Question

ok! If I have a table with X & P(x), how would I find the expected value??

x| 23|25|26|31|34|38
P(X)| 0.16|.09|0.18|0.12|0.24|0.21

Answer 1

@FutureMathProfessor

Answer 2

@campbell_st @Loujoelou

Answer 3

@mathmate

Answer 4

would you do:
X * P(X) and do it for all the values and ADD THEM all together to get the expected value??

Answer 5

HELP guys?

Answer 6

well my understanding of expected value from probability is

sum x*P(x) so

23*0.16 + 25*0.09 + .....

Answer 7

yea that's what I was asking. so is that how u do it?

Answer 8

cuz that's how I got 30.47

Answer 9

as its shown above

I haven't done the calculation, you are summing 6 terms

and doing a quick approximation, your answer seems reasonable

Answer 10

ok

Answer 11

thanks

Answer 12

also:

Answer 13

@FutureMathProfessor

Answer 14

Conditional probability:
A=first throw, odd
B=second throw, 1 or 6
\( P(B|A)=P(B\cap A) / P(A) \)
You can see from the contingency table that there are 6 outcomes for \( A \cap B \) and 18 outcomes for A, so the probability is (6/36)/(18/36)=1/3