a) What is the probability the sample will contain exactly 5 defective drives? b

Question

a) What is the probability the sample will contain exactly 5 defective drives?
b) What is the probability the sample will contain more than 5 defective drives?
c) What is the probability the sample will contain less than 5 defective drives?
d) What is the expected number of defective drives in the sample?
e) What is the variance of the number of defective drives in the sample?
f) What is the standard deviation of the number of defective drives in the sample?
g) Each defective drive costs the company 11 dollars. What is the expected cost to the company for the defective drives in the sample?
h) Each defective drive costs the company 11 dollars. What is the standard deviation of the cost to the company for the defective drives in the sample?

Explanation / Answer

a) P(x=5) = (53,5) (0.93)^48 (0.07)^5 = 00.1481 b) P(x > 5) = 1- P(x<=5) = 1-0.8355 = 0.1645 c) P(x<5) = P(x=1)+...+P(x=5) = 00.68745 d) E(x)=np = 54(0.07)=3.71 e) Var(x) = np(1-p) = 53(0.07)(0.93)=3.4503 f) SD = square.root(Var(x)) = sqroot(3.4503)=1.857 g) Use (d) where we have 3.71 ~ 4 expected defective drive and each costs $11. So total expected cost is 4 x 11 = 44 g) Use (f), 11 x 1.857=20.427

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