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MATH 115,
PROBABILITY & STATISTICS
Ch. 5.1 - HOMEWORK (SOLUTIONS)
Section 5.1
5.9 Generating a sampling distribution.the case of a very small
sample from a very small population. The population is the 10
scholarship players currently on your men’s basketball team. For
convenience, the 10 players have been labeled with the integers 0
to 9. For each player, the total amount of time spent (in minutes)
on Facebook during the last month is recorded in the table
below.
The parameter of interest is the average amount of time on
Facebook. The sample is an SRS of size n = 3 drawn from this
population of players. Because the players are labeled 0 to 9, a
single random digit from Table B chooses one player for the
sample.
(a) Find the mean of the 10 p(b) Use Table B to draw an SRS of
size 3 from this population (Note: you may sample the
same player’s time more than once).calculate the sample mean
(c) Repeat this process 10 times using different parts of values
of . You are constructing the sampling distribution of
(d) Is the center of your histogram close to repeated this
sampling process? Explain.
Solution
(a) Mean time for ten players =338.8
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distribution. Let’s illustrate the idea of a sampling
distribution in the case of a very small sample from a very small
population. The population is the 10 scholarship players currently
on your men’s basketball team. For convenience, the 10 players
ith the integers 0 to 9. For each player, the total amount of
time spent (in minutes) on Facebook during the last month is
recorded in the table below.
The parameter of interest is the average amount of time on
Facebook. The sample is an SRS of 3 drawn from this population of
players. Because the players are labeled 0 to 9, a single
chooses one player for the sample.
Find the mean of the 10 players in the population. This is the
population mean to draw an SRS of size 3 from this population
(Note: you may sample the
same player’s time more than once). Write down the three times
in your sample and calculate the sample mean . This statistic is an
estimate of µ. Repeat this process 10 times using different parts
of Table B. Make a histogram of the 10
. You are constructing the sampling distribution of . Is the
center of your histogram close to µ? Would it get closer to µ the
more times you repeated this sampling process? Explain.
8.8
N. PSOMAS
Let’s illustrate the idea of a sampling distribution in the case
of a very small sample from a very small population. The population
is the 10 scholarship players currently on your men’s basketball
team. For convenience, the 10 players
ith the integers 0 to 9. For each player, the total amount of
time spent (in
The parameter of interest is the average amount of time on
Facebook. The sample is an SRS of 3 drawn from this population of
players. Because the players are labeled 0 to 9, a single
layers in the population. This is the population mean µ. to draw
an SRS of size 3 from this population (Note: you may sample the
Write down the three times in your sample and
ake a histogram of the 10
the more times you
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(b) , (c), (d)
Average of 10 sample means =338.0
5.10 Total sleep time of college college students was
approximately Normally distributed with mean deviation σ = 1.15
hours. Suppose you plan to take an SRS of size average total sleep
time.
(a) What is the standard deviation for the average time?(b) Use
the 95 part of the 68–(c) What is the probability that your average
will be below 6.9 hours?
Solution
(a) 1.15/sqrt(200) = 0.078842
(b) 95% of sample means fall in the interval
µ±2σ/sqrt(200) or 7.02 ± 2*(0.078842) = (6.8623, 7.1777)
(c) P[x-bar < 6.9] = 0.064002 (why?)
Sample 1 Sample 2 Sample 3 Sample 4
370 319 319
366 358 327
309 370 327
348.33 349.00 324.33
338.0
students. In Example 5.1, the total sleep time per night among
college students was approximately Normally distributed with mean µ
= 7.02 hours and standard
= 1.15 hours. Suppose you plan to take an SRS of size n = 200
and compute the
the standard deviation for the average time? –95–99.7 rule to
describe the variability of this sample mean.
What is the probability that your average will be below 6.9
hours?
ple means fall in the interval
0.078842) = (6.8623, 7.1777)
(why?)
Sample 4 Sample 5 Sample 6 Sample 7 Sample 8 Sample 9
358 290 319 327 323
358 370 368 327 309
358 323 319 358 290
358.00 327.67 335.33 337.33 307.33
, the total sleep time per night among = 7.02 hours and
standard
= 200 and compute the
99.7 rule to describe the variability of this sample mean.
Sample 9 Sample 10
309 370
309 366
358 366
325.33 367.33
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5.11 Determining sample size. Recall the previous exercise.
Suppose you want to use a sample size such that about 95% of the
averages fall within ±5 minutes of the true mean µ = 7.02.
(a) Based on your answer to part (b) in Exercise 5.8, should the
sample size be larger or smaller than 200? Explain.
