1. The manager of a large retail store in the city determined the amount of mone

Question

1. The manager of a large retail store in the city determined the amount of money customers spend is normally distributed with a mean of $156 and a standard deviation of $32. It was believed 85% of customers paid using a credit card.

(a) What is the probability a randomly selected customer will spend less than $110?

(b) If 5% of all customers spend less than a given amount, what is the maximum amount this group of customers would spend?

(c) If a random sample of 22 customers is selected, what is the probability that the average amount spent is between $146 and $160?

(d) From a sample of 60 customers, what is the probability that more than 96% of these customers paid using a credit card?

(e) The manager has doubts that 85% of customers pay using a credit card. It is decided to estimate, with 90% level of confidence, the proportion of customers who pay with credit card so as to be within 5% of the true proportion. How large a sample should be selected? Justify your final answer. Do not simply give the numerical value.

Explanation / Answer

Question 1

Mean amount of customers spend = $ 156

Standard deviation = $ 32

Pr(Credit Card User) = 0.85

(a) Pr(x < $ 110) = NORM(x < $ 110 ; $ 156 ; $ 32)

Z = (110 - 156)/32 = -1.4375

Pr(x < $ 110) = Pr(Z < -1.4375) = 0.0753

(b) Here if this value is x0

Pr(x < x0 ; 156 ; 32) = 0.05

Here

Z = -1.645

(X0 - 156)/32 = -1.645

x0 = 156 - 32 * 1.645 = $103.36

(c) Here if

n = 22

standard error of sample mean = 32/sqrt(22) = 6.8224

Pr($ 146 < x < $ 160) = Pr(x < $ 160 ; 156 ; 6.8224) - Pr(x < $ 146 ; 156 ; 6.8224)

Z2 = (160 - 156)/6.8224 = 0.5863

Z1 = (146 - 156)/6.8224 = -1.4658

Pr($ 146 < x < $ 160) = Pr(Z < 0.5863) - Pr(Z < -1.4658) = 0.7212 - 0.0714 = 0.6498

(d) Here n = 60

standard error of proportion = sqrt (0.85 * 0.15/60) = 0.0461

Pr(p^ > 0.96; 0.85 ; 0.0461) = 1 - Pr(p^ < 0.96 ; 0.85 ; 0.0461)

Z = (0.96 - 0.85)/ 0.0461 = 2.386

Pr(p^ > 0.96; 0.85 ; 0.0461) = 1 - Pr(Z < 2.386) = 1 - 0.9915 = 0..085

(e) Here sample size = n

Here margin of error = 0.05

so here level of confidence = 90%

so critical value = 1.645

Here standard error of proportion = sqrt (0.85 * 0.15/n)

so,

0.05 = 1.645 * sqrt (0.85 * 0.15/n)

sqrt(n) = (1.645/0.05) * sqrt(0.85 * 0.15)

n = 138

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