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 between32​% and 42​% 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 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_%.

​(Round to one decimal place as​ needed.)

Answer 1

Ans:

mean=0.36

standard deviation=sqrt(0.36*(1-0.36)/100)=0.048

a)

z(0.32)=(0.32-0.36)/0.048=-0.833

z(0.42)=(0.42-0.36)/0.048=1.25

P(-0.833<z<1.25)=P(z<1.25)-P(z<-0.833)

=0.8943-0.2023=0.6920

b)

z=+/-1.036 for middle 70%

lower limit=0.36-1.036*0.048=0.310 or 31.0%

upper limit=0.36+1.036*0.048=0.410 or 41.0%

The probability is 70​% that the sample percentage will be contained 31.0% and below 41.0 %.

c)

z=+/-2.9677 for middle 70%

lower limit=0.36-2.9677*0.048=0.218 or 21.8%

upper limit=0.36+2.9677*0.048=0.502 or 50.2%

The probability is 99.7​% that the sample percentage will be contained 21.8% and below 50.2 %.