Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 225.1-cm and a standard deviation of 1.3-cm. For shipment, 10 steel rods are bundled together. Find P95, which is the average length separating the smallest 95% bundles from the largest 5% bundles. P95 = _____-cm Enter your answer as a number accurate to 2 decimal place.

Answer 1

Solution :

Given that ,

mean = = 225.1

standard deviation = = 1.3

n = 10

= = 225.1 and

= / n = 1.3 / 10 = 0.4111

The z - distribution of the 95% is,

P( Z < z ) = 95 %

P( Z < z ) = 0.95

P( Z < 1.645) = 0.95

z = 1.645

Using z - score formula,

X = z * +

= 1.645 * 0.4111 + 225.1  

= 225.78

P95 = 225.78