The following table gives the joint probability distribution between employment status and college graduation among those...

Solution:

From the given table, we know that Pobability of X equal 0, P(X=0) = 0.045 + 0.621 = 0.666

And, P(X=1) = 0.026 + 0.308 = 0.334

Similarly, P(Y=0) = 0.045 + 0.026 = 0.071

P(Y =1) = 0.621 + 0.308 = 0.929

(a) Expected value (mean) of X, E(X) = sum of X*P(X)

So, E(X) = X*P(X=0) + X*P(X=1)

E(X) = 0*0.666 + 1*0.334 = 0.334

Same way, expected value (mean) of Y, E(Y) = sum of (Y*P(Y))

So, E(Y) = Y*P(Y =0) + Y*P(Y=1)

E(Y) = 0*0.071 + 1*0.929 = 0.929

(b) Variance of X, V(X) = E(X²) - (E(X))²

E(X²) = sum of (X²*P(X))

E(X²) = X²*P(X=0) + X²*P(X=1)

E(X²) = (0)²*0.666 + (1)²0.334 = 1*0.334 = 0.334

V(X) = 0.334 - (0.334)² = 0.334 - 0.112 = 0.222

Similarly, variance of Y, V(Y) = E(Y²) - (E(Y))²

E(Y²) = (0)²*0.071 + (1)²*0.929 = 0.929

V(Y) = 0.929 - (0.929)² = 0.929 - 0.863 = 0.066

(c) Covariance of X and Y,

cov(X,Y) = E(X*Y) - E(X)*E(Y)

E(X,Y) = sum of [(X*Y)*P(X and Y)]. For this, we will require the joint probability of X and Y.

E(X,Y) = X*Y*P(X=0 and Y=0) + X*Y*P(X=0 and Y=1) + X*Y*P(X=1 and Y=0) + X*Y*P(X=1 and Y=1)

E(X,Y) = (0*0*0.045) + (0*1*0.621) + (1*0*0.026) + (1*1*0.308) = 0.308

So, covariance between X and Y, cov(X,Y) = 0.308 - (0.334)*(0.929)

Cov (X,Y) = 0.308 - 0.310 = -0.002

Correlation coefficient between X and Y:

corr(X,Y) = Cov (X,Y)/(V(X)*V(Y))^1/2

Corr (X,Y) = -0.002/(0.222*0.066)^1/2

Corr(X,Y) = -0.002/0.121 = -0.017

Since, the covariance (and so the correlation coefficient) between X and Y is negative, it means that X and Y are negatively related.