Simple stats: Finding standard deviation?

This is a hypergeometric distribution.

2 people out of 15 can be selected in 15C2 = (15*14)/(1*2) = 105 ways

x (number of men ) can take on values from 0,1,2

P(x=0) = C(5,0)*C(10,2) /C(15,2) = 9/21

P(x=1) = C(5,1)*C(10,1) /C(15,2) = 10/21

P(x=2) = C(5,2)*C(10,0) /C(15,2) = 2/21

Distribution of x:

x p(x)

0 9/21

1 10/21

2 2/21

E(x) = ( (0)(9/21) + (1)(10/21) + (2)(2/21)) = 2/3

V(x) = ( (0-2/3)^2 (9/21) + (1-2/3)^2 (10/21) + (2-2/3)^2 (2/21)) =26/63 = 0.412698

Standard deviation of X = sqrt(var(x)) = sqrt(0.412698) = 0.6424

You weren't clear whether the sample was with or without replacement. I'm going to assume the sample is selected without replacement, and therefore this process is a hypergeometric distribution (otherwise, with replacement, it would be a binomial distribution).

You can verify that for a hypergeometric dist, the variance of X is np(1-p)[(N-n)/(N-1)], where n is the sample size, p is the prob of selection, and N is the population size. (If binomial, you dont have the last factor above).

So the variance is 2(1/3)(2/3)[(15-2)/(15-1)]=.412698.

So the std dev is sqrt(.412698)=.6424.

What could happen? MM, MF, FM, FF

What is the probability of each?

MM (5/15)(4/14) = 20/210 = 2/21

MF (5/15)(10/14) = 50/210 = 5/21

FM (10/15)(5/14) = 50/210 = 5/21

FF (10/15)(9/14) = 90/210 = 9/21

Number of men: 2 occurs 2/21

Number of men: 1 occurs 10/21

Number of men: 0 occurs 9/21

Can you take from there?