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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...

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  • $\begingroup$ Well...are we to assume that the only options are Arts vs. Science? No History, say? If there are just the two options then saying that we have at least three Science majors is the same as saying we have at most one Arts major. Also, in your variance formula you should always subtract the mean. That is, it's $\sum (x-1.79)^2p_x$, trusting you got the $1.79$ right. $\endgroup$ – lulu Sep 21 '16 at 15:15
  • $\begingroup$ In #1 I guess you must assume a student is either an art major or a science major. In #2 your mean sounds reasonable (though I did not check it). Your mean is towards the middle but biased a bit to the left. Your variance is wrong, which is probably caused by using $x$ as your index. The $E [X]$ in all four terms of the variance is the expected value you computed before i.e. 1.79. Similarly you should write $p_X (x) $. $\endgroup$ – Ian Sep 21 '16 at 15:15
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    $\begingroup$ Worth noting: your probabilities do not add to $1$, they add to $.89$ ... Presumably that's because there is a non-zero probability that there are $0$ Arts majors. $\endgroup$ – lulu Sep 21 '16 at 15:19
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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.

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  • $\begingroup$ What does "expected value" mean here? The number of students expected to have an art major? $\endgroup$ – basil Sep 21 '16 at 17:08
  • $\begingroup$ Yes, that is correct. $\endgroup$ – K. Brix Sep 21 '16 at 18:17

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