Math 115 Probability & Statistics
Section 1.3 HW
Section 1.3 Exercises (Solutions)
1.109, 1.110, 1.111, 1.114*, 1.115, 1.119*, 1.122, 1.125, 1.127*, 1.128*, 1.131*,
1.133*, 1.135*, 1.137*, 1.139*, 1.145*, 1.146 - 148.
1.109 Sketch some normal curves.
(a) Sketch a normal curve that has mean 10 and standard deviation 3.
(b) On the same x axis, sketch a normal curve that has mean 20 and standard deviation 3.
(c) How does the normal curve change when the mean is varied but the standard deviation
stays the same.
Solution
(a) It is easiest to draw the curve first, and then mark the scale on the axis. (b) Draw a
copy of the first curve, with the peak over 20. (c) The curve has the same shape, but is
translated left or right.
The curve has the same shape, but is translated left or right.
1.110 The effect of changing the standard deviation.
(a) Sketch a normal curve that has mean 10 and standard deviation 3.
(b) On the same x axis, sketch a normal curve that has mean 10 and standard deviation 1.
(c) How does the normal curve change when the standard deviation is varied but the mean
stays the same.
Solution
(a) As in the previous exercise, draw the curve first, and then mark the scale on the axis.
(b) In order to have a standard deviation of 1, the curve should be 1/3 as wide, and three
times taller. (c) The curve is centered at the same place (the mean), but its height and
width change. Specifically, increasing the standard deviation makes the curve wider and
shorter; decreasing the standard deviation makes the curve narrower and taller.
Math 115 Probability & Statistics
Section 1.3 HW
1.111 Know your density. Sketch density curves that might describe distributions with the
following shapes:
(a) Symmetric, but with two peaks (that is, two strong clusters of observations).
(b) Single peak and skewed to the left.
Solution
1.111. Sketches will vary.
1.114 Total scores. Here are the total scores of 10 students in an introductory statistics course.
Previous experience with this course suggests that these scores should come from a distribution
that is approximately normal with mean 70 and standard deviation 10.
(a) Using these values for µ and σ, standardize the first exam scores of these 10 students.
(b) If the grading policy is to give grades of A to the top 15% of scores based on the normal
distribution with mean 70 and standard deviation 10, what is the cut-off for an A in terms
of a standardized score?
(c) Which students earned a grade of A on the final exam for this course?
Math 115 Probability & Statistics
Section 1.3 HW
Solution
1.114. (a) For example, (68−70)/10 = −0.2. The complete list is given below.
(b) The cut-off for an A is the 85th percentile for the N(0, 1) distribution. From Table A,
this is approximately 1.04; software gives 1.0364. (c) The top two students (with scores
of 92 and 98) received A’s.
68
-0.2
54
-1.6
92
2.2
75
0.5
73
0.3
98
2.8
64
-0.6
55
-1.5
80
1
70
0
1.115 Assign more grades. Refer to the previous exercise. The grading policy says the cut-offs
for the other grades correspond to the following: bottom 5% receive F, next 10% receive D, next
40% receive C, and next 30% receive B. These cut-offs are based on the N(70, 10) distribution.
(a) Give the cut-offs for the grades in this course in terms of standardized scores.
(b) Give the cut-offs in terms of actual total scores.
(c) Do you think that this method of assigning grades is a good one? Give reasons for your
answer.
Solution
1.115. (a) We need the 5th, 15th, 55th, and 85th percentiles for a N(0, 1) distribution.
These are given in the table below. (b) To convert to actual scores, take the standardscore cut-off z and compute 10z + 70. (c) Opinions will vary.
Note: The cut-off for an A given in the previous solution is the lowest score that gets an
A—that is, the point where one’s grade drops from an A to a B. These cut-offs are the
points where one’s grade jumps up. In practice, this is only an issue for a score that falls
exactly on the border between two grades.
1.116 A uniform distribution. If you ask a computer to generate “random numbers” between 0
and 1, you will get observations from a uniform distribution. Figure 1.34 graphs the density
curve for a uniform distribution. Use areas under this density curve to answer the following
questions.
(a) Why is the total area under this curve equal to 1?
(b) What proportion of the observations lie below 0.35?
Math 115 Probability & Statistics
Section 1.3 HW
(c) What proportion of the observations lie between 0.35 and 0.65?
