*This preview shows pages 1–4. Sign up to view the full content.*

MATH 115,
PROBABILITY & STATISTICS
N. PSOMAS
Ch. 5.1 - HOMEWORK (SOLUTIONS)
Section 5.1
5.9
Generating a sampling 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
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 players in the population. This is the population mean
μ
.
(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). Write down the three times in your sample and
calculate the sample mean
. This statistic is an estimate of
μ
.
(c)
Repeat this process 10 times using different parts of
Table B
. Make a histogram of the 10
values of
. You are constructing the sampling distribution of
.
(d)
Is the center of your histogram close to
μ
? Would it get closer to
μ
the more times you
repeated this sampling process? Explain.
Solution
(a) Mean time for
ten players =338.8

This ** preview** has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*(b) , (c), (d)
Average of 10 sample means =338.0
5.10
Total sleep time of college 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
deviation
σ
= 1.15 hours. Suppose you plan to take an SRS of size
n
= 200 and compute the
average total sleep time.
(a)
What is the standard deviation for the average time?
(b)
Use the 95 part of the 68–95–99.7 rule to describe the variability of this sample mean.
(c)
What is the probability that your average will be below 6.9 hours?
Solution
(a)
1.15/sqrt(200) = 0.08132
(b) 95% of sample means fall in the interval
μ±2σ/sqrt(200) or 7.02 ± 2*(
0.081332) = (6.8657, 7.1826)
(c) P[x-bar < 6.9] = 0.07002 (why?)
Sample 1
Sample 2
Sample 3
Sample 4
Sample 5
Sample 6
Sample 7
Sample 8
Sample 9
Sample 10
370
319
319
358
290
319
327
323
309
370
366
358
327
358
370
368
327
309
309
366
309
370
327
358
323
319
358
290
358
366
348.33
349.00
324.33
358.00
327.67
335.33
337.33
307.33
325.33
367.33

5.11
Determining sample size.
Recall the previous exercise. Suppose you want to use a sample

This ** preview** has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This is the end of the preview. Sign up to
access the rest of the document.