Study Guide
2.1a 6
8
9
Chapter 2 Study Guide Solutions
Ideal Response
Peter’s time was slower than 80% of his previous race times that season, but it was slower than only 50% of the racers at the league championship meet.
(a) Maryland has about 13% foreign-born residents placing it at about the 70 th percentile.
(b) Locate 30% on the y -axis, read over to the points and then find the relevant place on the x axis. The 30 th percentile is approximately 4.5% foreign-born
(a) First find the quartiles. The first quartile is the 25th percentile. Find 25 on the y-axis, read over to the line and then down to the x-axis to get about $19. The 3 rd quartile is the 75 th percentile. Find 75 on the y -axis, read over to the line and then down to the x -axis to get about $50. So the interquartile range is $50 $19

$31 .
(b) The person who spent $19.50 is just above what we have called the 25 th percentile. It appears that $19.50 is at about the 26 th percentile.
(c) The graph is below:
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11
14
15
16
(a) To find the 60 th percentile, find 60 on the y -axis, read over to the line and then read down to the x -axis to find approximately 1000 hours.
(b) To find the percentile for the lamp that lasted 900 hours, find 900 on the x -axis, read up to the line and across to the y -axis to find that it is approximately the 35th percentile.
Eleanor’s standardized score, z


100 z


6

1.5
.

1.8
, is higher than Gerald’s standardized score,
(a) Cary’s z -score (0.5) indicates that her bone density score is about half a standard deviation above the average score for all women her age. Even though the two bone density scores are exactly the same, Cary is 10 years older so her z -score is higher than Judy’s (−1.45).
Judy’s bones are healthier when comparisons are made to other women in their age groups.
(b) If we let σ denote the standard deviation of the bone density in Cary’s reference population, then we can solve for σ in the equation 0.5

 . Thus,
 
8 grams/cm 2 . There is more variability in the bone densities for older women, which is not surprising.
(a) Since 22 salaries were less than Lidge’s salary, his salary is at the 22
29

75.86
percentile.
(b) z

3,767,484

0.79
. Lidge’s salary was 0.79 standard deviations above the mean salary of $3,388,617.
Madson’s salary in 2008 was $1,400,000. Since there were 14 salaries less than Madson’s, his salary is at the
14
29

48 percentile. Note that we could also argue that since Madson’s salary was the median, his percentile would be 50%. Madson’s z -score is z

3,767,484
 
0.53
.
Madson had a typical salary compared to the rest of the team since his percentile was 50%, but there were some players who made much more than he did since his z-score is negative, indicating a salary below the mean team salary.

2.1b
2.1 MC
2.2a
20
22
24
27
29
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Chapter 2 Study Guide Solutions
(a) The mean and median salaries will each increase by $1000 (the distribution of salaries just shifts by $1000).
(b) The extremes and quartiles will also each increase by $1000. The standard deviation will not change. Nothing has happened to affect the variability of the distribution. The center has shifted location, but the spread has not changed.
(a) The mean and median will each increase by 5% since all of the observations will increase by 5%.
(b) The 5% raise will increase the distance of the quartiles from the median. The quartiles and the standard deviation will each increase by 5% (the original values will be multiplied by
1.05).
To get a correct measurement in inches we need to subtract the 0.2 inches that Clarence mistakenly added. Then to transform that same measurement to cm we need to multiply by 2.54.
So the new mean is
  
7.62
cm. The change in location does not affect the standard deviation so we just multiply the old standard deviation by 2.54, 0.1(2.54)

