quality engineer at a manufacturing company is keeping track of the number of pa

Question

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)

Explanation / Answer

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

PX(x) = P(X=x) =  PX,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

PX(x) = P(X=x) =  PX,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

PX,Y(x, y) = PX(x)*PY(y), for all x, y

Here for all pairs of X and Y,

PX,Y(x, y) PX(x)*PY(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

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