# Solution Manual for Statistics for the Life Sciences, 5th Edition

Preview Extract

Chapter 2 27
CHAPTER 2
Description of Samples and Populations
2.1.1 (a) i) Molar width
ii) Continuous variable
iii) A molar
iv) 36
(b) i) Birthweight, date of birth, and race
ii) Birthweight is continuous, date of birth is discrete (although one might say categorical and
ordinal), and race is categorical
iii) A baby
iv) 65
โข 2.1.2 (a) i) Height and weight
ii) Continuous variables
iii) A child
iv) 37
(b) i) Blood type and cholesterol level
ii) Blood type is categorical, cholesterol level is continuous
iii) A person
iv) 129
2.1.3 (a) i) Number of leaves
ii) Discrete variable
iii) A plant
iv) 25
(b) i) Number of seizures
ii) Discrete variable
iii) A patient
iv) 20
2.1.4 (a) i) Type of weather and number of parked cars
ii) Weather is categorical and ordinal, cars is discrete
iii) A day
iv) 18
(b) i) pH and sugar content
ii) Both variables are continuous
iii) A barrel of wine
iv) 7
2.1.5 (a) i) body mass and sex
ii) body mass is continuous, sex is categorical
iii) A blue jay
iv) 123
(b) i) lifespan, thorax length, and percent of time spent sleeping
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28 Solutions to Exercises
ii) Lifespan is discrete, thorax length and sleeping time are continuous
iii) A fruit fly
iv) 125
โข 2.2.1 (a)There is no single correct answer. One possibility is:
Molar
width
[5.4, 5.6)
[5.6, 5.8)
[5.8, 6.0)
[6.0, 6.2)
[6.2, 6.4)
[6.4, 6.6)
[6.6, 6.8)
Total
Frequency
(no.
specimens)
1
5
7
12
8
2
1
36
(b) The distribution is fairly symmetric.
2.2.2
2.2.3 There is no single correct answer. One possibility is
MAO
4.0-5.9
6.0-7.9
8.0-9.9
10.0-11.9
12.0-13.9
14.0-15.9
Frequency
(no. patients)
2
5
4
3
1
2
Copyright (c) 2016 Pearson Education, Inc.
Chapter 2 29
16.0-17.9
18.0-19.9
Total
0
1
18
2.2.4
2.2.5 There is no single correct answer. One possibility is
Branches
20-24
25-29
39-34
35-39
40-44
45-49
50-54
55-59
Total
Frequency
(no. cells)
6
9
7
5
3
2
3
1
36
Copyright (c) 2016 Pearson Education, Inc.
30 Solutions to Exercises
2.2.6 There is no single correct answer. One possibility is
Protein
production
375-399
400-424
425-449
450-474
475-499
500-524
525-549
550-574
Total
Frequency
(no. cows)
1
2
6
2
7
4
5
1
28
โข 2.2.7 There is no single correct answer. One possibility is
Glucose
(%)
70-74
75-79
80-84
85-89
90-94
95-99
100-104
105-109
110-114
115-119
120-124
125-129
130-134
Total
Frequency
(no. of dogs)
3
5
10
5
2
2
1
1
0
1
0
0
1
31
Copyright (c) 2016 Pearson Education, Inc.
Chapter 2 31
2.2.8 (a)
(b)
The distribution is very slightly skewed to the right.
2.2.9 (a)
(b) The distribution is bimodal.
(c) The histogram with only 6 classes obscures the bimodal nature of the distribution.
Copyright (c) 2016 Pearson Education, Inc.
32 Solutions to Exercises
โข 2.3.1 Any sample with ฮฃyi = 100 would be a correct answer. For example: 18, 19, 20, 21, 22.
2.3.2 Any sample with ฮฃyi = 100 and median 15 would be a correct answer. For example: 13, 14, 15, 28,
30.
2.3.3 y = ฮฃyi/n =
6.3 + 5.9 + 7.06.9 + 5.9
= 6.40 nmol/gm. The median is the 3rd largest value (i.e., the
5
third observation in the ordered array of 5.9 5.9 6.3 6.9 7.0), so the median is 6.3 nmol/gm.
