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The number of items sold on any one day in the traditional shop is a random variable X and the corresponding number of items sold via the Internet is a random variable Y. The joint distribution of X and Y is described by the probability function p(x,y) below:

X Y

0 1 2 3

0 0.01 0.07 0.05 0.00

1 0.09 0.15 0.15 0.06

2 0.00 0.13 0.10 0.19

Find Px(x) the marginal distribution of X

Find E(X), the expected number of items sold per day in the traditional shop

Find V(X) the variance of the number of items sold per day in the shop

Given E(Y) = 1.70 and V(Y) = 0.91 find the Co-variance between X and Y?

Owing to reduced costs, items sold on the Internet attract a higher profit margin.

The daily profit from sales of the item, P, is determined by the equation: P = 6X + 8Y - 5

Find E(P), the expected daily profit from the sales and also the standard deviation of P.

The marginal distribution for x means the probabilities for each value of x no matter what y may be. Therefore, you need to sum (integrate) the probabilities over their y values. These are the marginal distributions. I get

∑P(x,y) over y =

0.13

0.45

0.58

∑P(x,y) over x =

0.1 0.35 0.3 0.25.

The expected value is the sum of each value times its probability density: I get

E(x) 1.61 E(y) 1.7

var(x) 0.1779 var(y) 0.91.

The covariance is calculated like the variance in that the (value - mean) is squared and divided by the variance,

covar(x,y) = ∑(x-E(x))(y-E(y)) / ( var(x)var(y) ) ^1/2.

Since you have all the numbers, you should be able to calculate it.

For E(P) = E(6x +8y -5) = E(6x) + E(8y) - E(5) = 6E(x) + 8E(y) -5.

I'm glad to help but please don't just send me homework problems without trying to do some of the work yourself or, better yet, ask a question about something you are unsure about.

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