Question

Wires manufactured for a certain computer system are specified to have a resistance of between 0.12 and 0.14 ohm. The actual measured resistances of the wires produced by Company A have a normal probability distribution with a mean of 0.13 ohm and a standard deviation of 0.005 ohm. The computer firm orders 70% of the wires used in its systems from Company A and 30% from Company B.What is the probability that a randomly selected wire from Company B’s production lot will meet the specifications?

Answer 1

0.9544

Explanation:

Let the measured resistances of the wire production by Company B = x

Define the standard normal variable
`Z=(x-0.13)/(0.005)`

The probability that a randomly selected wire from Company B's production lot will meet the specification is :

P = (0.12 ≤ X ≤ 0.14)

=
`P((0.12-0.13)/(0.005)\leq Z\leq (0.14-0.13)/(0.005))`

=
`P(-2\leq Z\leq 2)`

=
`P(Z\leq 2)-P(Z\leq -2)`

= 0.9772 - (1 - 0.9972)

= 0.9544