2.3.4 Yes, the data are consistent with the claim that the typical liver tissue concentration is 6.3 nmol/gm.
The value of 6.3 fits comfortably near the center of the sample data.
โข 2.3.5 y = 293.8 mg/dl; median = 283 mg/dl.
โข 2.3.6 y = 309 mg/dl; median = 292 mg/dl.
2.3.7 y = 3.492 lb; median = 3.36 lb.
2.3.8 Yes, the data are consistent with the claim that, in general, steers gain 3.5 lb/day; the value of 3.5
fits comfortably near the center of the sample data. However, the data do not support the claim that
4.0 lb/day is the typical amount that steers gain. Both the mean and the median are less than 4.0;
indeed, the maximum in the sample is less than 4.0.
2.3.9 y = 3.389 lb; median = 3.335 lb.
2.3.10 There is no single correct answer. One possibility is
Resistant
Frequency
bacteria
(no. aliquots)
12-14
4
15-17
3
18-20
1
21-23
1
24-26
1
Total
10
Copyright (c) 2016 Pearson Education, Inc.
Chapter 2 33
(b) y = 16.7, median =
15 + 15
= 15.
2
โข 2.3.11 The median is the average of the 18th and 19th largest values. There are 18 values less than or
equal to 10 and 18 values that are greater than or equal to 11. Thus, the median is
10 + 11
= 10.5 piglets.
2
2.3.12 y = 375/36 = 10.4.
โข 2.3.13 The distribution is fairly symmetric so the mean and median are roughly equal. It appears that
half of the distribution is below 50 and half is above 50. Thus, mean โ median โ 50.
2.3.14 Mean โ 35, median โ 40
2.4.1 (a) Putting the data in order, we have
13 13 14 14 15 15 16 20 21 26
The median is the average of observations 5 and 6 in the ordered list. Thus, the median is
= 15. The lower half of the distribution is
13 13 14 14 15
The median of this list is the 3rd largest value, which is 14. Thus, the first quartile of the
distribution is Q1 = 14. Likewise, the upper half of the distribution is
15 16 20 21 26
The median of this list is the 3rd largest value, which is 20. Thus, the third quartile of the
distribution is Q3 = 20.
(b) IQR = Q3 – Q1 = 20 – 14 = 6
Copyright (c) 2016 Pearson Education, Inc.
15 + 15
2
34 Solutions to Exercises
(c) To be an outlier at the upper end of the distribution, an observation would have to be larger than
Q3 + 1.5(IQR) = 20 + 1.5(6) = 20 + 9 = 29, which is the upper fence.
โข 2.4.2 (a) The median is the average of the 9th and 10th largest observations. The ordered list of the data
is
4.1 5.2 6.8 7.3 7.4 7.8 7.8 8.4 8.7 9.7 9.9 10.6 10.7 11.9 12.7 14.2 14.5 18.8
Thus, the median is
8.7 + 9.7
= 9.2.
2
To find Q1 we consider only the lower half of the data set:
4.1 5.2 6.8 7.3 7.4 7.8 7.8 8.4 8.7 9.7
Q1 is the median of this half (i.e., the 5th largest value), which is 7.4.
To find Q3 we consider only the upper half of the data set:
9.7 9.9 10.6 10.7 11.9 12.7 14.2 14.5 18.8.
Q3 is the median of this half (i.e., the 5th largest value in this list), which is 11.9.
(b) IQR = Q3 – Q1 = 11.9 – 7.4 = 4.5.
(c) Upper fence = Q3 + 1.5 ร IQR = 11.9 + 6.75 = 18.65.
(d)
2.4.3 (a) Median = 82.6, Q1 = 63.7, Q3 = 102.9.
(b) IQR = 39.2.
(c)
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Chapter 2 35
2.4.4 (a) Q1 = 35, median = 50, Q3 = 65.
(b) Q1 = 20, median = 35, Q3 = 50.
