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 34​% and 42​% of companies in Country A that have three or more female board​ directors? The probability is . 5559. ​(Round to four decimal places as​ needed.) b. The probability is 90​% 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 90​% that the sample percentage will be contained above __% and below __​%. ​(Round to one decimal place as​ needed.)

Answer 1

p=0.36

n=100

For p,

Mean = 0.36

Standard deviation = sqrt (0.36 * 0.64)/sqrt(100) = 0.048

p ~ N(0.36 , 0.048²)

What is the probability that the sample will have between 34 % and 42 % of companies in Country A that have three or more female board directors(four decimal places)

P (0.34 < p < 0.37)

= P{ (.34-.36)/0.048 < Z <(0.42-0.36)/0.048}

= P(-0.417 < Z < 1.25)

= 0.5559 Answer

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? (one decimal place)

The 95% symmetric limits are