Question

A corporation is considering a new issue of convertible bonds. Management believes that the offer terms will be found attractive by 20% of all its current stockholders. Suppose that this belief is correct. A random sample of 130 current stockholders is taken. a. What is the standard error of the sample proportion who find this offer attractive? b. What is the probability that the sample proportion is more than 0.15? c. What is the probability that the sample proportion is between 0.18 and 0.22?

provide a step-by-step explanation

Answer 1

Solution

Given that,

p = 0.20

1 - p = 1 - 0.20 = 0.80

n = 130

= p = 0.20

a) =  [p( 1 - p ) / n] = [(0.20 * 0.80 ) / 130] = 0.0351

b) P( > 0.15) = 1 - P( < 0.15)

= 1 - P(( - ) / < (0.15 - 0.20) / 0.0351 )

= 1 - P(z < -1.42)

Using z table

= 1 - 0.0778

= 0.9222

c) P(0.18 < < 0.22)

= P[(0.18 - 0.20) / 0.0351 < ( - ) / < (0.22 - 0.20) / 0.0351]

= P(-0.57 < z < 0.57)

= P(z < 0.57) - P(z < -0.57)

Using z table,   

= 0.7157 - 0.2843

= 0.4314