2.4.5 The histogram is centered at 40. The minimum of the distribution is near 25 and the maximum is
near 65. Thus, boxplot (d) is the right choice.
2.4.6 Yes, it is possible that there is no data value of exactly 42. The first quartile does not need to equal
a data value. For example, it could be that the first quartile is the average of two points that have
the values 41 and 43.
2.4.7 (a) The IQR is 127.42 โ 113.59 = 13.83.
Copyright (c) 2016 Pearson Education, Inc.
36 Solutions to Exercises
(b) For a point to be an outlier it would have to be less than 113.59 โ 1.5*13.83 = 92.845 or else
greater than 127.42 + 1.5*13.83 = 148.165. But the minimum is 95.16 and the maximum is 145.11,
so there are no outliers present.
2.4.8
2.5.1 (a)
(b)
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Chapter 2 37
(c) The relative frequency chart in part (b) is more useful because it accounts for the different
samples sizes for each claw configuration.
2.5.2 (a)
Copyright (c) 2016 Pearson Education, Inc.
38 Solutions to Exercises
(b)
(c) Because of the small sample size, the dotplot is a simple and adequate way to represent these
data.
2.5.3
Copyright (c) 2016 Pearson Education, Inc.
Chapter 2 39
2.5.4 (a)
(b) The womenโs distribution is skewed to the right, with the first quartile close to the minimum and
the median close to the first quartile, but the third quartile farther away and the two largest values
being outliers. The distribution for the men is close to being symmetric except for a high outlier.
The median for the men is greater than that for the women but the first and third quartiles are
similar for the sexes. The men have a lower minimum and a higher maximum.
โข 2.6.1 (a) y = 15,
โ (y โ y ) = 18, s = 18 / 3 = 2.45.
2
i
โ (y โ y ) = 44, s =
44 / 4 = 3.32.
(c) y = 1,
โ (y โ y ) = 24, s =
24 / 3 = 2.83.
(d) y = 3,
โ (y โ y ) = 28, s =
28 / 4 = 2.65.
(b) y = 35,
2
i
2
i
2
i
2.6.2 (a) y = 7,
โ (y โ y ) = 16, s = 16 / 4 = 2.
2
i
(b) y = 5,
โ (y โ y ) = 6, s = 6 / 3 = 1.4.
(c) y = 6,
โ (y โ y ) = 26, s =
2
i
2
i
26 / 4 = 2.5.
2.6.3 Any sample for which the deviations ( yi โ y) are equal to -3, -1, 0, 2, 2 would be a correct answer.
Example: 17, 19, 20, 22, 22; in this case y = 20.
Copyright (c) 2016 Pearson Education, Inc.
40 Solutions to Exercises
โข 2.6.4 y = 33.10 lb; s = 3.444 lb.
2.6.5 y = 1.190; s = 0.1840.
2.6.6 The data are -13, -29, -7, 2, -10, -43, 4, 15, -13, -30. y = -12.4 mm Hg; s = 17.6 mm Hg.
2.6.7 (a) y = 6.343; s = 0.7020.
(b) Median = 6.2; Q1 = 5.9, Q3 = 6.8, IQR = 6.8 – 5.9 = .9.
(c) New y = 6.77; new s = 1.68; new median = 6.2, new IQR = 0.9. The median and interquartile
range display resistance in that they do not change. The standard deviation changes greatly,
showing its lack of resistance. The mean changes a modest amount.
2.6.8 (a) The first quartile is the 4th largest observation, which is 26.4. The third quartile is the 12th
largest observation, which is 37.5. The interquartile range is 37.5 – 26.4 = 11.1.
(b) The range is 45.5 – 18.4 = 27.1.
โข 2.6.9 (a) y ยฑ s is 32.23 ยฑ 8.07, or 24.16 to 40.30; this interval contains 10/15 or 67% of the
observations.
(b) y ยฑ 2s is 16.09 to 48.37; this interval contains 15/15 or 100% of the observations.
