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CHAPTER 18 PART 4
#1
A museum offers several levels of membership, as shown in the table.
Member Category
Individual
Family
Amount of Donation ($)
50
100
Percent of Members
41
37
Sponsor
Patron
Benefactor
250
500
1000
14
7
1
a) Find the mean (expected value) and standard deviation of the donations.
b) During their annual membership drive, they hope to sign up 50 new members each day.
Would you expect the distribution of the donations for a day to follow a Normal model?
Explain.
c) Consider the mean donation of the 50 new members each day. Describe the sampling
model for these means (shape, center, and spread).
a) 𝜇 = 50 .41 + 100 .37 + 250 .14 + 500 .07 + 1000 .01 = $137.50
𝜎=
50 − 137.5
2
.41 + 100 − 137.5
2
.37 +∙∙∙ + 1000 − 137.5 2 (.01) = $148.56
b) The distribution of donations is most likely skewed to the right because a few
people will donate $500 or $1000.
c) The sample of 50 is large enough and the other conditions are satisfied. The
sampling distribution will be Normal (symmetric). The mean is 137.5. The
148.56
standard deviation for the model is 𝜎𝑥 =
= 21.01. The sampling
50
distribution model is N(137.5, 21.01).
#2 One of the museum’s phone volunteers sets a personal goal of
getting an average donation of at least $100 from the new members
she enrolls during the membership drive. If she gets 80 new members
and they can be considered a random sample of all the museum’s
members, what is the probability that she can achieve her goal?
100 − 137.5
𝑧=
= −2.26
16.61
𝜇𝑥 = 137.5
𝜎𝑥 =
148.56
80
= 16.61
𝑃 𝑧 > −2.26 = 1 − 𝑃 𝑧 < −2.26 = 1 − 0.0119 = 0.9881
According to the sampling distribution model, there is a
98.81% probability that the average donation for 80
new members is at least $100.
#3 Carbon monoxide (CO) emissions for a certain kind of car
vary with mean 2.9 g/mi and standard deviation of 0.4 g/mi. A
company has 80 of these cars in its fleet. Let 𝑥 represent the
mean CO level for the company’s fleet.
a)What’s the approximate model for the distribution of 𝑥?
b)Estimate the probability that 𝑥 is between 3.0 and 3.1 g/mi.
c)There is only a 5% chance that the fleet’s mean CO level is
greater than what value?
a) The conditions are met to use a Normal model. The mean for the model is 2.9.
0.4
The standard deviation is
= 0.045.
80
b) Find the z-score for 3.0 and for 3.1:
3.1−2.9
3.0−2.9
𝑧1 =
= 4.44
𝑧2 =
= 2.22
0.045
0.045
𝑃 𝑧 < 4.44 = 0.9999
𝑃 𝑧 < 2.22
= 0.9868
𝑃 2.22 < 𝑧 < 4.44 = 0.9999 − 0.9868 = 0.0131
c) The upper 5% can’t be found on the z table, so use lower 95%. This
corresponds to a z-score of 1.645.
𝑥−2.9
1.645 =
→ 0.074 = 𝑥 − 2.9 → 𝑥 = 2.97
0.045
According to the normal model, there is only a 5% chance that the fleet’s mean
CO level is greater than approximately 2.97g/mi.