Question

If out of all American workers 45% are completely satisfied with their jobs, In a random sample of 1000 American workers, find the probability that the sample proportion of who will say that they are completely satisfied with their jobs is greater than 0.51

Answer 1

Solution

Given that,

p = 45%=0.45

1 - p = 1-0.45=0.55

n = 1000

= p =0.45

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

P( >0.51 ) = 1 - P( < 0.51)

= 1 - P(( - ) / < (0.51 - 0.45) / 0.0157)

= 1 - P(z <3.82 )

Using z table

= 1 -0.9999

probability=0.0001 approximately