Figure 1.34 The density curve of a uniform distribution, for Exercise 1.116.
1.118 Find the mean, the median, and the quartiles. What are the mean and the median of the
uniform distribution in Figure 1.34? What are the quartiles?
Solutions (1.116 - 1.118)
1.119 Three density curves. Figure 1.35 displays three density curves, each with three points
marked on it. At which of these points on each curve do the mean and the median fall?
Math 115 Probability & Statistics
Section 1.3 HW
Figure 1.35 Three density curves, for Exercise 1.119.
Solution
1.119. (a) Mean is C, median is B (the right skew pulls the mean to the right). (b) Mean
A, median A. (c) Mean A, median B (the left skew pulls the mean to the left).
1.120 Length of pregnancies. The length of human pregnancies from conception to birth varies
according to a distribution that is approximately Normal with mean 266 days and standard
deviation 16 days. Draw a density curve for this distribution on which the mean and standard
deviation are correctly located.
Solution
Math 115 Probability & Statistics
Section 1.3 HW
1.122 Pregnancies and the 68–95–99.7 rule. The length of human pregnancies from conception
to birth varies according to a distribution that is approximately Normal with mean 266 days and
standard deviation 16 days. Use the 68–95–99.7 rule to answer the following questions.
(a) Between what values do the lengths of the middle 95% of all pregnancies fall?
(b) How short are the shortest 2.5% of all pregnancies? How long do the longest 2.5% last?
Solution
1.122. See the sketch of the curve in the solution to Exercise 1.120. (a) The middle 95%
fall within two standard deviations of the mean: 266 ± 2(16), or 234 to 298 days. (b) The
shortest 2.5% of pregnancies are shorter than 234 days (more than two standard
deviations below the mean).
1.123 Horse pregnancies are longer. Bigger animals tend to carry their young longer before
birth. The length of horse pregnancies from conception to birth varies according to a roughly
Normal distribution with mean 336 days and standard deviation 3 days. Use the 68–95–99.7 rule
to answer the following questions.
(a) Almost all (99.7%) horse pregnancies fall in what range of lengths?
(b) What percent of horse pregnancies are longer than 339 days?
Solution
1.123. (a) 99.7% of horse pregnancies fall within three standard deviations of the mean:
336 ± 3(3), or 327 to 325 days. (b) About 16% are longer than 339 days since 339
days or more corresponds to at least one standard deviation above the mean.
1.125 Acidity of rainwater. The Normal quantile plot in Figure 1.32 (page 66) shows that the
acidity (pH) measurements for rainwater samples in Exercise 1.36 are approximately Normal.
How well do these scores satisfy the 68–95–99.7 rule? To find out, calculate the mean and
standard deviation s of the observations. Then calculate the percent of the 105 measurements that
fall between − s and + s and compare your result with 68%. Do the same for the intervals
covering two and three standard deviations on either side of the mean. (The 68–95–99.7 rule is
exact for any theoretical Normal distribution. It will hold only approximately for actual data.)
Solution
1.125. The mean and standard deviation are x = 5.4256 and s = 0.5379. About 67.62%
(71/105 = 0.6476) of the pH measurements are in the range x ± s = 4.89 to 5.96. About
95.24% (100/105) are in the range x ± 2s = 4.35 to 6.50. All (100%) are in the range
x ± 3s = 3.81 to 7.04.
Math 115 Probability & Statistics
Section 1.3 HW
1.126 Find some proportions. Using either Table A or your calculator or software, find the
proportion of observations from a standard Normal distribution that satisfies each of the
following statements. In each case, sketch a standard Normal curve and shade the area under the
curve that is the answer to the question.
(a) Z > 1.65
(b) Z < 1.65
(c) Z > −0.76
(d) −0.76 < Z < 1.65
Solution
1.126. Using values from Table A: (a) Z > 1.65: 0.0495. (b) Z < 1.65: 0.9505.
(c) Z > −0.76: 0.7764. (d) −0.76 < Z <1.65: 0.9505 − 0.2236 = 0.7269.
1.127 Find more proportions. Using either Table A or your calculator or software, find the
proportion of observations from a standard Normal distribution for each of the following events.
In each case, sketch a standard Normal curve and shade the area representing the proportion.
