Question

A grocery store's receipts show that Sunday customer purchases have a skewed distribution with a mean of $33 and a standard deviation of $19. Suppose the store had 312 customers this Sunday. a) Estimate the probability that the store's revenues were at least $10,800 b) If, on a typical Sunday, the store serves 312 customers, how much does the store take in on the worst 1% of such days?

​a) The probability is

(Round to four decimal places as​ needed.)

B) the store takes in at most $. ?

(Round to two decimal places as​ needed.)

Answer 1

Ans:

a)

sampling distribution of sample sums:

mean=33*312=10296

standard deviation=sqrt(312)*19=335.61

z=(10800-10296)/335.61

z=1.502

P(z>=1.502)=0.0665

(if rounded off z value is used,then probability is 0.0668)

b)

P(Z<=z)=0.01

z=-2.326

At most revenue=10296-2.326*335.61=9515.37

(if exact a value is taken,then revenue is=9515.26)