A corporation is considering a new issue of convertible bonds, mandagemnt belives that the offer terns will be founf attractiv by 20% of all its current stockholders, suppoer that the belif is correct, a random sample of 130 stockholderss 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?
d. Suppose that a sample of 500 current stockholders had been taken. Without doing the calculations, state whether the probabilities in parts (b) and (c) would have been higher, lower, or the same as those found.

a. What is the standard error of the sample proportion who find this offer attractive? =  0.0351

b. What is the probability that the sample proportion is more than 0.15? = 0.9222

c. What is the probability that the sample proportion is between 0.18 and 0.22? =  0.4314

Step-by-step explanation:

see attachment for explanation

Step-by-step explanation:

Given that,

p = 0.20

1 - p = 1 - 0.20 = 0.80

n = 130

\mu\hat p = p = 0.20

A) \sigma \hat p =  \sqrt[p( 1 - p ) / n] = \sqrt [(0.20 * 0.80 ) / 130] = 0.0351

B) P( \hat p > 0.15) = 1 - P( \hat p < 0.15)

= 1 - P(( \hat p - \mu \hat p ) / \sigma \hat p < (0.15 - 0.20) / 0.0351 )

= 1 - P(z < -1.42)

Using z table

= 1 - 0.0778

= 0.9222

C) P(0.18 < \hat p < 0.22)

= P[(0.18 - 0.20) / 0.0351 < ( \hat p - \mu \hat p ) / \sigma \hat p < (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

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