$f ( x ) = \left\{ \begin{array} { l l }
\frac { 1 } { 6 } & \text { if } 2 \leq x \leq 8 \
0 & \text { otherwise }
\end{array} \right.$ is a probability density function for a continuous random variable X. The expected value is 5.

Question

$f ( x ) = \left\{ \begin{array} { l l } 
\frac { 1 } { 6 } & \text { if } 2 \leq x \leq 8 \
0 & \text { otherwise }
\end{array} \right.$ is a probability density function for a continuous random variable X. The expected value is 5.

Quizplus · Accepted Answer

The expected value is not 5. The expected value is $$E(X) = \int_{-\infty}^{\infty} x f(x) dx = \int_{2}^{8} x \frac{1}{6} dx = \frac{1}{6} \left[ \frac{x^2}{2} \right]_{2}^{8} = \frac{1}{6} \left( \frac{64}{2} - \frac{4}{2} \right) = \frac{1}{6} (30) = 5$$

f(x)={16 if 2≤x≤80 otherwise f ( x ) = \left\{ \begin{array} { l l } \frac { 1 } { 6 } & \text { if } 2 \leq x \leq 8 \\0 & \text { otherwise }\end{array} \right.f(x)={61​0​ if 2≤x≤8 otherwise ​

$f ( x ) = \left\{ \begin{array} { l l } \frac { 1 } { 6 } & \text { if } 2 \leq x \leq 8 \\0 & \text { otherwise }\end{array} \right.$