Question

A telephone exchange operator assumes that 8% of the phone calls are wrong numbers. If the operator is correct, what is the probability that the proportion of wrong numbers in a sample of 528 phone calls would be greater than 10%? Round your answer to four decimal places.

Answer 1

Solution

Given that,

p = 0.08

1 - p =1 -0.08 =0.92

n = 528

= p =0.08

=  [p( 1 - p ) / n] = [(0.08*0.92) / 528 ] = 0.011806521

P( > 0.10) = 1 - P( <0.10 )

= 1 - P(( - ) / < (0.10 -0.08) /0.011806521 )

= 1 - P(z < 1.69)

Using z table

= 1 -0.9545

=0.0455

probability=0.0455