4.(TCO B) At a local supermarket the monthly customer expenditure follows a norm

Question

4.(TCO B) At a local supermarket the monthly customer expenditure follows a normal distribution with a mean of $495 and a standard deviation of $121.

a. Find the probability that the monthly customer expenditure is less than $300 for a randomly selected customer.
b. Find the probability that the monthly customer expenditure is between $300 and $600 for a randomly selected customer.
c. The management of a supermarket wants to adopt a new promotional policy giving a free gift to every customer who spends more than a certain amount per month at this supermarket. Management plans to give free gifts to the top 8% of its customers (in terms of their expenditures). How much must a customer spend in a month to qualify for the free gift? (Points : 18)

5. (TCO C) DJ Car Rental wants to estimate the average number of miles traveled per day by each of its cars rented in California. A random sample of 110 cars rented in California yields the following results.

Sample Size = 110
Sample Mean = 85.5 mi
Sample Standard Deviation = 19.3 mi

a. Construct the 99% confidence interval for the average number of miles traveled per day by each of its cars rented in California.
b. Interpret this interval.
c. How many cars should be sampled if we wish to construct a 99% confidence interval for the average number of miles traveled per day that is accurate to within 2 mi? (Points : 18)

Explanation / Answer

4) MEAN = 495

STANDARD DEVIATION = 121

A) For x = 300, the z-value z = (300 - 495) /121 = -1.61

Hence P(x < 300) = P(z < -1.61) = [area to the left of -1.61] = 0.0567

B)

For x = 300 , z = (300 - 495) / 121 = -1.61 and for x = 600, z = (600 - 495) / 121 = 0.86

Hence P(300 < x < 495) = P(-1.61 < z < 0.86) = [area to the left of z = 0.86] - [area to the left of -1.61]

= 0.1949 - 0.0567 = 0.1382

C) FOR TOP 8% = 0.92

THE Z SCORE FOR 0.92 = 1.41

X = MEAN + Z*STANDARD DEVIATION

= 495 +1.41*121 = 665.61

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