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?
for normal distribution z score =(p̂-p)/σ_{p} | |
here population proportion= p= | 0.3600 |
sample size =n= | 100 |
std error of proportion=σ_{p}=√(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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