Question

2. Sharphotos sends photographers around to various shopping malls in the Midwest to take pictures of children. The company charges only $0.99 for a photo shoot, which consists of six poses. The company then creates a package of 3 photos that are offered to the parents, who have a choice of buying zero, one, two, or all three of the packages. Based on his experience in the business, Belinda Shar has assessed the following probabilities of the number of packages Number of Packages that might be purchased by a parent. X-00.30 X-0.40 X-2 0.20 X-3 0.10 (a) What is the expected number of packages to be purchased by each parent? b) What is the standard deviation for the number of packages to be purchased by each parent? c) Suppose all of the packages are to be priced at the same level. How much should they be priced if Sharphotos wants to break even in expectation? Assume that the production costs are $3.00 per photo, and remember that the sitting charge is SO.99.

Answer 1

Question 2

(a) Here for every discrete  probability distribution

Expected Value = E(x) = $\sum xf(x)$

= 0 * 0.30 + 1 * 0.40 + 2 * 0.20 + 3 * 0.10 = 1.1 Packages

(b) Here

variance for the number of packages to be purchased is

VaR[X] = E[X²] - E[X]²

E[X²] = 0 * 0.30 + 1 * 1 * 0.40 + 2 * 2 * 0.20 + 3 * 3 * 0.10 = 2.1

VaR[X] = 2.1 - 1.1² = 0.89

Standard deviation = sqrt(Var(x)) = sqrt(0.89) = 0.9434

(c) Here production cost = $ 3.00 per photo

Sitting charge = $ 0.99

For the breakeven we will take the cost of package from the expected number of packages purchased.

so here

Total costs = Cost of pictures

= 3 * 3 = $ 9

Total photos purchased expected = 1.1

Photoshoot cost = 0.99

Total revenue = 0.99 + 1.1 * Price of a package

So here

Total revenue = Total Cost

So, breakeven price for profit = (9 - 0.99)/1.1 = $ 7.28 per package