Question

The following table shows the probability distribution for a discrete random variable.
X 11 14 16 19 21 23 24 29
P(X) 0.07 0.21 0.17 0.25 0.05 0.04 0.13 0.08

The mean of the discrete random variable X is 18.59. What is the variance of X? Round your answer to the nearest hundredth.

Answer 1

do you know how to find these things? if not, I can help.

Answer 2

yeah I just add them like this right?

\[11(0.07)+14(0.21)+16(0.17)+19(0.25+21(0.05)+23(0.04)+24(0.13)+29(0.08)\]

Answer 3

yeah... that's E(X). You need E(X^2)

Answer 4

\[E \left( X ^{2} \right)= 11^{2}\left( 0.07 \right)+14^{2}\left( 0.21 \right)+ ...\]

Answer 5

okay and do I subtract that from E(x)= 18.59

Answer 6

Then

\[Var \left( X \right)=E \left( X ^{2} \right)-\left[ E \left( X \right) \right]^{2}\]

Answer 7

\[E \left( X \right)=18.59\]

Answer 8

And then E(x^2)= 368.77 right?

Answer 9

or would it be E(x^2)= 23.1819

Answer 10

hold on... i stepped away for a few.

Answer 11

\[E \left( X ^{2} \right)=368.77 \text{, } \left[ E \left( X \right) \right]^{2}=18.59^{2}=345.5881\]

so
\[Var \left( X \right)= 368.77 - 345.5881 = 23.1819\]

Answer 12

perfect @pgpilot326