Question #80380

4) Suppose the manufacturer’s specifications for the length of a certain type of computer cable are 200010 millimeters. In this industry, it is known that short cable is just as likely to be defective (not meeting specifications) as long cable. That is, the probability of randomly producing a cable with length exceeding 2010 millimeters is equal to the probability of producing a cable with length smaller than 1990 millimeters. The probability that the production procedure meets specifications is known to be 0.99.
(a) What is the probability that a cable selected randomly is too long?
(b) What is the probability that a randomly selected cable is longer than 1990 millimeters?

Expert's answer

Answer on Question #80380 – Math – Statistics and Probability Question

4) Suppose the manufacturer’s specifications for the length of a certain type of computer cable are 2000±102000 \pm 10 millimeters. In this industry, it is known that short cable is just as likely to be defective (not meeting specifications) as long cable. That is, the probability of randomly producing a cable with length exceeding 2010 millimeters is equal to the probability of producing a cable with length smaller than 1990 millimeters. The probability that the production procedure meets specifications is known to be 0.99.

(a) What is the probability that a cable selected randomly is too long?

(b) What is the probability that a randomly selected cable is longer than 1990 millimeters?

Solution

Let AA be the event that a cable meets specifications. Let SS and LL be the events that the cable is too short and too long, respectively. Then

(a) P(A)=0.99,P(S)=P(L)=>P(A) = 0.99, P(S) = P(L) =>

=>P(S)=P(L)=1P(A)2=10.992=0.005.=> P(S) = P(L) = \frac{1 - P(A)}{2} = \frac{1 - 0.99}{2} = 0.005.

The probability that a cable selected randomly is too long

P(L)=0.005P(L) = 0.005

(b) Denoting by XX the length of a randomly selected cable, we have

P(1990X2010)=P(A)=0.99P(1990 \leq X \leq 2010) = P(A) = 0.99

P(X>2010)=P(L)=0.005P(X > 2010) = P(L) = 0.005

P(X>1990)=P(A)+P(L)P(X > 1990) = P(A) + P(L)

P(X>1990)=0.99+0.005=0.995P(X > 1990) = 0.99 + 0.005 = 0.995

The probability that a randomly selected cable is longer than 1990 millimeters is 0.995.

Answer: a) 0.005; b) 0.995.

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