Question

The following table shows the probability distribution for a discrete random variable.
X- 12,15,17,20,22,24,25,30
P(X)- 0.07,0.21,0.17,.025,0.05,0.04,0.13,0.08

The mean of the discrete random variable X is 19.59. What is the variance of X? Round your answer to the nearest hundredth.

Answer 1

n*p*q ?? or is that for a binomial

Answer 2

I have no idea that is all the pieces of problem i got,..

Answer 3

http://berlin.csie.ntnu.edu.tw/Courses/Probability/2011Lectures/PROB2011F_Lecture-06-Discrete%20Random%20Variables%20-Expectation,%20Mean%20and%20Variance.pdf

im reading thru this to refresh my memory

Answer 4

\[var(X)=\sum_{n}{(n-E(x))^2*p_X(n)}\] ..... whatever that means

Answer 5

please do not bypass the filters in order to use profanity. We try to keep this a family freindly place to interact in.

Answer 6

http://www.statisticslectures.com/topics/variancestandarddeviationdiscrete/

this looks more readable to me

Answer 7

lets use this setup to see what we can come up with:

x P(x) x^2 x^2*P(x)
12 0.07 144
15 0.21 225
17 0.17 289
20 .025 400
22 0.05 484
24 0.04 576
25 0.13 625
30 0.08 900


is that P(20) correct?

Answer 8

that looks good

Answer 9

0.07* 144 +0.21* 225 +0.17* 289 + .025* 400
+ 0.05* 484+ 0.04* 576+ 0.13* 625+ 0.08* 900 = 316.95

if i assume P(20) is a typo i get


0.07* 144 +0.21* 225 +0.17* 289 + 0.25* 400
+ 0.05* 484+ 0.04* 576+ 0.13* 625+ 0.08* 900 = 406.95


but im not sure if we are done lol

Answer 10

looks like we need to subtract the square of the E(x) from that as well

Answer 11

19.56^2 = 382.5936

so subtract that from the correct version of P(20) and it should be good

Answer 12

yea. it was 0.25

Answer 13

whew!! good :)

the process seems a bit long and cumbersome; but a calculator should help

406.95
- 382.5936
-----------
var(x)

Answer 14

24.356

Answer 15

with any luck, yes

Answer 16

outstanding.

Answer 17

dbl chk my math to be sure ;)