Ok, I am working on sample problems to help me better understand simple stats/prob. These are probably overly simple to most but I the whole domain of stats/prob is, to me, counter intuitive.
Moving forward:
There are 4 students with declared major. X denotes the number with an art major.
P(X=1) = 0.36
P(X=2) = 0.25
P(X=3) = 0.19
P(X=4) = 0.09
From this, the complements (the number of students whose major is not art?), would be:
P(X=1') = 0.64
P(X=2') = 0.75
P(X=3') = 0.81
P(X=4') = 0.91
1) Find the probability that at least three of the students are science majors
Here is where I am confused. From what I can tell, we can say how many are not art majors, but how can we say anything about who is a science major? This is all the information that is given. From above, it looks like the complements of X=3 and X=4 would give the number of those who aren't art majors.
2) Find the expected value of X and the variance of X
Now, the expected value should be:
E[X] = $\sum_{x=1}^4 xp_x(x)$ = 1(0.36) + 2(0.25) + 3(0.19) + 4(0.09) = 1.79
And the variance should be:
var(x) = $\sum_{x=1}^4(x-E[x])^2p_x(x)$
$A = (1-E[1])^2(0.36) = (1-0.36)^2(0.36) \approx 0.15$
$B = (2-E[2])^2(0.25) = (2-0.25)^2(0.25) \approx 0.77$
$C = (3-E[3])^2(0.19) = (3-0.19)^2(0.19) \approx 1.5$
$D = (4-E[4])^2(0.09) = (4-0.09)^2(0.09) \approx 1.38$
So, $\sum_{x=1}^4(x-E[x])^2p_x(x) = A + (B + A) + (C + B + A) = (D + C + B + A) = 4A + 3B + 2C + D \approx 4(0.15) + 3(0.77) + 2(1.5) + (1.38) \approx 7.29$
Those don't really seem right to me but I can't make heads or tails of it...
The expectation is correct:
$EX = \sum_{i=1}^4 x \cdot P(X=x) = 1 \cdot 0.36 + 2 \cdot 0.25 + 3 \cdot 0.19 + 4 \cdot 0.09 = 1.79$,
but your variance is not:
$VX = EX^2 - (EX)^2$, where
$EX^2 = \sum_{i=1}^4 x^2 \cdot P(X=x) = 1^2 \cdot 0.36 + 2^2 \cdot 0.25 + 3^2 \cdot 0.19 + 4^2 \cdot 0.09 = 4.51$.
Collecting the terms you get $VX = 4.51 - 1.79^2 = 1.3059$.
I think you need more information to solve the first problem.