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QB_EM_ADV_069
Question Bank • QB_EM_ADV_069
NAT • 2 Marks
Section 1: Engineering Mathematics
Probability and Statistics
Subtopic: Random variables
Question Statement:
A continuous random variable $X$ has probability density function (PDF):
$f(x) = \begin{cases} c(1 - x^2), & 0 \le x \le 1 \\ 0, & \text{otherwise} \end{cases}$
The expected value $E[X]$ is ________ (round off to three decimal places).
Step-by-Step Solution & Explanation
**Step 1: Normalization to find $c$**
$\int_0^1 c(1 - x^2) \, dx = 1 \implies c \left[ x - \frac{x^3}{3} \right]_0^1 = c\left(1 - \frac{1}{3}\right) = \frac{2}{3}c = 1 \implies c = \frac{3}{2}$
**Step 2: Compute expectation $E[X]$**
$E[X] = \int_{-\infty}^\infty x f(x) \, dx = \frac{3}{2} \int_0^1 x(1 - x^2) \, dx = \frac{3}{2} \int_0^1 (x - x^3) \, dx$
$= \frac{3}{2} \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_0^1 = \frac{3}{2} \left( \frac{1}{2} - \frac{1}{4} \right) = \frac{3}{2} \left( \frac{1}{4} \right) = \frac{3}{8} = 0.375$
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