Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 211.4-cm and a standard deviation of 1.3-cm. For shipment, 5 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is greater than 211.5-cm.
P(M > 211.5-cm) =

Answer 1

Solution :

Given that,

mean = = 211.4

standard deviation = = 1.3

n=5

= =211.4

= / n = 1.3 / 5 = 0.58

P(M > 211.5) = 1 - P(M <211.5 )

= 1 - P[(M - ) / < (211.5 - 211.4) / 0.58]

= 1 - P(z <0.17)

Using z table

= 1 - 0.5675

probability= 0.4325