(b) What standard deviation of the average do you need such that
about 95% of all samples will have a mean within 5 minutes of
µ?
(c) Using the standard deviation calculated in part (b),
determine the number of students you need to sample.
Solution
(a) Larger
(b) 5' = (5/60) hr = 0.083333 hrs ==> sample mean SD should
be 1/2 of 0.083333
(c) 2*(1.15)/sqrt(n) = 0.083333 == > sqrt(n) = 2*(1.15)/
0.083333 ==> n = (27.6)^2 = 761.8 or 762
5.12 Songs on an iPod. An iPod has about 10,000 songs. The
distribution of the play time for these songs is highly skewed.
Assume that the standard deviation for the population is 280
seconds.
(a) What is the standard deviation of the average time when you
take an SRS of 10 songs from this population?
(b) How many songs would you need to sample if you wanted the
standard deviation of to be 15 seconds?
Solution
(a) 280/sqrt(10) = 88.544 sec
(b) 280/sqrt(n) = 15 sec ==> sqrt(n) = 280/15 = 18.66667
==> n = (18.7)^2 = 348.4444 or 349 songs
5.13 Bottling an energy drink. A bottling company uses a filling
machine to fill cans with an energy drink. The cans are supposed to
contain 250 milliliters (ml). The machine, however, has some
variability, so the standard deviation of the size is σ = 3 ml. A
sample of 6 cans is inspected each hour for process control
purposes, and records are kept of the sample mean volume. If the
process mean is exactly equal to the target value, what will be the
mean and standard deviation of the numbers recorded?
Solution
mean of x-bar = 250 ml & SD of x-bar = 3/sqrt(6) =
1.224745
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5.14 Play times for songs on anindividual measurements. Suppose
the true mean duration of the play time for the songs in the iPod
of Exercise 5.12 is 350 seconds.
(a) Sketch on the same graph the two Normal curves, for sampling
a single song and for the mean of 10 songs.
(b) What is the probability that the sample mean differs from
the population mean by more than 19 seconds when only 1 song is
sampled?
(c) How does the probability tha10 songs?
Solution
(a)
(b) Find P[ x < 350 - 19 OR x > 350 + 19]
= P[ x < 350 - 19] + P[ x > 350 + 19]
(c) Find P[ < 350 - 19 OR > 350 + 19]
= P[ < 350 - 19] + P[ > 350 + 19]
an iPod. Averages of several measurements are less variable than
individual measurements. Suppose the true mean duration of the play
time for the songs in the
is 350 seconds.
n the same graph the two Normal curves, for sampling a single
song and for the
What is the probability that the sample mean differs from the
population mean by more than 19 seconds when only 1 song is
sampled? How does the probability that you calculated in part (b)
change for the mean of an SRS of
19 OR x > 350 + 19] ; Here X=N(350, 280)
19] + P[ x > 350 + 19] = ..... = 0.94586 or 94.6%
> 350 + 19] ; Here X-bar = N(350, 280/sqt(10))
> 350 + 19] = ..... =0 .8301 or 83.0%
Averages of several measurements are less variable than
individual measurements. Suppose the true mean duration of the play
time for the songs in the
n the same graph the two Normal curves, for sampling a single
song and for the
What is the probability that the sample mean differs from the
population mean by more
t you calculated in part (b) change for the mean of an SRS
of
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5.15 Can volumes. Averages are less variable than individual
observations. Suppose that the can volumes in Exercise 5.13 vary
according to a Normal distribution. In that case, the mean of an
SRS of cans also has a Normal distribution.
(a) Make a sketch of the Normal curve for a single can. Add the
Normal curve for the mean of an SRS of 6 cans on the same
sketch.
(b) What is the probability that the volume of a single randomly
chosen can differs from the target value by 1 ml or more?
(c) What is the probability that the mean volume of an SRS of 6
cans differs from the target value by 1 ml or more?
Solutions
Work like in the previous exercise (5.14)
5.16 Number of friends on Facebook. Facebook provides a variety
of statistics on their Web site that detail the growth and
popularity of the site. One such statistic is that the average user
has 130 friends. This distribution only takes integer values, so it
is certainly not Normal. We’ll also assume it is skewed to the
right with a standard deviation σ = 85. Consider an SRS of 30
Facebook users.
(a) What are the mean and standard deviation of the total number
of friends in this sample? (b) What are the mean and standard
deviation of the mean number of friends per user? (c) Use the
central limit theorem to find the probability that the average
number of friends in
30 Facebook users is greater than 140.
Solutions
Use the results from the Central Limit theorem and work like in
the previous exercises
5.18 ACT scores of high school seniors. The scores of high
school seniors on the ACT college entrance examination in a recent
year had mean µ = 19.2 and standard deviation σ = 5.1. The
distribution of scores is only roughly Normal.
