Question

Wires manufactured for use in a computer system are specified to have resistances between.12 and .14 ohms. The actual measured resistances of the wires produced by company A have a normal probability distribution with mean .13 ohm and standard deviation .005 ohm.

a) what resistance value is the cut-off for the top 5% of resistance of these wires?

Answer 1

Solution:

Given, X follows Normal distribution with,

   = 0.13

= 0.005

a)

For top 5% data , let x be the required cut-off.

P(X > x) = 5%

P(X > x) = 0.05

P(X < x) = 1 - 0.05

P(X < x) = 0.95

For the standard normal variable z , P(Z < z) = 0.95

Use z table , see where is 0.95 probability and then see the corresponding z value.

P(Z < 1.645) = 0.95

So z = 1.645

Now using z score formula ,

x = + (z * ) = 0.13 + (1.645 * 0.005) = 0.138225

Answer: 0.138225 ohm