(a) Z ≤ −1.8
(b) Z ≥ −1.8
(c) Z > 1.6
(d) −1.8 < Z < 1.6
Solution
1.127. Using values from Table A: (a) Z ≤ −1.8: 0.0359. (b) Z ≥ −1.8: 0.9641.
(c) Z > 1.6: 0.0548. (d) −1.8 < Z < 1.6: 0.9452 − 0.0359 = 0.9093.
Math 115 Probability & Statistics
Section 1.3 HW
1.128 Find some values of z. Find the value z of a standard Normal variable Z that satisfies each
of the following conditions. (If you use Table A, report the value of z that comes closest to
satisfying the condition.) In each case, sketch a standard Normal curve with your value of z
marked on the axis.
(a) 22% of the observations fall below z.
(b) 40% of the observations fall above z.
Solution
1.128. (a) 22% of the observations fall below −0.7722. (This is the 22nd percentile of the
standard Normal distribution.) (b) 40% of the observations fall above 0.2533 (the 60th
percentile of the standard Normal distribution).
Math 115 Probability & Statistics
Section 1.3 HW
1.129 Find more values of z. The variable Z has a standard Normal distribution.
(a) Find the number z that has cumulative proportion 0.65.
(b) Find the number z such that the event Z > z has proportion 0.45.
Solution
1.129. (a) z = 0.3853 has cumulative proportion 0.65 (that is, 0.3853 is the 65th
percentile of the standard Normal distribution). (b) If z = 0.1257, then Z > z has
proportion 0.45 (0.1257 is the 55th percentile).
1.130 Find some values of z. The Wechsler Adult Intelligence Scale (WAIS) is the most
common IQ test. The scale of scores is set separately for each age group and is approximately
Normal with mean 100 and standard deviation 15. People with WAIS scores below 70 are
considered mentally retarded when, for example, applying for Social Security disability benefits.
What percent of adults are retarded by this criterion?
Solution
1.130. 70 is two standard deviations below the mean (that is, it has standard score z =
−2), so about 2.5% (half of the outer 5%) of adults would have WAIS scores below 70.
1.131 High IQ scores. The Wechsler Adult Intelligence Scale (WAIS) is the most common IQ
test. The scale of scores is set separately for each age group and is approximately Normal with
mean 100 and standard deviation 15. The organization MENSA, which calls itself “the high IQ
society,” requires a WAIS score of 130 or higher for membership. What percent of adults would
qualify for membership?
There are two major tests of readiness for college, the ACT and the SAT. ACT scores are
reported on a scale from 1 to 36. The distribution of ACT scores are approximately Normal with
mean µ = 21.5 and standard deviation σ = 5.4. SAT scores are reported on a scale from 600 to
2400. The SAT scores are approximately Normal with mean µ = 1509 and standard deviation σ
= 321. Exercises 1.132 to 1.141 are based on this information.
Math 115 Probability & Statistics
Section 1.3 HW
Solution
1.131. 130 is two standard deviations above the mean (that is, it has standard
score z = 2), so about 2.5% of adults would score at least 130.
1.132 Compare an SAT score with an ACT score. Tonya scores 1820 on the SAT. Jermaine
scores 29 on the ACT. Assuming that both tests measure the same thing, who has the higher
score? Report the z-scores for both students.
Solution
1.132. Tonya’s score standardizes to z = (1820−1509)/321 = 0.9688, while Jermaine’s
score corresponds to z = (29−21.5)/5.4 = 1.3889. Jermaine’s score is higher.
1.133 Make another comparison. Jacob scores 16 on the ACT. Emily scores 1020 on the SAT.
Assuming that both tests measure the same thing, who has the higher score? Report the z-scores
for both students.
Solution
1.133. Jacob’s score standardizes to z = (16−21.5)/5.4 = −1.0185, while Emily’s score
corresponds to z = (1020−1509)/321 = −1.5234. Jacob’s score is higher.
1.134 Find the ACT equivalent. Jose scores 2080 on the SAT. Assuming that both tests
measure the same thing, what score on the ACT is equivalent to Jose’s SAT score?
Solution
1.134. Jose’s score standardizes to z = (2080−1509)/321 = 1.7788, so an equivalent
ACT score is 21.5 + 1.7788 × 5.4 = 31.1. (Of course, ACT scores are reported as whole
numbers, so this would presumably be a score of 31.)
1.135 Find the SAT equivalent. Maria scores 30 on the ACT. Assuming that both tests measure
the same thing, what score on the SAT is equivalent to Maria’s ACT score?
Solution
1.135. Maria’s score standardizes to z = (30−21.5)/5.4 =1.5741, so an equivalent SAT
score is 1509 + 1.5741 × 321 = 2014.
Math 115 Probability & Statistics
Section 1.3 HW
1.136 Find an SAT percentile. Reports on a student’s ACT or SAT usually give the percentile
as well as the actual score. The percentile is just the cumulative proportion stated as a percent:
the percent of all scores that were lower than this one. Maria scores 2090 on the SAT. What is
her percentile?
Solution
1.136. Maria’s score standardizes to z = (2090−1509)/321 = 1.81, for which Table A
gives 0.9649. Her score is the 96.5 percentile.
1.137 Find an ACT percentile. Reports on a student’s ACT or SAT usually give the percentile
as well as the actual score. The percentile is just the cumulative proportion stated as a percent:
the percent of all scores that were lower than this one. Jacob scores 19 on the ACT. What is his
percentile?
Solution
1.137. Jacob’s score standardizes to z = (19−21.5)/5.4 = −0.4630, for which Table A
gives 0.3228. His score is the 32.3 percentile.
1.138 How high is the top 10%? What SAT scores make up the top 10% of all scores?
Solution
1.138. 1920 and above: The top 10% corresponds to a standard score of z = 1.2816,
which in turn corresponds to a score of 1509 + 1.2816 × 321 = 1920 on the SAT.
.
1.139 How low is the bottom 20%? What SAT scores make up the bottom 20% of all scores?
Solution
1.139. 1239 and below: The bottom 20% corresponds to a standard score of
z = −0.8416, which in turn corresponds to a score of 1509 − 0.8416 × 321 =1239 on the
SAT.
1.140 Find the ACT quartiles. The quartiles of any distribution are the values with cumulative
proportions 0.25 and 0.75. What are the quartiles of the distribution of ACT scores?
Solution
1.140. The quartiles of a Normal distribution are ±0.6745 standard deviations from the
mean, so for ACT scores, they are 21.5 ± 0.6745 × 5.4 = 17.9 to 25.1.
Math 115 Probability & Statistics
Section 1.3 HW
1.141 Find the SAT quintiles. The quintiles of any distribution are the values with cumulative
proportions 0.20, 0.40, 0.60, and 0.80. What are the quintiles of the distribution of SAT scores?
Solution
1.141. The quintiles of the SAT score distribution are 1509 − 0.8416 × 321 = 1239,
1509−0.2533×321 = 1428, 1509+0.2533×321 = 1590, and 1509+0.8416×321 = 1779.
1.142 Do you have enough “good cholesterol?” High-density lipoprotein (HDL) is sometimes
called the “good cholesterol” because low values are associated with a higher risk of heart
disease. According to the American Heart Association, people over the age of 20 years should
have at least 40 mg/dL of HDL cholesterol.36 U.S. women aged 20 and over have a mean HDL
of 55 mg/dL with a standard deviation of 15.5 mg/dL. Assume that the distribution is Normal.
(a) What percent of women have low values of HDL (40 mg/dL or less)?
(b) HDL levels of 60 mg/dL are believed to protect people from heart disease. What percent
of women have protective levels of HDL?
(c) Women with more than 40 mg/dL but less than 60 mg/dL of HDL are in the intermediate
range, neither very good or very bad. What proportion are in this category?
Solution
1.142. For a Normal distribution with mean 55 mg/dl and standard deviation 15.5 mg/dl:
(a) 40 mg/dl standardizes to z = (40−55)/15.5 = −0.9677. Using Table A, 16.60% of
women fall below this level (software: 16.66%).
(b) 60 mg/dl standardizes to z = (60−55)/15.5 = 0.3226. Using Table A, 37.45
(c) Subtract the answers from (a) and (b) from 100%: Table A gives 45.95% (software:
45.99%), so about 46% of women fall in the intermediate range.
1.143 Men and HDL cholesterol. HDL cholesterol levels for men have a mean of 46 mg/dL
with a standard deviation of 13.6. Answer the questions given in the previous exercise for the
population of men.
Solution
1.143. For a Normal distribution with mean 46 mg/dl and standard deviation 13.6 mg/dl:
(a) 40 mg/dl standardizes to z = (40−46)/13.6 = −0.4412. Using Table A, 33% of men fall
below this level (software: 32.95%).
