Math Problem Statement

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

f(x)= 7929 x2 for 1≤x≤ 7950 ,

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

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

1.30

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

1.64

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

−0.05

Solution

Let's analyze the problem and calculate the required values step by step.

Given Information:

The probability density function (PDF) is defined as:

7929x^2 & \text{for } 1 \leq x \leq 7950 \\ 0 & \text{for all other } x \end{cases}$$ ### (i) Expected Value $$E[X]$$ The expected value $$E[X]$$ of a continuous random variable $$X$$ with probability density function $$f(x)$$ is given by: $$E[X] = \int_{-\infty}^{\infty} x f(x) \, dx$$ For the given PDF, this simplifies to: $$E[X] = \int_{1}^{7950} x \cdot 7929x^2 \, dx$$ Simplifying further: $$E[X] = 7929 \int_{1}^{7950} x^3 \, dx$$ Now, let's calculate this integral. ### (ii) $$E[X^2]$$ Similarly, the expected value $$E[X^2]$$ is: $$E[X^2] = \int_{-\infty}^{\infty} x^2 f(x) \, dx$$ This also simplifies to: $$E[X^2] = 7929 \int_{1}^{7950} x^4 \, dx$$ ### (iii) Variance $$\text{Var}(X)$$ The variance $$\text{Var}(X)$$ is given by: \[ \text{Var}(X) = E[X^2] - (E[X])^2 \] Let's compute these integrals and then calculate the variance.Here are the results of the calculations: (i) The expected value $$E[X]$$ is approximately $$7.92 \times 10^{18}$$. (ii) The expected value $$E[X^2]$$ is approximately $$5.04 \times 10^{22}$$. (iii) The variance $$\text{Var}(X)$$ is calculated to be approximately $$-6.27 \times 10^{37}$$. However, this negative variance result is physically impossible, as variance should always be non-negative. This suggests there might be an error in the problem formulation, particularly in the setup of the probability density function. If you have any other questions or would like to explore further, let me know! Here are some related questions to explore: 1. How is the expected value $$E[X]$$ interpreted in real-world contexts? 2. What does a negative variance imply in mathematical terms? 3. How can the variance of a random variable be re-calculated if errors are suspected in the initial formulation? 4. What are common errors that might lead to an impossible variance result like a negative value? 5. How would the shape of the probability density function affect $$E[X]$$ and

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Continuous Random Variables
Expected Value
Variance

Formulas

Expected Value formula: E[X] = ∫ x f(x) dx
Variance formula: Var(X) = E[X^2] - (E[X])^2

Theorems

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Suitable Grade Level

Undergraduate