A metal fabricating plant currently has five major pieces under contract each wi
ID: 2907629 • Letter: A
Question
A metal fabricating plant currently has five major pieces under contract each with a deadline for completion. Let Y be the number of pieces completed by their deadlines. Suppose that Y is a random variable with pmf given by 1 3 4 p(y) 0.150.150.25 0.30.1 (a) Check and state whether the given table is a valid pmf. Provide reason(s) (b) If probability of Y 4 was a typo in the table, then find the correct value of p(4) so that the table is a valid pmf. (c) Compute the probability that less than 3 pieces completed by the deadline. (d) Compute the average number of pieces completed by the deadline. (e) Compute the standard deviation of number of pieces completed by the deadline.Explanation / Answer
a) No, it is not valid pmf as the su, of all probabilities is not equal to 1 (0.15+0.15+0.25+0.30+0.1 = 0.95)
b) For it's to be valid pmf
0.15+0.15+0.25+p(4)+0.1 = 1
p(4) = 0.35
c) P(X<3)
= P(1)+p(2)
= 0.15+0.15
= 0.30
4) E(X) = 1*0.15+2*0.15+3*0.25+4*0.35+5*0.1 = 3.1