Question

Suppose the manufacturer’s specifications for the length of a certain type of computer cable are 1000 ± 10 millimeters. It is known that small cable is just as likely to be defective as large cable. That is, the probability of randomly producing a cable with length exceeding 1010 millimeters is equal to the probability of producing a cable with length smaller than 990 millimeters. The probability that the production procedure meets specifications is known to be 0.97.

Answer 1

Let A be the event that a cable meets specification. Let S and L be the events that the cable is too short and too long, respectively. Then

$P(A)=0.97, \ P(S)=P(L)\\\Rightarrow P(S)=P(L)=\dfrac{1-P(A)}{2}=\dfrac{1-0.97}{2}=0.015$

So, The probability that a cable selected randomly is too long

$$P(L)=0.015$$

Now,

Denoting by X the length of a randomly selected cable, we have

$$P(990\le X\le 1010)=P(A)=0.97\\P(X>1010)=P(L)=0.015\\P(X>990)=P(A)+P(L)=0.97+0.015=0.985$$

The probability that a randomly selected cable is longer than 990 millimeters is 0.985

Answers: 0.015 ; 0.985