Question

A study reports that 36​% of companies in Country A have three or more female board...

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.

What is the probability that the sample will have between 30​% and 37​% of companies in Country A that have three or more female board​ directors?

The probability is 80​% 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 68​% 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?

Homework Answers

Answer #1
for normal distribution z score =(p̂-p)/σp
here population proportion=     p= 0.3600
sample size       =n= 100
std error of proportion=σp=√(p*(1-p)/n)= 0.0480

1)

probability that the sample will have between 30​% and 37​% of companies in Country A that have three or more female board​ directors:

probability =P(0.3<X<0.37)=P((0.3-0.36)/0.048)<Z<(0.37-0.36)/0.048)=P(-1.25<Z<0.21)=0.5832-0.1056=0.4776

2)

for middle 80% values, critical z =1.28

therefore corresponding 80% interval = 0.36 -/+1.28*0.0480 =0.2986 to 0.4214

3)

for middle 68% values, critical z =0.99

therefore corresponding 80% interval = 0.36 -/+0.99*0.0480 =0.3125 to 0.4075

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