(a) What is the approximate probability that a single student
randomly chosen from all those taking the test scores 23 or
higher?
(b) Now take an SRS of 25 students who took the test. What are
the mean and standard deviation of the sample mean score of these
25 students?
(c) What is the approximate probability that the mean score of
these students is 23 or higher?
(d) Which of your two Normal probability calculations in parts
(a) and (c) is more accurate? Why?
Solution
Use the results from the Central Limit theorem and work like in
the previous exercises
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5.20 Grades in an English course.
for its courses online. Students in one section of English 210
in the Fall 2008 semester received
33% A’s, 24% B’s, 18% C’s, 16% D’s, and 9% F’s.
(a) Using the common scale A = 4, B = 3, C = 2, D = 1, F = 0,
take randomly chosen English 210 student. Use the definitions of
the mean (standard deviation (page 269standard deviation σ of
grades in this course.
(b) English 210 is a large course. We can take the grades of an
SRS of 50 students to be independent of each other. If standard
deviation of ?
(c) What is the probability P(better? What is the approximate
probability randomly chosen English 210 students is a B or
better?
Solution
(a)
Letter
Grade
Point
equivalent
(X)
A 4
B 3
C 2
D 1
F 0
µX = Σ (x) * p(x) = 2.56
σ2
X = Σ(x - µ)2 * p(x) = 1.7664
σX = sqrt(11.568) = 1.32906
(c)
P(X ≥ 3) = 0.24 + 0.33 = 0.57 (Why?)
= normalcdf(3, ∞, 2.56, 1.32906/sqrt(50)
course. North Carolina State University posts the grade
distributions
Students in one section of English 210 in the Fall 2008 semester
received
33% A’s, 24% B’s, 18% C’s, 16% D’s, and 9% F’s.
Using the common scale A = 4, B = 3, C = 2, D = 1, F = 0, take X
to be the grade of a randomly chosen English 210 student. Use the
definitions of the mean (page 261
page 269) for discrete random variables to find the mean of
grades in this course.
English 210 is a large course. We can take the grades of an SRS
of 50 students to be ch other. If is the average of these 50
grades, what are the mean and
(X ≥ 3) that a randomly chosen English 210 student gets a B
or
better? What is the approximate probability that the grade point
averagrandomly chosen English 210 students is a B or better?
(b)
Probability
P(x)
0.33
0.24
0.18
0.16
0.09
= sample mean in a sample of 50 students
��̅ = 2.56
��̅ = 3.401176/√50 = 0.480999
1.7664
1.32906
57 (Why?)
1.32906/sqrt(50)) = .0096
North Carolina State University posts the grade
distributions
Students in one section of English 210 in the Fall 2008 semester
received
to be the grade of a page 261) and
) for discrete random variables to find the mean µ and the
English 210 is a large course. We can take the grades of an SRS
of 50 students to be is the average of these 50 grades, what are
the mean and
≥ 3) that a randomly chosen English 210 student gets a B or that
the grade point average for 50
= sample mean in a sample of 50 students
480999
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5.22 A lottery payoff. A $1 bet in a state lottery’s Pick 3 game
pays $500 if the three-digit number you choose exactly matches the
winning number, which is drawn at random. Here is the distribution
of the payoff X:
Each day’s drawing is independent of other drawings.
(a) What are the mean and standard deviation of X? (b) Joe buys
a Pick 3 ticket twice a week. What does the law of large numbers
say about the
average payoff Joe receives from his bets? (c) What does the
central limit theorem say about the distribution of Joe’s average
payoff
after 104 bets in a year? (d) Joe comes out ahead for the year
if his average payoff is greater than $1 (the amount he
spent each day on a ticket). What is the probability that Joe
ends the year ahead?
5.25 Weights of airline passengers. In response to the
increasing weight of airline passengers, the Federal Aviation
Administration told airlines to assume that passengers average 190
pounds in the summer, including clothing and carry-on baggage. But
passengers vary: the FAA gave a mean but not a standard deviation.
A reasonable standard deviation is 35 pounds. Weights are not
Normally distributed, especially when the population includes both
men and women, but they are not very non-Normal. A commuter plane
carries 25 passengers. What is the approximate probability that the
total weight of the passengers exceeds 5200 pounds? (Hint: To apply
the central limit theorem, restate the problem in terms of the mean
weight.)