Question

The mean of discrete random variance variable X is 19.59. What is the variance?

Answer 1

@camerondoherty

Answer 2

mean = np

variance = np(1-p)

but there are way to meany versions of np for the factors of 19.59 to determine that with from what i see ...

Answer 3

Do you know what kind of random variable it is?

Answer 4

lol, its just some random variable you find lying in the street :)

Answer 5

(x,p(x)) = (12,0.07) (15,0.21) (17,0.17) (20,0.25) (22,0.05) (24,0.04) (25,0.13) (30,0.08)

Answer 6

Well what is it's distribution. We can't assume the mean is np unless it's binomial

Answer 7

discrete random variable

Answer 8

hmm, i do like making assumptions, but ill concede this point

Answer 9

\[ E(X^2)=\sum_{x}x^2p(x)\]
and
\[Var(X)=E(X^2)-[E(X)]^2 \]

Answer 10

\[E(X)=\sum xp(x)\]
\[VAR(X)=\sum [x-E(X)]^2p(x)\]

Answer 11

yeah, something like that :)

Answer 12

they are the same :)

Answer 13

It says...The following table shows the probability distribution for a discrete variable. Then the chart i listed above, and then the question i asked

Answer 14

we have presented you with a few ways to approach the solution .... any questions about these methods?

Answer 15

I have no idea what that formula means

Answer 16

or what to do with it?

Answer 17

have you read your lesson materials?

Answer 18

There is no lesson material

Answer 19

it is an online credit recovery program for highschool and it just gives quizzes from algebra 2 that i am supposed to know how to do already

Answer 20

can you use excel?

Answer 21

the process is simple enough, its just time consuming with so many data points

Answer 22

They give you a pair of (x, p(x)), so to find
\[E(X^2)=\sum_{x}x^2p(x)\], you multiply the square of the x-value with p(x). Do this for every pair they gave you, then add them up.
\[E(X^2)=12^2(0.07)+15^2(0.21)+...+30^2(0.08) \]

The the variance is
\(Var(X)=E(X^2)-[E(X)]^2\). Now \(E(X)\) is just the mean that they gave you above (19.59)
So once you calculate \(E(X^2)\) from above, just do:
\(Var(X)=E(X^2)-(19.59)^2\)