. From part (a), we have P(x<10.25)=P(z<0.5) = 0.6915. To
find the percent of bags weighing less than 9.5 pounds, we find the area
under the standard Normal curve corresponding to z < -1.
Step 3:
Use the table.
From Table A, we see that the percent of
observations less than -1 is 0.1587. Thus, the percent of bags weighing
between 9.5 and 10.25 pounds is 0.6915-0.1587=0.5328.
Step 4:
Conclusion.
Approximately 53% of the bags in the shipment weighed
between 9.5 and 10.25 pounds.
3.38
We want to find μ so that 40% of the distribution is less than 60 seconds.
Area = 0.4
The z value with an area of 0.4 to the left is -0.25. We can then find μ as
follows:
−0.25 =
60 − μ
⇒ μ = 60 + 0.25 * 20 = 65.
20
The mean of the distribution is 65 seconds.
Chapter 3
45
3.39
99.7%
95%
68%
57.5
60
62.5
65
67.5
70
72.5
The picture shows that
68%
of
women
are
between 62.5 and 67.5
inches
tall,
95%
are
between 60 and 70 inches
tall,
and
99.7%
are
between 57.5 and 72.5
inches tall.
3.40
The mean is 28 and the standard deviation is about 1. Most of the values are
between 25 and 31 which, in a normal distribution, would be about six
standard deviations. Since the range is about six, the standard deviation
must be about 1.
3.41
In a normal distribution, 68% of the values are within one standard deviation
of the mean. Since the distribution is symmetric, 34% of these values must
be below the mean. That leaves 16% of values that fall more than one
standard deviation below the mean.
Since the percentile rank of an
observation is the percent of observations below it, the percentile rank of an
observation one standard deviation below the mean is 16. Similarly, 95% of
the observations are within two standard deviations of the mean, leaving
97.5% below a value that is two standard deviations above the mean.
3.42
(a)
The area to the right of z=-1.81
is 1- 0.0351=0.9649.
(b)
The area to the left of z=2.29 is
0.9890.
Chapter 3
46
(c)
The area between z=-1.81 and
z= -0.47 is 0.3192-0.0351 =
0.2841.
(d)
The area between z=-1.02 and
z = 0.65 is 0.7422–0.1539 =
0.5883.
3.43
(a)
The area to the left
of z is 32%.
The value of z with an area of
0.32 to the left is z=-0.47.
(b)
The value of z with an area of
0.6 (1-0.4) to the left is
z=0.25.
The area to the right
of z is 0.4.
(c)
The area to the
left of z is 0.10.
The value of z with an area of
0.10 to the left is z=-1.28.
Chapter 3
47
(d)
The area to the
of z is 0.83.
3.44
z =
The value of z with an area of
0.83 to the left is z=0.95.
820 − 1017
= − 0.93. From the calculator [normalcdf(−1000,−0.93)] we
211
see that about 17.6% of the scores were less than 820.
3.45
(a)
(b)
A score of 130 is two standard deviations above the mean. From the
calculator [normalcdf(2,100)] we see that about 2.28% of children had very
superior scores in 1932.
Now we have a distribution with a mean of 120 and a standard deviation of
15.
z =
130 − 120
= 0.67 ⇒ about 25.1% [normalcdf(0.67,100)] would
15
have very superior scores.
3.46
We want to find the score such that 98% of the scores are less than or equal
to its value.
Area to the
left of z is
0.98.
The z value with an area of 0.98 to the left is 2.05.
corresponding IQ score, we solve the following:
2.05 =
3.47
(a)
(b)
To find the
x −100
⇒ x = 2.05 *15 + 100 = 130.75.
15
$215 and $395 represent one standard deviation above and below the mean
of $305. Thus, about 68% spent an amount in this range.
z =
287 − 305
= − 0.2 ⇒ about 42% [normalcdf(−100,-.2)] spent less than
90
$287 on textbooks.
48
3.48
(a)
(b)
Chapter 3
Joey’s score on the national standardized math test was as high or higher
than 97% of students who took the test. His score on the reading portion of
the test was as high or higher than 72% of students who took the test.
