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 %.**

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