Question

1) 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 11.07 pounds?

2) According to one pollster, 46% of children are afraid of the dark. Suppose that a sample of size 20 is drawn. Find the value of standard error , the standard deviation of the distribution of sample proportions.

Answer 1

Solution :

Given that ,

mean = = 11.5

standard deviation = =2.7

n = 36

= 11.5

=  / n = 2.7/ 36=0.45

P( <11.07 ) = P[( - ) / < (11.07 -11.5) / 0.45]

= P(z <-0.96 )

Using z table  

= 0.1685

probability=0.1685

(B)

Solution

Given that,

p = 0.46

1 - p = 1-0.46=0.54

n = 20

mean = = p =0.46

standard error=standard deviation =  [p ( 1 - p ) / n] =   [(0.46*0.54) / 20 ] = 0.1114