With steps please Automobile engines and transmissions are produced on assembly

Question

With steps please Automobile engines and transmissions are produced on assembly lines, and are inspected for defects after they come off their assembly lines. Those with defects are repaired. Let X represent the number of engines, and Y the number of transmissions that required repairs in a one-hour time interval. The joint probability mass function of X and Y as follows: Find the marginal probability mass function p_x (x). Find the marginal probability mass function p_y(y) Find mu_x. Find mu_y. Find sigma_x. Find sigma_y. Find Cov(x, y). Find rho (X, Y).

Explanation / Answer

Let X : number of engines

Y : number of transmissions that required repairs in a one hour time interval.

We have given joint pobability mass function of X and Y.

If we summing all the rows and columns then we get the following table as :

Marginal pmf is defined as,

P(x) = P(x,y) over y

P(y) = P(x,y) over x

Marginal pmf of X :

Similarly marginal pmf of y :

FInd µx and µy :

µx = xp(x)

µy = yp(y)

µx = 0*0.33 + 1*0.4 + 2*0.2 + 3*0.07 = 1.01

µy = 0*0.28 + 1*0.34 + 2*0.25 + 3*0.13 = 1.23

Now we have to find x and y.

The formulaes are :

x = sqrt [E(X^2) - (E(X))^2]

y = sqrt [E(Y^2) - (E(Y))^2]

E(X^2) = x^2p(x)

E(Y^2) = y^2p(y)

E(X^2) =  0^2*0.33 + 1^2*0.4 + 2^2*0.2 + 3^3*0.07 = 1.83

E(Y^2) = 0^2*0.28 + 1^2*0.34 + 2^2*0.25 + 3^2*0.13 = 2.51

x = sqrt(1.83 - 1.01^2) = 0.8999

y = sqrt(2.51 - 1.23^2) = 0.9985

y x 0 1 2 3 total 0 0.13 0.1 0.07 0.03 0.33 1 0.12 0.16 0.08 0.04 0.4 2 0.02 0.06 0.08 0.04 0.2 3 0.01 0.02 0.02 0.02 0.07 total 0.28 0.34 0.25 0.13 1

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