At supermarkets sales of “Chicken of the Sea” canned tuna
vary from week to week. Marketing...

Question

At supermarkets sales of “Chicken of the Sea” canned tuna
vary from week to week. Marketing researchers have determined that there is a
relationship between sales of canned tuna and the price of canned tuna.
Specifically, SALES = 50000 − 100 PRICE. SALES is measured as the number
of cans per week and PRICE is measured in cents per can. Suppose PRICE over the
year can be considered (approximately) a normal random variable with mean
μ = 248 cents and standard deviation σ = 10 cents

a. Find the expected value of SALES. 

b. Find the variance of SALES. 

c. Find the probability that more than 24,000 cans are sold
in a week. Draw a sketch illustrating the calculation. 

d. Find the PRICE such that SALES is at its 95th percentile
value. That is, let SALES0.95 be the 95th percentile of SALES. Find the value
PRICE0.95 such that P (SALES > SALES0.95) = 0.05.

Komalavalli · Accepted Answer

a. Expected value of sales = µ=50000-100*248 = 50000-24800= 25200 cans
b. standard deviation of sales = 50000-100*10= 50000-1000 =49000 cans
Variance of sales = (standard deviation)2 = 490002 = 2,40,10,00,000 cans
Variance of sales is 2,40,10,00,000 cans.
c. P(x>24000) = 1-P(x

At supermarkets sales of “Chicken of the Sea” canned tuna vary from week to week. Marketing researchers have determined that there is a relationship between sales of canned tuna and the price of...

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