Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 91.1-cm and a standard deviation of 0.5-cm. For shipment, 25 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is greater than 90.8-cm.
P(M > 90.8-cm) =

Answer 1

Solution :

Given that ,

mean = = 91.1

standard deviation ==0.5

n = 25

m = 91.1

m = / n = 0.5/ 25 = 0.1

P(M 90.8> ) = 1 - P(M <90.8 )

= 1 - P[( M- m ) / m < (90.8-91.1) /0.1 ]

= 1 - P(z <-3 )

Using z table

= 1 - 0.0013

= 0.9987

probability= 0.9987