0.254
cm.
(a) It is on or above the horizontal axis everywhere, and because it forms a
1
3
by 3 rectangle, the area beneath the curve is 1.
(b) One third of accidents occur in the first mile: this is a
1
3
1
3
.
by 1 rectangle, so the proportion is
(c) One-tenth of accidents occur next to Sue’s property: this is a
1
3 proportion 0.1.
by 0.3 rectangle, so the
Both are 1.5. The mean is 1.5 because this is the obvious balance point of the rectangle. The median is also 1.5 because the distribution is symmetric (so that median = mean) and because half of the area lies to the left and half to the right of 1.5.
33. C 34. B 35. C 36. B 37. D
The Normal distribution for the weights of 9-ounce bags of potato chips is shown below.
38. E
The interval containing weights within 1 standard deviation of the mean goes from 9.07 to 9.17.
The interval containing weights within 2 standard deviations of the mean goes from 9.02 to
9.22. The interval containing weights within 3 standard deviations of the mean goes from 8.97 to 9.27.

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45
46
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Chapter 2 Study Guide Solutions
(a) Approximately 2.5% of bags weigh less than 9.02 ounces, because 9.02 is two standard deviations below the mean.
(b) Approximately 68% of bags have weights between 9.12 0.05

9.07
9.12 0.05

9.17
ounces.
ounces and
(c) Approximately 84% of bags have weights less than 9.17 ounces, because 9.17 is one standard deviation above the mean. Approximately 0.15% of bags have weight less than
8.97 ounces, because 8.97 is three standard deviations below the mean. So approximately
84% 0.15%

83.85% of bags have weight between 8.97 and 9.17 ounces.
(d) The value 9.07 is one standard deviation below the mean. This means that the area to the left of 9.07 is 0.16. In other words, 9.07 is the 16 th percentile of the weights of these potato chip bags.
The standard deviation is approximately 0.2 for the tall, more concentrated one and 0.5 for the short, less concentrated one.
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.
(a) 0.0069. The graph is show below (left).
(b) 1 − 0.9931 = 0.0069. The graph is shown below (right).
(c) 0.9931 − 0.8133 = 0.1798. The graph is shown below (left).
(d)
0.1020 − 0.0016 = 0.1004. The graph is shown below (right).
This distribution is roughly symmetric. There are no clear outliers, though some may suggest that the approximate 10 minute drive times are outliers.

50
Chapter 2 Study Guide Solutions
(a) The graph is shown below (left).
(b) The graph is shown below (right).
2.2b 51
52
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(a) The value that is closest to 0.1000 in Table A is 0.1003. This corresponds to a value of

1.28
for z .
(b) The point where 34% of observations are greater is also the
 
66 th percentile. The value that is closest to 0.6600 in Table A is 0.6591 which corresponds to a z -score of 0.41.
(a) The value that is closest to 0.6300 in Table A is 0.6293 which corresponds to a z -value of 0.33.
(b) If 75% of values are greater than z , then 25% are lower. The value that is closest to
0.2500 in Table A is 0.2514 which corresponds to a z -score of

0.67
.
(a) State: Let x
 the IQ scores of people aged 20 to 34 distribution with
. The variable x has a Normal
 
110 and
 
25 . We want the proportion of IQ scores that are less than 150.
Plan: The proportion of IQ scores less than 150 is shown in the graph below (left).
Do: For we have Using Table A we see that the proportion of observations less than 1.6 is 0.9452 or approximately 94.5%.
Conclude: An IQ score of 150 is approximately the 95th percentile.
(b) State: Let x
 the IQ scores of people aged 20 to 34 distribution with
. The variable x has a Normal
 
110 and
 
25 . We want the proportion of IQ scores that are between 125 and 15.
Plan: The proportion of IQ scores between 125 and 150 is shown in the graph above
(right).
Do: From part (a), we have that for z
 
25

0.6
x

150 , so z

1.6
. For x

125 , we have that
. So the proportion of observations between 0.6 and 1.6 using
Table A is 0.9452 0.7257

0.2195
, or about 22%.
Conclude: About 22% of people aged 20-34 have IQ scores between 125 and 150.

