Question

Q#1: Let X denote the number of times a certain numerical control machine will malfunction: 1,
2, or 3 times on any given day. Let Y denote the number of times a technician is called on an
emergency call. Their joint probability distribution is given as: [0.5+0.5+1+1+1+1]
f x y ( , )
x
1 2 3
y
1 0.05 0.05 0.10
2 0.05 0.10 0.35
3 0.00 0.20 0.10
a) Evaluate the marginal distribution of X;
b) Evaluate the marginal distribution of Y;
c) Find the mean of X and Y;
d) Find the variance of X and Y;
e) Find the correlation coefficient between X and Y;
f) Interpret the result find in part (e).

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c} f(x,y) & & && x \\ & & & 1 & 2 & 3 \\ \hline & 1 & & 0.05 & 0.05 & 0.10 \\ y & 2 & & 0.05 & 0.10 & 0.35 \\ & 3 & & 0.00 & 0.20 & 0.10 \\ \end{array}$$

a) Evaluate the marginal distribution of X

$$g(x)=\sum_yf(x, y)$$ $$g(1)=f(1, 1)+f(1,2)+f(1,3)=0.05+0.05+0.00=0.10$$

$$g(2)=f(2, 1)+f(2,2)+f(2,3)=0.05+0.10+0.20=0.35$$

$$g(3)=f(3, 1)+f(3,2)+f(3,3)=0.10+0.35+0.10=0.55$$

b) Evaluate the marginal distribution of Y

$$h(y)=\sum_xf(x, y)$$ $$h(1)=f(1, 1)+f(2,1)+f(3,1)=0.05+0.05+0.10=0.20$$

$$h(2)=f(1, 2)+f(2,2)+f(3,2)=0.05+0.10+0.35=0.50$$

$$h(3)=f(1, 3)+f(2,3)+f(3,3)=0.00+0.20+0.10=0.30$$

c) Find the mean of X and Y

$$E(X)=1(0.10)+2(0.35)+3(0.55)=2.45$$

$$E(Y)=1(0.20)+2(0.50)+3(0.30)=2.1$$

d) Find the variance of X and Y

$$E(X^2)=1^2(0.10)+2^2(0.35)+3^2(0.55)=6.45$$

$$V(X)=E(X^2)-(E(X))^2=6.45-(2.45)^2=0.4475$$

$$E(Y^2)=1^2(0.20)+2^2(0.50)+3^2(0.30)=4.90$$

$$V(Y)=E(Y^2)-(E(Y))^2=4.90-(2.1)^2=0.49$$

e) Find the correlation coefficient between X and Y

$$E(XY)=1(1)(0.05)+1(2)(0.05)+1(3)(0.10)+$$ $$+2(1)(0.05)+2(2)(0.10)+2(3)(0.35)+$$ $$+3(1)(0.00)+3(2)(0.20)+3(3)(0.10)=5.15$$

$$cov_{xy}=\sigma_{xy}=E(XY)-E(X)E(Y)=$$ $$=5.15-2.45(2.1)=0.005$$

$$\rho_{XY}={\sigma_{XY}\over \sigma_X \sigma_Y}={0.005\over\sqrt{0.4475(0.49)}}\approx0.0107$$

f) Interpret the result find in part (e).

$$0<\rho_{XY}<0.3$$

Very weak positive correlation.