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 29% and 44% of companies in Country A that have three or more female board directors? The probability is ....... (Round to four decimal places as needed.) 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? The probability is 70% that the sample percentage will be contained above ..... % and below ..... %. (Round to one decimal place as needed.) c. The probability is 95% 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 95% that the sample percentage will be contained above ...... % and below .....%. (Round to one decimal place as needed.)

Answer 1

To find the probability of a sample having a certain percentage of companies with three or more female board directors, we use the normal approximation to the binomial distribution. Probability = 0.8464

Step-by-step explanation:

To find the probability that the sample will have between 29% and 44% of companies in Country A that have three or more female board directors, we can use the normal approximation to the binomial distribution. First, calculate the mean and standard deviation of the sample distribution:

Mean = sample proportion = 0.36

Standard deviation = sqrt((p(1-p))/n) = sqrt((0.36(1-0.36))/100) = 0.048979

Next, convert the given percentages into z-scores:

Z1 = (0.29 - 0.36) / 0.048979 = -1.4289

Z2 = (0.44 - 0.36) / 0.048979 = 1.6313

Using a standard normal distribution table or calculator, we can find the probabilities:

Probability = P(Z1 < Z < Z2) = P(-1.4289 < Z < 1.6313)

= 0.9211 - 0.0747

= 0.8464