Question

A survey of all medium- and large-sized corporations showed that 70% of them offer retirement plans to their employees. Let p^ be the proportion in a random sample of 50 such corporations that offer retirement plans to their employees. Find the probability that the value of p^ will be between 0.62 and 0.64.

Round your answer to four decimal places.

P(0.62<p^<0.64)= Enter your answer in accordance to the question statement

Answer 1

Solution

Given that,

p = 0.70

1 - p = 1 - 0.70 = 0.30

n = 50

= p = 0.70

  [p ( 1 - p ) / n] = [(0.70 * 0.30) / 50 ] = 0.0648

P( 0.62 < < 0.64 )

= P[(0.62 - 0.70) / 0.0648 < ( - ) / < (0.64 - 0.70) / 0.0648]

= P(- 1.23 < z < 0.93)

= P(z < 0.93 ) - P(z < - 1.23)

Using z table,   

= 0.8238 - 0.1093

= 0.7145