(b) 60 mg/dl standardizes to z = (60−46)/13.6 = 1.0294. Using Table A, 15.15
(c) Subtract the answers from (a) and (b) from 100%: Table A gives 51.85% (software:
51.88%), so about 52% of men fall in the intermediate range.
Math 115 Probability & Statistics
Section 1.3 HW
1.145 Length of pregnancies. The length of human pregnancies from conception to birth varies
according to a distribution that is approximately Normal with mean 266 days and standard
deviation 16 days.
(a) What percent of pregnancies last less than 240 days (that’s about 8 months)?
(b) What percent of pregnancies last between 240 and 270 days (roughly between 8 months
and 9 months)?
(c) How long do the longest 20% of pregnancies last?
Solution
1.145. (a) About 5.2%: x < 240 corresponds to z < −1.625. Table A gives 5.16% for
−1.63 and 5.26% for −1.62. Software (or averaging the two table values) gives 5.21%.
(b) About 54.7%: 240 < x < 270 corresponds to −1.625 < z < 0.25. The area to the
left of 0.25 is 0.5987; subtracting the answer from part (a) leaves about 54.7%.
(c) About 279 days or longer: Searching Table A for 0.80 leads to z > 0.84, which
corresponds to x > 266 + 0.84(16) = 279.44. (Using the software value z > 0.8416 gives
x > 279.47.)
1.146
Quartiles for Normal distributions. The quartiles of any distribution are the values
with cumulative proportions 0.25 and 0.75.
(a) What are the quartiles of the standard Normal distribution?
(b) Using your numerical values from (a), write an equation that gives the quartiles of the
N(µ, σ) distribution in terms of µ and σ.
(c) The length of human pregnancies from conception to birth varies according to a
distribution that is approximately Normal with mean 266 days and standard deviation 16
days. Apply your result from (b): what are the quartiles of the distribution of lengths of
human pregnancies?
Solution
1.146. (a) The quartiles for a standard Normal distribution are ±0.6745.
(b) For a N(µ, σ) distribution, Q1 = µ − 0.6745σ and Q3 = µ + 0.6745σ.
(c) For human pregnancies, Q1 = 266 − 0.6745 × 16 .= 255.2 and Q3 = 266 + 0.67455 ×
16 = 276.8 days.
Math 115 Probability & Statistics
Section 1.3 HW
1.147
IQR for Normal distributions. Continue your work from the previous exercise. The
interquartile range IQR is the distance between the first and third quartiles of a distribution.
(a) What is the value of the IQR for the standard Normal distribution?
(b) There is a constant c such that IQR = cσ for any Normal distribution N(µ, σ). What is the
value of c?
Solution
1.147. (a) As the quartiles for a standard Normal distribution are ±0.6745,
we have IQR = 1.3490.
(b) c = 1.3490: For a N(µ, σ) distribution, the quartiles are Q1 = µ − 0.6745σ and
Q3 = µ + 0.6745σ.
1.148
Outliers for Normal distributions. Continue your work from the previous two
exercises. The percent of the observations that are suspected outliers according to the 1.5 × IQR
rule is the same for any Normal distribution. What is this percent?
Solution
1.148. In the previous two exercises, we found that for a N(µ, σ) distribution,
Q1 = µ − 0.6745σ, Q3 = µ + 0.6745σ, and IQR = 1.3490σ.
Therefore, 1.5 × IQR = 2.0235σ, and the suspected outliers are below
Q1 − 1.5 × IQR = µ − 2.698σ, and above Q3 + 1.5 × IQR = µ + 2.698σ.
The percentage outside of this range is 2 × 0.0035 = 0.70%.
1.149 Deciles of Normal distributions. The deciles of any distribution are the 10th, 20th,…,
90th percentiles. The first and last deciles are the 10th and 90th percentiles, respectively.
(a) What are the first and last deciles of the standard Normal distribution?
(b) The weights of 9-ounce potato chip bags are approximately Normal with mean 9.12
ounces and standard deviation 0.15 ounce. What are the first and last deciles of this
distribution?
Solution
1.149. (a) The first and last deciles for a standard Normal distribution are ±1.2816.
(b) For a N(9.12, 0.15) distribution, the first and last deciles are µ − 1.2816σ = 8.93 and
µ + 1.2816σ = 9.31 ounces.