2. Problem 2: Let $ X $ be a continuous random variable with the following probability density function\[f(x) = \begin{cases} 3x^2, & \text{for } x \in [0, 1] \\0, & \text{otherwise}\end{cases}\]- Compute $ E[X] $ and $ Var(X) $- Compute $ P[0.25 \leq X \leq 0.75] $

PDF of X is given by, f(x) = 3x2for x∈[0,1]0otherwise So, E(X)=∫-∞∞x f(x)dx=∫01x 3x2…

Answered: Problem 2: Let X be a continuous random…

Problem 2: Let X be a continuous random variable with the followi probability density function S3a², for æ € [0, 1] 0 otherwise f(x) = • Compute E[X] and Var(X) • Compute P[0.25 < X < 0.75]

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Question

2. Problem 2: Let $ X $ be a continuous random variable with the following probability density function

\[
f(x) =
\begin{cases}
3x^2, & \text{for } x \in [0, 1] \\
0, & \text{otherwise}
\end{cases}
\]

- Compute $ E[X] $ and $ Var(X) $
- Compute $ P[0.25 \leq X \leq 0.75] $

Expert Solution

Step 1

PDF of X is given by,

$f (x) = \{\begin{cases} 3 x^{2} & f o r x \in [0, 1] \\ 0 & o t h e r w i s e \end{cases}$

So,

$E (X) = \int_{- \infty}^{\infty} x f (x) d x = \int_{0}^{1} x 3 x^{2} d x = 3 \int_{0}^{1} x^{3} d x = 3 \times {[\frac{x^{4}}{4}]}_{0}^{1} = 3 \times \frac{1}{4} = 0.75$

Now,

$E (X^{2}) = \int_{0}^{1} x^{2} 3 x^{2} d x = 3 \int_{0}^{1} x^{4} d x = 3 \times {[\frac{x^{5}}{5}]}_{0}^{1} = \frac{3}{5} = 0.6$

We know,

Var(X)

= E(X²) - E(X)²

=0.6 - 0.75²

=0.0375

Step by step

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