Calculate Expected Value and Variance of Continuous Random Variable

Math Problem Statement

Define a continuous random variable by the following probability distribution function:

f(x)=f(x)= 7929 x27929 x2 for 1≤x≤1≤x≤ 79507950 ,

and f(x)=0f(x)=0 for all other xx.

(i) Find E[X]E[X], giving the answer accurate to at least 2 decimal places. (ii) Find E[X2]E[X2], giving the answer accurate to at least 2 decimal places.

(iii) Hence find Var(X)Var(X), giving the answer accurate to at least 2 decimal places.

Solution

Let's work through the problem systematically, calculating the expected values and the variance for the given probability density function (PDF).

Given:

The probability density function (PDF) is: $f(x) = \frac{79}{29} x^2 \quad \text{for } 1 \leq x \leq \frac{79}{50}$ $f(x) = 0 \quad \text{for all other } x$

Part (i) - Finding $E[X]$ :

The expected value $E[X]$ for a continuous random variable ( X \

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Math Problem Analysis

Mathematical Concepts

Probability
Continuous Random Variables
Expected Value
Variance

Formulas

Probability Density Function (PDF)
Expected Value (E[X])
Variance (Var(X))

Theorems

Suitable Grade Level

Advanced undergraduate level

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