2.6.10 According to the Empirical Rule, we expect 68% of the data to be within one SD of the mean; this
is quite close to the observed 67%. We expect 95% of the data to be within two SDs of the mean.
In fact, 100% of the observations are in this interval.
2.6.11 (a) y ยฑ s is 57.9 to 138.7; this interval contains 26/36 or 72% of the observations.
(b) y ยฑ 2s is 17.5 to 179.1; this interval contains 34/36 or 94% of the observations.
(c) y ยฑ 3s is -22.9 to 219.5; this interval contains 36/36 or 100% of the observations.
2.6.12 According to the Empirical Rule, we expect 68% of the data to be within one SD of the mean; this
is close to the observed 72%. We expect 95% of the data to be within two SDs of the mean; this is
quite close to the observed 94%. We expect over 99% of the data to be within three SDs of the
mean; in fact, 100% of the observations are in this interval.
2.6.13 We would expect the coefficient of variation for weight to change more from age 2 to age 9. This
is due to the fact that genetic factors tend to influence height more than weight and these genetic
factors do not change in the course of life. On the other hand, environmental factors such as food,
etc., can vary significantly.
โข 2.6.14 Coefficient of variation =
s
6.8
=
= 0.04 or 4%.
y 166.3
Copyright (c) 2016 Pearson Education, Inc.
Chapter 2 41
โข 2.6.15 The mean is about 45. The length of the interval that covers the middle 95% of the data is
approximately equal to 70 – 20 = 50. An estimate of s is (length of interval)/4 = 50/4 โ 12.
2.6.16 The mean is about 100. The length of the interval that covers the middle 95% of the data is
approximately equal to 150 – 60 = 90. An estimate of s is (length of interval)/4 = 90/4 โ 22.
2.6.17 (a) The two SDs should be about the same. (The sample size doesnโt affect the SD.)
(b) (ii) The sample of basketball players will include mostly tall men while the sample of college men
will include tall, short, and average height men, leading to a larger SD.
(c) (ii) The jockeys will all be close to the same height, leading to a small SD for them.
โข 2.7.1 y’ = (y – 7)*100. Thus, the mean of y’ is ( y – 7)*100 = (7.373 – 7)*100 = 37.3. The SD of y’ is
s ร 100 = 0.129*100 = 12.9.
2.7.2 (a) Mean = (36.497)(1.8) + 32 = 97.695; SD = (0.172)(1.8) = 0.310
(b) Coefficient of variation = .310/97.695 = 0.003 or 0.3%. [Remark: This is not the same as the
original coefficient of variation, which is 0.172/36.497 = 0.005 or 0.5%.]
2.7.3 (a) Mean = (1.461)(2.20) = 3.214 lb/day; SD = (0.178)(2.20) = 0.392 lb/day.
(b) (i) Coefficient of variation = 0.178/1.461 = 0.122 or 12.2%
(ii) Coefficient of variation = 0.392/3.214 = 0.122 or 12.2%
2.7.4 New mean = (1.461-1)(100) = 46.1; new SD = (.178)(100) = 17.8.
2.7.5 Both a log transformation and a square root transformation will pull in the right-hand tail and push
out the left-hand tail, but a log transformation is more severe than is a square root transformation.
Thus, the first histogram, (i) is for the square root transformation and (ii) is the histogram that
results from a log transformation.
2.7.6 Both log(Y) and 1/sqrt(Y) do a good job, with 1/sqrt(Y) being slightly better (but either of these two
could be considered acceptable).
2.S.1 If the mean of five observations is 181, then the sum of the five observations is 5 ร 181 = 905. The
sum of the first four observations is 180 + 182 + 179 + 176 = 717. Thus, the height of the fifth
student must be 905 โ 717 = 188 cm.
2.S.2 (a)
(b) IQR = 13.2 – 12 = 1.2
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42 Solutions to Exercises
(c) To be a low outlier a value must be less than 12 โ 1.5*1.2 = 12 โ 1.8 = 10.2, so there are no
outliers on the low end. To be a high outlier a value must be greater than 13.2 + 1.5*1.2 = 13.2 +
1.8 = 15, so there are no outliers on the high end.
