Question

If X is a uniformly distribution random variabe whose possible value are 5,6,7,8,9 are, find the mean and variance of X

Answer 1

There are 5 values that \(X\) can take on, which means each event has a \(\dfrac{1}{5}\) probability of occurring (because it's uniformly distributed).

The mean will be
\[E(X)=\large\sum_{x=5}^{9}x~p(x)=\frac{1}{5}\sum_{x=5}^9 x\]
and the variance is given by
\[V(X)=E(X^2)-[E(X)]^2=\large\frac{1}{5}\sum_{x=5}^9x^2-\left(\frac{1}{5}\sum_{x=5}^9 x\right)^2\]

Answer 2

I sse thx @SithsAndGiggles