  A grocery​ store's receipts show that Sunday customer purchases have a skewed distribution with a mean of \$25 and a standard deviation of ​\$14. Suppose the store had 322 customers this Sunday. ​a) Estimate the probability that the​ store's revenues were at least ​\$8 comma 200 ​b) If, on a typical​ Sunday, the store serves 322 ​customers, how much does the store take in on the worst 1​% of such​ days?

Question

A grocery store's receipts show that Sunday customer purchases have a skewed distribution with a mean of \$25 and a standard deviation of \$14. Suppose the store had 322 customers this Sunday.

a) Estimate the probability that the store's revenues were at least
\$8 comma 200

b) If, on a typical Sunday, the store serves
322 customers, how much does the store take in on the worst 1% of such days?
Step 1

given that sunday customers purchases have a skewed distribution with mean of \$25 and standard deviation of \$14. There are 322 customers this sunday.

Estimate probability that stores revenue is at least 8200.

The purchases of 322 customers is a sample large enough to consider Central limit theorem. So the distribution of aveage purchases of 322 customers is normal distributed with mean purchase \$25 and standard deviation = 14/√322

For revenue to be at least 8200 from 322 customers each customer should spend atleast of 8200/322 = \$25.47. So here we need P(Xbar > 25.47), Here Xbar is the mean purchase of each customer.

Step 2

To find the Z score of X = 25.47. The formula for Z score is shown below. By substituting values we get  Z score = 0.6024. So P(Xbar > 25.47) = P(Z > 0.6024)

Step 3

Using the Z tables as shown below we get P( Z > 0.6024) = P(Z >0.60) = 0.2743 as sho...

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