There are 500 employees in a firm, 45% are female. a sample of 60 employees is selected randomly. the probability that the sample proportion of females is between 0.40 and 0.55 is

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 =

Here we have p=0.45

So mean = p = 0.45 and

standard deviation =

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) =

= 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

-10i^2-4i

step-by-step explanation:

evaluate for p=2,q=5i

(2−4)(2+5i)i

(2−4)(2+5i)i

=−10i2−4i

answer: the correct answer is -32

step-by-step explanation: