Let X be the number of female employee. Let n be the sample size, p be the probability that selected employee is female.
It is given that 45% employee are female it mean p=0.45
Sample size n=60
From given information X follows Binomial distribution with n=50 and p=0.45
For large value of n the Binomial distribution approximates to Normal distribution.
Let p be the proportion of female employee in the given sample.
Then distribution of proportion P is normal with parameters
mean =p and standard deviation = [tex] \sqrt{\frac{p(1-p)}{n}} [/tex]
Here we have p=0.45
So mean = p = 0.45 and
standard deviation = [tex] \sqrt{\frac{0.45(1-0.45)}{60}} [/tex]
standard deviation = 0.0642
Now probability that sample proportions of female lies between 0.40 and 0.55 is
P(0.40 < P < 0.45) = [tex] P(\frac{0.40 - 0.45}{0.0642} < \frac{P-mean}{standard deviation} < \frac{0.55- 0.45}{0.0642} ) [/tex]
= P(-0.7788 < Z < 1.5576)
= P(Z < 1.5576) - P(Z < -0.7788)
= P(Z < 1.56) - P(Z < -0.78)
= 0.9406 - 0.2177
= 0.7229
The probability that the sample proportion of females is between 0.40 and 0.55 is 0.7229
