More Examples Chapter 18 PART 4

#1 A museum offers several levels of membership, as shown in the table. Member Category Amount of Donation ($) Percent of Members Individual 50 41 Family 100 37 Sponsor 250 14 Patron 500 7 Benefactor 1000 1 Find the mean (expected value) and standard deviation of the donations. 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. Consider the mean donation of the 50 new members each day. Describe the sampling model for these means (shape, center, and spread).

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 The distribution of donations is most likely skewed to the right because a few people will donate $500 or $1000. 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 standard deviation for the model is 𝜎 𝑥 = 148.56 50 =21.01. The sampling 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? 𝑧= 100−137.5 16.61 =−2.26 𝜇 𝑥 =137.5 𝜎 𝑥 = 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. What’s the approximate model for the distribution of 𝑥 ? Estimate the probability that 𝑥 is between 3.0 and 3.1 g/mi. There is only a 5% chance that the fleet’s mean CO level is greater than what value?

The conditions are met to use a Normal model The conditions are met to use a Normal model. The mean for the model is 2.9. The standard deviation is 0.4 80 =0.045. Find the z-score for 3.0 and for 3.1: 𝑧 1 = 3.1−2.9 0.045 =4.44 𝑧 2 = 3.0−2.9 0.045 =2.22 𝑃 𝑧<4.44 =0.9999 𝑃 𝑧<2.22 =0.9868 𝑃 2.22<𝑧<4.44 = 0.9999−0.9868=0.0131 The upper 5% can’t be found on the z table, so use lower 95%. This corresponds to a z-score of 1.645. 1.645= 𝑥 −2.9 0.045 →0.074= 𝑥 −2.9 → 𝑥 =2.97 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.