[43 points] Quality audit records are kept on the numbers of major and minor fai
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Question
[43 points] Quality audit records are kept on the numbers of major and minor failures that occur to a certain type of circuit pack used during the burn-in period of large electronic switching devices. Let . X = the number of major failures Y = the number of minor failures Suppose that the random variables X and Y can be described, at least approximately, by the joint probability mass function given below 0 23 4 0 0.15 0.10 0.10 0.10 0.05 x 0.05 0.08 0.14 0.08 0.05 2 0.01 0.01 0.02 0.03 0.03 (a) Find the probability that a randomly selected circuit pack will have 1 major and 2 minor failures. [1 point] (b) Find PX andY s 1). [2 points] (c) Find the probability that a randomly selected circuit pack will have fewer major failures than minor failures, i.e. find P(X Y). [2 points] (d) Suppose that demerits are assigned to a circuit pack according to the formula D-5X+ Y. Find the probability that a randomly selected circuit pack scores 7 or fewer demerits, i.e. find P(D s 7). 3 points] (e) (i) Give the marginal probability mass function of X. [2 points] (ii Find the mean value of X, i.e. find E(X). [3 points] (iii) Find the variance of X. [4 points] (f) (i) Give the marginal probability mass function of Y. [2 points] (ii Find the mean value of Y. [3 points] (g) (i) Find E(XY). [4 points] (h) Are X and Y independent random variables? Clearly answer yes or no and explain why (i) Find Cov(X,Y). [3 points] (ii) Find the variance ofY. [4 points] ii Find the expected number of demerits for a circuit pack, i.e. find E(D). [3 points] or why not. [4 points G) Find Corr(X,Y). [3 points]Explanation / Answer
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(a) P(a randomly selected circuit pack will have 1 major and 2 minor failures) = P(X=1 and Y=2)
The and within the probability refer to intersection of events. So we basically want to find the intersection probability of X=1 and Y=2, which we obtain from the cell in row 2 and column 3. So required prob = 0.14.
(b) P(X<=1 and Y<=1) = sum of probability of all points which satisfy the condition that X<=1 and Y<=1.
=P(X=0,Y=0) + P(X=0,Y=1) + P(X=1,Y=0) + P(X=1,Y=1)
=0.15 + 0.10 + 0.05 + 0.08
=0.38 (Ans)
(c) P(X<Y) = sum of probability of all possible pairs (X,Y) which satisfy the criteria X<Y.
=P(X=0,Y=1) + P(X=0,Y=2) + P(X=0,Y=3) + P(X=0,Y=4)
+ P(X=1,Y=2) + P(X=1,Y=3) + P(X=1,Y=4)
+ P(X=2,Y=3) + P(X=2,Y=4)
= 0.1 + 0.1 + 0.1 + 0.05
+ 0.14 + 0.08 + 0.05
+ 0.03 + 0.03
= 0.68 (Ans)
(d) The Demerit D = 5X + Y. We need to find the probability that P(D<=7) = P(5X+Y<=7)
=sum of probability of all points which satisfy the above relation.
We pick each possible pair of (X,Y) ad check if the relation 5X+Y<=7 holds true or not and add up the probabilities of the pairs for which the inequality holds true.
= P(X=0,Y=0) + P(X=0,Y=1) + P(X=0,Y=2) + P(X=0,Y=3) + P(X=0,Y=4)
+ P(X=1,Y=0) + P(X=1,Y=1) + P(X=1,Y=2)
= 0.15 + 0.1 + 0.1 + 0.1 + 0.05
+ 0.05 + 0.08 + 0.14
= 0.77 (Ans.)
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