Only if we assume that the distribution of scores nationally follows a Normal
distribution. If so, his scores can be found by determining the z values that
represent the 97th and 72nd percentiles of the standard Normal distribution,
1.88 and 0.58, respectively.
3.49
(a)
The area to the right of z=-2.25
is 1-0.0122=0.9878.
(b)
The area between z=-2.25 and
z=1.77 is 0.9616-0.0122 =
0.9494.
(c)
The area to the left of z=2.89 is
0.9981.
(d)
The area between z=1.44 and
z=2.89 is 0.9981-0.9251 =
0.0730.
Chapter 3
49
3.50
(a)
The area to the
left of z is 0.70.
The z value with an area of 0.7
to the left is 0.52.
(b)
The z value with an area
of 0.1 (1-0.9) to the left
is -1.28.
The area to the right
of z is 0.90.
(c)
The area to the
left of z is 0.46.
3.51
The z value with an area of 0.46
to the left is -0.10.
2500 − 3668
= −2.29. The z value that corresponds to a baby weighing less
511
(a)
z=
(b)
than 2500 grams at birth is -2.29. The percent of babies weighing less than
this is the area to the left, 0.01. 1% of babies will be identified as low birth
weight.
The z-values corresponding to the quartiles (25th percentile, median, 75th
percentile) are -0.67, 0 and 0.67. The median is equal to the mean of the
distribution, 3668. To find the x values corresponding to the 25th and 75th
percentiles, we can solve the z equations for x as follows:
−0.67 =
x − 3668
⇒ x = −0.67 * 511+ 3668 = 3325.63;
511
0.67 =
x − 3668
⇒ x = 0.67 * 511+ 3668 = 4010.37.
511
The quartiles of the birth weight distribution are 3325.6, 3668 and 4010.4
grams.
Chapter 3
50
3.52
(a)
(b)
(c)
3.53
(a)
(b)
3.54
Based on the histogram, the distribution of coliform counts appears to be
approximately normal.
The mean is calculated to be 5.88 and the standard deviation is calculated to
be 2.026.
Observations between 3.854 and 7.906 are within one standard deviation of
the mean. 65 out of 100, or 65%, of the observations are within one
standard deviation of the mean. Observations between 1.828 and 9.932 are
within two standard deviations of the mean. 95 out of 100, or 95%, of the
observations are within two standard deviations of the mean. Observations
between -0.198 and 11.958 are within three standard deviations of the
mean. All of the observations, or 100%, are within three standard deviations
of the mean. Note how closely this follows the 68-95-99.7 rule.
The overall shape of the distribution is bi-modal. Because of the symmetry
of the curve, the mean and the median are the same and both are equal to
B.
The overall shape of the distribution is skewed to the right. The mean is
pulled toward the tail of the distribution. Accordingly, the median is at A and
the mean is at B.
We want to find the head sizes that correspond to the 5th and 95th percentiles
of the distribution. The corresponding z values are -1.645 and 1.645. We
solve for the x values as follows:
x − 22.8
⇒ x = −1.645 *1.1+ 22.8 = 20.99
1.1
x − 22.8
1.645 =
⇒ x = 1.645 *1.1+ 22.8 = 24.61.
1.1
−1.645 =
Head sizes that are less than 20.99 inches or greater than 24.61 inches get
custom-made helmets.
Chapter 3
3.55
(a)
(b)
51
The middle 95% of all yearly returns lie within 2 standard deviations of the
mean or between 12-2*16.5=-21 and 12+2*16.5=45. The middle 95% of
all yearly returns are between -21% and 45%.
z=
0 −12
= −0.73.
16.5
The area to the left of z=-0.73 is 0.2327. The market is down for about 23%
of the years.
(c)
z=
15 −12
25 −12
= 0.18 and z =
= 0.79.
16.5
16.5
The area between z =0.18 and z=0.79 is 0.7852-0.5714=0.2138.
3.56
(a)
(b)
The density curve is skewed to the right. This makes sense because you
would expect most people to play the game in a relatively short amount of
time with some taking much longer (they aren’t as good at the game or they
must leave their computer for some reason).
The author scored in the top 19%. Thus, the time it took the author to play
the game was less than 81% of all of the game times for the week.