Question

# A grocery? store's receipts show that Sunday customer purchases have a skewed distribution with a mean...

A grocery? store's receipts show that Sunday customer purchases have a skewed distribution with a mean of \$31 and a standard deviation of \$21. Suppose the store had 292 customers this Sunday.

?a) Estimate the probability that the? store's revenues were at least \$9,500.

?b) If, on a typical? Sunday, the store serves 292 customers, how much does the store take in on the worst 10?% of such? days?

(a) = Population mean = 31

= Population SD = 21

n = sample size = 292

= sample mean = 9500/292 = 32.5342

SE = /

= 21/ = 1.2147

To find P(32.5342):

Z = ( - )/SE

= (32.5342 - 31)/1.2147 = 1.2630

Table of Area Under Standard Normal Curve gives area = 0.3962

So,

P(>32.5342) = 0.5 - 0.3962 = 0.1038

So,

0.1038

(b)

Worst 10% corresponds to area = 0.5 - 0.10 = 0.40 from mid value tro Z on LHS.

Table gives Z score - 1.28

So,

Z = - 1.28 = ( - 31)/1.2147

So,

= 31 - (1.28 X 1.2147)

= 31 - 1.5548

= 29.45

So,

29.45 X 292 = \$ 8,597.99

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