quality engineer at a manufacturing company is keeping track

quality engineer at a manufacturing company is keeping track of the number of parts that are produced during first 3 A during frst and second shift that do not fall within the required tolerance bounds. Let X the number of parts out of the tolerance bounds from first shift and let the number of parts out of the tolerance bounds from the second shift. A joint probability table is given below 10 10 a. Find the marginal distribution for X. (8 points b. Find the marginal distribution for Y. (8 points) c. Find P(Y-21X-0). (7 points) d. Are Xand Y independent? Why or why not? (6 points)

Solution

Let X = the number of parts out of the tolerance bounds from first shift.

and Y = the number of parts out of the tolerance bounds from second shift.  

The probability distribution is,

a) Find marginal distribution of X.

Marginal probability of X is the probability that RV X has the value x regardless of the value of Y. That is

P_X(x) = P(X=x) =  P_X,Y (x,y) (summation over y)

Find marginal distribution of Y.

Marginal probability of Y is the probability that RV Y has the value y regardless of the value of X. That is

P_X(x) = P(X=x) =  P_X,Y (x,y) (summation over y)

Find P(Y=2 / X=0) :

This is conditional probability.

P(Y=2 / X=0) = P(Y=2, X=0) / P(X=0)

= 0.02 / 0.2 = 0.1

Are X and Y independent?

The discrete r.v.s X, Y are called independent if

P_X,Y(x, y) = P_X(x)*P_Y(y), for all x, y

Here for all pairs of X and Y,

P_X,Y(x, y) P_X(x)*P_Y(y)

that is,

0.02 0.07*0.2 = 0.014

0.06 0.36*0.2 = 0.072

0.02 0.36*0.2 = 0072

and so on ....

Therefore X and Y are not independent.

			Y
		0	1	2	3	total
X	0	0.02	0.06	0.02	0.1	0.2
	1	0.04	0.15	0.2	0.1	0.49
	2	0.01	0.15	0.14	0.01	0.31
	total	0.07	0.36	0.36	0.21	1