Question

Find the mean and the variance of the random variable X with probability function that corresponds to uniform distribution on [0, 4].

Answer 1

Because of the uniform distribution the mean is the middle value, i.e. 2

Answer 2

That probability function would be
\[P(x) = 1/4\]
So the mean would be
\[\bar{x}= \int_0^4xP(x)dx = \frac{1}{4}\int_0^4xdx = 2\]
(of course) and the variance is
\[\sigma^2= \int_0^4x^2P(x)dx = \frac{1}{4}\int_0^4x^2dx = \frac{16}{3}\]

Answer 3

Ahhh, I see now. Thanks you guys! I appreciate it.

Answer 4

that is not how you calculate the variance.

Answer 5

\[\sigma^2=\int\limits_\Omega(x-\mu)^2 f(x)dx\]

Answer 6

\[\sigma^2=\int\limits_{0}^4(x-2)^2 \frac{1}{4}dx\]

Answer 7

\[=\frac{4}{3}\]

Answer 8

you can also do it this way

\[\sigma^2=\mathbb{E}(X^2)-\mathbb{E}(X)^2=\frac{16}{3}-2^2=\frac{4}{3}\]

Answer 9

or just use the formula ..\[\frac{(b-a)^2}{12}\]

\[\frac{(4-0)^2}{12}=\frac{16}{12}=\frac{4}{3}\]

Answer 10

Oh, I see now. Thanks for the correction.

Answer 11

no problem

Answer 12

Oh man, brain fart. Sorry about that, thanks Zarkon :)