Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 195.2-cm and a standard deviation of 0.8-cm. For shipment, 22 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is less than 195.1-cm.
P(M < 195.1-cm) =

Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Solution :

Given that ,

mean = = 195.2

standard deviation = = 0.8

n = 22

m = 195.2

m =  / n = 0.8 / 22=0.1706

P(M <195.1 ) = P[( - m ) / m < (195.1-195.2) / 0.1706]

= P(z <-0.586 )

Using z table  

= 0.2789

probability=0.2789