There are 500 employees in a firm, 45% are female. A sample of 70 employees is s
ID: 3182426 • Letter: T
Question
There are 500 employees in a firm, 45% are female. A sample of 70 employees is selected randomly.
a.
Determine the standard error of the proportion.
b.
What is the probability that the sample proportion of females is between 0.40 and 0.55?
c.
What is the probability that the sample proportion of females is less than 10%.
d.
What is the probability that the sample proportion of males is less than 10%
a.
Determine the standard error of the proportion.
b.
What is the probability that the sample proportion of females is between 0.40 and 0.55?
c.
What is the probability that the sample proportion of females is less than 10%.
d.
What is the probability that the sample proportion of males is less than 10%
Explanation / Answer
Here p=0.45 and n=70
a. Standard error=sqrt(pq/n)=0.059
b. P(0.40<p<0.50)=P(0.40-0.45/0.059<z<0.50-0.45/0.059)=P(-0.85<z<0.85)=P(0<z<0.85)-P(0<z<-0.85)=0.3023+0.3023=0.6046
c. Now as p=0.45 and n=70, np=31.5 and sd=sqrt(npq)=4.16
So P(x<22.5)=P(z<22.5-31.5/4.16)=P(z<-2.16)=0.5+P(0<z<-2.16)=0.5-0.4846=0.0154
d. P(x<27.5)=P(z<-0.96)=0.5+P(0<z<-0.96)=0.5-0.3315=0.1685
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