Question

Consider the continuous random variable x, which has a uniform distribution over the interval from 10 to 18. the mean of x is _____. a. 13 b. 0 c. 14 d. 0.125

Answer 1

The PDF of
`X` is

`f_X(x) = \begin{cases} \frac1{18-10} = \frac18 & \text{if } 10 \le x \le 18 \\\\ 0 & \text{otherwise} \end{cases}`

Then the mean is

`\displaystyle \int_(-\infty)^\infty x f_X(x) \, dx = \frac18 \int_(10)^(18) x \, dx = \frac1{16} (18^2 - 10^2) = \boxed{14}`