Chapter 2 Study Guide Solutions
(c) State: Let x with
 the IQ scores of people aged 20 to 34 . The variable x has a Normal distribution
 
110 and
 
25 . We want to find the score such that 98% of people aged 20-34 have an IQ that is smaller.
Plan: The 98 th percentile of the IQ scores is shown in the graph below.
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Do: Using Table A, the 98 th percentile for the standard Normal distribution is closest to 2.05.
Therefore the 98 th
2.05
 x

110
25
for
percentile for the IQ scores can be found by solving the equation x . Thus x

.
Conclude: In order to qualify for MENSA membership a person must score 162 or higher.
(a) If a lid is too small to fit, that means that its diameter is less than 3.95. Let x
 diameter of the lid . We wish to find the proportion of lids for which showing this area is below (on the left). For x

3.95
, we have z
 
0.02
x

Table A, the proportion of observations less than
3.95
. The graph
 
1.5
. Using

1.5
is 0.0668. So, approximately 6.7% of lids are too small.
(b) If a lid is too big to fit, that means that the diameter is more than 4.05. Let x
 diameter of the lid . We wish to find the proportion of lids for which showing this area is below (on the right). For x

4.05
, we have z
 x


0.02
4.05
. The graph

3.5
.Using
Table A, the proportion of observations less than 3.5 is approximately 1, so the proportion of observations greater than 3.5 is approximately 0. In other words, it is extremely rare for a lid to be too big.
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(c) It makes more sense to have a larger proportion of lids too small rather than too large. The company wants to make sure that the fit is snug. If more lids are too large, there will be more spills. If lids are too small, customers will just try another lid. But if lids are too large, the customer may not notice and then spill the drink.
(a) We want to solve for the appropriate mean so that less than 1% of lids have diameter less than
3.95. First, find the z -value for which 1% of observations are lower. Using Table A, we find z
 
2.33
. Since we want this z-value to correspond to a diameter of 3.95, we solve z
 3.95

0.02

 
2.33
for μ. In other words,
  
3.95 2.33(0.02) inches. So the mean lid diameter should be 4.00 inches.
(b) If the mean diameter stays at 3.98, then solve
 


2.33
z


 
2.33
for σ. In other words

0.013
inches. So the standard deviation of the lid diameters should be 0.013 in.
(c) We prefer reducing the standard deviation. This will not increase the number of lids that are too big, but will reduce the number of lids that are too small. If we just shift the mean to the right, we will reduce the number of lids that are too small, but we will increase the number of lids that are too big.

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Chapter 2 Study Guide Solutions
(a) Descriptive statistics and a histogram are provided below.
Variable N Mean StDev Minimum Q1 Median Q3 Maximum
Density 29 5.4479 0.2209 4.8800 5.2950 5.4600 5.6150 5.8500
The measurements of the earth’s density are roughly symmetric with a mean of 5.45
(b) and varies from 4.88 to 5.85.
The densities follow the 68−95−99.7 rule closely—75.86% (22 out of 29) of the
(c) densities fall within one standard deviation of the mean, 96.55% (28 out of 29) of the densities fall within two standard deviations of the mean, and 100% of the densities fall within 3 standard deviations of the mean.
Normal probability plot from MiniTab is shown below.
2.2 MC
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68
(d)
The Normal probability plot is roughly linear, indicating that the densities are approximately Normal.
The graphical display in (a), the 68-95-99.7 rule in (b) and the Normal probability plot in (c) all indicate that these measurements are approximately Normal.
The plot is nearly linear. Because heart rate is measured in whole numbers, there is a slight
“step” appearance to the graph.
The shape of the Normal probability plot suggests that the data are right-skewed. This can be seen in the steep, nearly vertical section in the lower left—these numbers were less spread out than they should be for Normal data—and the three apparent outliers that deviate from the line in the upper right; these were much larger than they would be for a Normal distribution.
Women’s weights are skewed to the right: This makes the mean higher than the median, and it is also revealed in the differences M

Q
  
pounds and
Q 3

M

157.3 133.2

24.1
pounds.
69. D 70. C 71. B 72. C 73. C 74. C