2.S.3 (a) The median is (29 + 31)/2 = 30. Q1 = (22 + 24)/2 = 23; Q3 = (46 + 48)/2 = 47.
(b) IQR = 47 – 23 = 24.
(c) Upper fence = 47 + (1.5)(24) = 47 + 36 = 83. Lower fence = 23 – (1.5)(24) = 23 – 36 = -13.
2.S.4 (a) There is no single correct answer. One possibility is
Absorbance
0.095-0.099
0.100-0.104
0.105-0.109
0.110-0.114
0.115-0.119
0.120-0.124
0.125-0.129
0.130-0.134
Total
Frequency
3
2
5
4
6
6
0
1
27
2.S.5 (a) Median = 0.114; Q1 = 0.107; Q3 = 0.120; IQR = 0.120 – 0.107 = 0.013.
Copyright (c) 2016 Pearson Education, Inc.
Chapter 2 43
(b) To be an outlier on the high end of the distribution, an observation must be greater than the upper
fence of 0.120 + (1.5)(0.013) = 0.120 + 0.0195 = 0.1395.
2.S.6 The midrange is not robust, because changes in the extreme values โ the minimum or the maximum
โ result in changes in the midrange.
2.S.7 (a) Median = (0 + 1)/2 = 0.5.
(b) Mean = 2.75.
(c)
(d) The distribution is strongly bimodal, with no data near the mean.
โข 2.S.8 The bars in the histogram that correspond to observations less than 45 represent roughly one-third
of the total area. Thus, about 30% of the observations are less than 45.
2.S.9 The bars in the histogram that correspond to observations greater than 25 represent roughly ninetenths of the total area. Thus, about 90% of the observations are greater than 25.
2.S.10 (a) y = 8,
(b) y = 24,
(c) y = 2,
โ (y โ y ) = 30, s = 30 / 4 = 2.74.
2
i
โ (y โ y ) = 36, s = 36 / 4 = 3.00.
2
i
โ (y โ y ) = 54, s = 54 / 4 = 3.67.
2
i
2.S.11 (a) (i) y ยฑ s is 0.76 to 3.70; this interval contains 34 + 50 + 18 = 102 of the observations. Thus,
102/144 or 71% of the observations are within 1 SD of the mean.
(ii) y ยฑ 2s is -.71 to 5.17; this interval contains 13 + 34 + 50 + 18 + 16 + 10 = 141 of the
observations. Thus, 141/144 or 98% of the observations are within 2 SDs of the mean.
(b) The total number of larvae is (0)(13) + (1)(34) + (2)(50) + … + (7)(1) = 321. The mean is y =
321/144.
Copyright (c) 2016 Pearson Education, Inc.
44 Solutions to Exercises
(c) The median is the average of the 72nd and 73rd largest observations. Both of these observations
are in the class “2”, so the median is (2 + 2)/2 = 2.
2.S.12 (a) y = 0.19; s = 4.22.
(b) Median = 1.0.
(c) New mean = 1.44; new SD = 2.08; new median = (1.0 + 1.5)/2 = 1.25. The median displays
resistance. The mean and the SD change greatly, showing lack of resistance.
2.S.13 (a) Median = 28; Q1 = 22; Q3 = 33; IQR = 33 – 22 = 11.
(b) Upper fence = 33 + (1.5)(11) = 49.5; lower fence = 22 – (1.5)(11) = 5.5.
2.S.14 Yes, the distribution appears to be reasonably symmetric and mound-shaped.
โข 2.S.15 (a) n = 119, so the median is the 60th largest observation. There are 32 observations less than or
equal to 37 and 44 observations less than or equal to 38. Thus, the median is 38.
(b) The first quartile is the 30th largest observation, which is 36. The third quartile is the 90th largest
observation, which is 41.
(c)
(d) The mean is 38.45 and the SD is 3.20. Thus, the interval โy ยฑ s is 38.45 ยฑ 3.20, which is 35.25 to
41.65. This interval includes 36, 27, 28, 29, 40, and 41. The number of flies with 36 to 41 bristles
Copyright (c) 2016 Pearson Education, Inc.
