Question

A grocery store’s reciepts show that Sunday customer purchases bave a skewed distribution with a mean of $34 and a standard deviation of $22. Suppose the store had 292 customers this Sunday.

a.) Estimate the probability that the store’s revenues were at least $10,200 (round to four decimal places as needed)

b.) The store takes in at most $_____

(round to two decimal places as needed)

for b.)

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

Answer 1

(a) \mu = Population mean = 34

\sigma = Population SD = 22

n = sample size = 292

\bar{x} = sample mean = 10200/292 = 34.9315

SE = \sigma /\sqrt{n}

= 22/\sqrt{292} = 1.2874

To find P(\bar{x}\geq34.9315):

Z = (\bar{x} - \mu )/SE

= (34.9315 - 34)/1.2874 = 0.7235

Table of Area Under Standard Normal Curve gives area = 0.4694

So,

P(\bar{x}>32.5342) = 0.5 - 0.4694 = 0.0306

So,

Answer is:

0.0306

(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 = (\bar{x} - 34)/1.2874

So,

\bar{x} = 31 - (1.28 X 1.2147)

= 31 - 1.5548

= 32.3521

So,

Answer is:

32.3521 X 292 = $ 9446.82