Question

If out of all American workers 55% are completely satisfied with their jobs. If a random sample of 1000 American workers were considered,

1. Find the mean and standard deviation of the sample proportion.

2. Find the probability that in a random sample of 1000 American workers, the proportion who will say that they are completely satisfied with their jobs is between 0.51 and 0.57

Answer 1

Solution

Given that,

p = 55%=0.55

1 - p = 1-0.55=0.45

n = 1000

a.

mean = = p =0.55

stnadrad devoation =  [p ( 1 - p ) / n] = [(0.55*0.45) /1000 ] = 0.0157

b

P( 0.51<< 0.57)= P[(0.51-0.55) / 0.0157< ( - ) / < (0.57-0.55) /0.0157 ]

= P( -2.55< z <1.27 )

= P(z < 1.27) - P(z < -2.55)

Using z table

=0.8980 - 0.0054

=0.8926

probability=0.8926