Question

A study reports that 36% of companies in Country A have three or more female board directors. Suppose you select a random sample of 100 respondents. Complete parts​ (a) through​ (c) below.

a. What is the probability that the sample will have between 33​% and 43​% of companies in Country A that have three or more female board​ directors?

b. The probability is 70​% that the sample percentage of Country A companies having three or more female board directors will be contained within what symmetrical limits of the population​ percentage?

c. The probability is 99.7​% that the sample percentage of Country A companies having three or more female board directors will be contained within what symmetrical limits of the population​ percentage?

The probability is 99.7​% that the sample percentage will be contained above _____ ​% and below _____ ​%.

Answer 1

P(A company in country A having three or more female board directors), p = 0.36

q = 1 - p = 0.64

Sample size, n = 100

a) Normal approximation for binomial distribution. P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np

= 100 x 0.36 = 36

Standard deviation =

= 4.8

P(33 < X < 43) = P(X < 43) - P(X < 33)

= P(Z < (43 - 36)/4.8) - P(X < (33 - 36)/4.8)

= P(Z < 1.46) - P(Z < -0.625)

= 0.9279 - 0.2660

= 0.6619

b) Let 70% be between M and N

P(X < M) = 0.5 + (0.70/2) = 0.85

P(X < N) = 0.5 - (0.70/2) = 0.15

P(Z < (M - 36)/4.8) = 0.85

(M - 36)/4.8 = -1.04 (from standard normal distribution table)

M = -1.04x4.8 + 36 = 31%

N = 1.04x4.8 + 36 = 41%

70​% sample percentage of Country A companies having three or more female board directors will be contained within 31% and 41%

c) According to empirical rule, 99.7% data values lie within 3 standard deviations of mean.

The probability is 99.7​% that the sample percentage will be contained above 36-3x4.8 = 21.6​% and below 36+3x4.8 = 50.4%.