Fi515

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1. (TCO B)  The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 29.00 inches and a standard deviation of 1.25 inches.  What is the probability that the average length of a steel sheet from a sample of 25 units is more than 28.50 inches long?

ANSWER

σx=σ√n=1.25/√25

σx=.25

P(Z≥ 28.50) = P(Z≥ (28.50-29)/.25)

P(z≥-2) from table for z=2 = 0.4772 +.5000 = .9772

2. (TCO B) Quality Progress, February 2005, reports on improvements in customer satisfaction and loyalty made by Bank of America.  A key measure of customer satisfaction is the response (on a scale from 1 to 10) to the question:  “Considering all the business you do with Bank of America, what is your overall satisfaction with Bank of America?”  Here, a response of 9 or 10 represents “customer delight.” 

Historically, the percentage of Bank of America customers expressing customer delight has been 48%.  Suppose that we wish to use the results of a survey of 350 Bank of America customers to justify the claim that more than 48% of all current Bank of America customers would express customer delight.  The survey finds that 175 of 350 randomly selected Bank of America customers express customer delight.  If, for the sake of argument, we assume that the proportion of customer delight is p = .48, calculate the probability of observing a sample proportion greater than or equal to 175/350 = .50.  That is, calculate P(ṗ≥.50).

ANSWER

n=350

P=.48...