Question

Wally runs a fruit & vege stall Wally’s VegeRama -at the local Sunday market. He can buy watermelons from his supplier for $5 each. He can sell watermelons for $10

Each. On any particular Sunday, demand for watermelons follows a Poisson distribution with mean 5. Any watermelons that are not sold on Sunday go bad before the next

weekend. Wally’s current policy is to stock 6 watermelons.

2. What is Wally’s expected and variance of the profit?

Answer 1

Solution

Back-up Theory

probability mass function (pmf) of X is given by P(X = x) = e ^{– λ}.λ^x/(x!) ………..........................................................…..(1)

where x = 0, 1, 2, ……. , ∞

This probability can also be obtained by using Excel Function, Statistical, POISSON …............................................. (1a)

If a discrete random variable, X, has probability function, p(x), x = x₁, x₂, …., x_n, then

Expected value = Mean (average) of X = E(X) = µ = Σ{x.p(x)} summed over all possible values of x….................…. (2)

E(X²) = Σ{x².p(x)} summed over all possible values of x……………………......................................…......…………..(2a)

Variance of X = Var(X) = σ² = E[{X – E(X)}²] = E(X²) – {E(X)}²…………............................................…….....………..(3)

Now, to work out the solution,

Given,

‘He can buy watermelons from his supplier for $5 each. He can sell watermelons for $10 each. Any watermelons that are not sold on Sunday go bad before the next weekend.’

For every watermelon sold, profit is $5 and for every watermelon not sold, profit is - $5 (i.e., loss)

Let X = demand

So, given the current policy of stocking 6 watermelons,

If X ≥ 6, all 6 watermelons would be sold and hence profit = 30............................................................................ (4)

If X < 6, only x watermelons would be sold, (6 - x) watermelons would not be sold and hence profit

= 5x – 5(6 – x)

= 10x - 30............................................................................................................................................................. (5)

Expected value of Profit = $15.07 Answer 1

Variance of Profit = 234.32 Answer 2

Details of calculations

x	Profit - P	Probability p	P x p	P2 x p
0	-30	0.0067	-0.2021	6.0642
1	-20	0.0337	-0.6738	13.4759
2	-10	0.0842	-0.8422	8.4224
3	0	0.1404	0.0000	0.0000
4	10	0.1755	1.7547	17.5467
5	20	0.1755	3.5093	70.1869
6	30	0.1462	4.3867	131.6005
7	30	0.1044	3.1333	94.0004
8	30	0.0653	1.9583	58.7502
9	30	0.0363	1.0880	32.6390
10	30	0.0181	0.5440	16.3195
11	30	0.0082	0.2473	7.4180
12	30	0.0034	0.1030	3.0908
13	30	0.0013	0.0396	1.1888
14	30	0.0005	0.0142	0.4246
15	30	0.0002	0.0047	0.1415
16	30	0.0000	0.0015	0.0442
	Total	1.0000	15.0664	461.3137
NOTE: All probabilities are obtained using Excel Function, Statistical, POISSON
Expected Profit - Vide (2)		15.0664
Variance Vide (3)		234.3164

DONE