Question

The weights of bowling balls are normally distributed with mean 11.5 pounds and standard deviation 2.7 pounds. A sample of 36 bowling balls is selected. What is the probability that the average weight of the sample is less than 11.03 pounds?

Answer 1

Solution :

Given that,

mean = = 11.5

standard deviation = = 2.7

n = 36

= 11.5

=  / n = 2.7 / 36=0.45

P( < 11.03) = P[( - ) / < (11.03-11.5) /0.45 ]

= P(z < -1.04)

Using z table  

= 0.1492   

probability= 0.1492