1. calculating binomials

I finished this homework question but wanted to see if it looks like I worked it out right. Here is the question: "64.9 % of students at a college are women. Suppose that 8 of those students are randomly selected: let X denote the number of women students among these eight. Find the expected value and standard deviation of X. Find the probability that exactly six of the selected students will be women. Also find the probability that at least six of the selected students will be women.

For values of the variables I got: x= 6, for n = 8, for q = .351, for p = .649

For mean, I got M= np or M= 8 x .649 = 5.2

For standard deviation I got lowercase sigma = square root of n x p x q or 8 x .649 x .351 = 1.35

For "at least six women", I got P (3) = .82 + P (4) = .188 + P(5) = .279 + P(6)= .258 + P(7)=.136 + P(8)= .031 = .974

For "exactly six students out of 8 as females" I got P(x) = nCx x p^x (q^n-x) = P(^) = 8 C 6 x (.649)^6 (.351) ^ 2 = .258

Just wondering how my calculations look? Am I way off here? Thanks.

2. I don't know what
For "at least six women", I got P (3) = .82 + P (4) = .188 + P(5) = .279 + P(6)= .258 + P(7)=.136 + P(8)= .031 = .974
means

I would obtain the sum, which you may have computed

$\displaystyle {8\choose 6}(.649)^6(.351)^2+ {8\choose 7}(.649)^7(.351)+ (.649)^8$

I'm confident that .031 does not equal .974.

3. For $\displaystyle pr(X\geq 6)$ we can take the sum of the probability of 6, 7, and 8 of the 8 selected students being female. Therefore:

$\displaystyle pr(X\geq 6)=pr(X=6)+pr(X=7)+pr(X=8)\approx .4254$