Chapter 2 45
is 11 + 12 + 18 + 13 + 10 + 15 = 79. Thus, the percentage of observations that fall within one
standard deviation of the mean is 79/119*100% = 0.664*100% = 66.4%.
(e) The quartiles are 36 and 41. Between 36 and 41 (inclusive) there are 79 observations out of 119
total so 66.4%.
2.S.16 (a) Mean increase = 3.870; SD of increase = 1.274.
(b) Mean before = 1.680, mean after = 5.550. Yes; 5.550 – 1.680 = 3.870.
(c) The median increase is (3.5 + 3.5)/2 = 3.5.
(d) Before: The median increase is (1.5 + 1.5)/2 = 1.5.
After: The median increase is (5.2 + 5.8)/2 = 5.5.
No; 5.5 – 1.5 โ 3.5.
2.S.17 (a)
Number of
Parasites
0-9,999
10,000-19,999
20,000-29,999
30,000-39,999
40,000-49,999
50,000-59,999
60,000-69,999
70,000-79,999
80,000-89,999
90,000-99,999
100,000-109,999
110,000-119,999
120,000-129,999
130,000-139,999
Total
Frequency
23
4
2
0
0
0
0
0
1
0
0
0
0
1
31
Log (Number of
Parasites)
Frequency
(b)
Copyright (c) 2016 Pearson Education, Inc.
46 Solutions to Exercises
2.000-2.499
2.500-2.999
3.000-3.499
3.500-3.999
4.000-4.499
4.500-4.999
5.000-5.499
Total
4
5
6
8
6
1
1
31
The original distribution is highly skewed; the distribution of the logs is much less skewed.
(c) The original mean is 12,889.6; the mean of the logs is 3.4854. The log of the original mean is
log(12,889.6) = 4.1102 โ 3.4854. No, they are not equal.
(d) The original median is 3672; the median of the logs is 3.5649. The log of the original median is
log(3673) = 3.5649. Yes, they are equal.
2.S.18 The first histogram is a rough histogram for the data, because the data imply that the distribution is
skewed to the right. The average is 3.6 and the SD is 1.6. If the distribution were symmetric and
mound shaped, like the middle histogram, then there would be data throughout the interval โy ยฑ 2s,
which is 3.6 ยฑ 3.2, or 0.4 to 6.8. However, the minimum is 1.2, which is greater than 0.4. The
situation for the third histogram is even worse (i.e., for this histogram there are values much lower
than the mean).
2.S.19 The fourth computer output has the largest SD and has the mean and median nearly equal. This
corresponds to histogram (b), which has the most spread. Histogram (c) has a mean which exceeds
the mean. It follows that this histogram corresponds to the first computer output. Histogram (a)
has the smallest SD and the mean and median are nearly equal, so the second computer output
corresponds to histogram (a). The third computer output has a lower mean than median, so it is not
used.
2.S.20 The low volume distribution is symmetric, centered at 20, with a minimum of 0 and a maximum of
40. The high volume distribution is shifted down from the low volume distribution, with a median
of about 18 and a maximum of 30, which is the third quartile for the low volume distribution.
Thus, one-fourth of the low volume hospitals have mortality rates greater than the highest mortality
rate among high volume hospitals.
2.S.21 The mean is 107.87, the median is 108.5, and the SD is 16.08.
Copyright (c) 2016 Pearson Education, Inc.
Chapter 2 47
The histogram is fairly symmetric.
2.S.22 The median is 28, which is consistent with any of the histograms. Likewise the minimum and
maximum values agree for all three histograms. However, the boxplot shows that the distribution
has a small IQR and is skewed to the right, which means that histogram (a) is correct.
2.S.23 The mean is pulled down by the long left-hand tail, which results in the mean being less than the
median.
2.S.24 Approximately 15% of the area in the histogram is to the left of 40.
Copyright (c) 2016 Pearson Education, Inc.

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