X= -3 -1 1 3
P(X)= .1 .2 .4 .3
2006-12-18 05:20:58 · 5 answers · asked by Shamia B 1 in Science & Mathematics ➔ Mathematics
When you have a sequence of values and their probabilities, you multiply the probabilities times their values and add them up.
For x = -3, -1, 1, 3 and respective probabilities of .1, .2, .4, .3, the expected value is:
E(x) = (-3 x .1) + (-1 x .2) + (1 x .4) + (3 x .3)
E(x) = -.3 + -.2 + .4 + .9
E(x) = .8
So the expected value is .8
2006-12-18 05:36:45 · answer #1 · answered by Puzzling 7 · 1⤊ 0⤋
Expected value is the sum of the values times probabilities:
-3*.1+-1*.2+1*.4+3*.3=.8
What this means: if you had such a random variable (a strange die, whatever) and you sampled it (rolled the die) many times (100, 1000, whatever) the average of all those values will be close to .8
You would need the deviation and the central limit theorem to figure out what "close" means.
2006-12-18 05:35:56 · answer #2 · answered by a_math_guy 5 · 0⤊ 0⤋
Everyone's answer basically says the same thing: apply the definition of expected value.
2006-12-18 06:00:25 · answer #3 · answered by modulo_function 7 · 0⤊ 0⤋
so, just X*P(X), then sum them:
-.3 + -.2 + .4 + .9 = 0.8
2006-12-18 05:34:27 · answer #4 · answered by TankAnswer 4 · 1⤊ 0⤋
Add up the x*P(x)'s and you'll find out!
2006-12-18 05:23:42 · answer #5 · answered by Anonymous · 1⤊ 0⤋