Question

According to a national poll, 41% of American workers said that
they were completely

satisfied with their job. Find the probability that in a random
sample of 1140 American workers, the

proportion of workers who are completely satisfied with their
current job is between 0.4 and 0.44.

Answer #1

Solution

Given that,

p = 0.41

1 - p = 1-0.41=0.59

n = 1140

= p =0.41

= [p ( 1 - p ) / n] = [(0.41*0.59) /1140 ] = 0.0146

= P( 0.4<<0.44 )= P[(0.4-0.41) / 0.0146< ( - ) / < (0.44-0.41) / 0.0146]

= P( -0.68< z <2.05 )

= P(z < 2.05) - P(z <-0.68 )

Using z table

=0.9798-0.2483

=0.7315

probability= 0.7315

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