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Chapter 12: Problem 5

Find the expected value and variance for each random variable whose probability density function is given. When computing the variance, use formula (5). \(f(x)=5 x^{4}, 0 \leq x \leq 1\)

Short Answer

Expert verified
Expected value \( E(X) = \frac{5}{6} \). Variance \( \text{Var}(X) = \frac{5}{252} \).

Step by step solution

01

- Understand the Probability Density Function (PDF)

The given probability density function (PDF) is \[ f(x) = 5x^4, \quad 0 \leq x \leq 1 \].This function defines how the probability is distributed over the interval \( [0, 1] \).
02

- Verify PDF Validity

For a function to be a valid PDF, the total area under the curve must be 1. Calculate the integral of \(f(x)\):\[ \int_{0}^{1} 5x^4 \, dx \].Compute it:\[ \int_{0}^{1} 5x^4 \, dx = 5 \left[ \frac{x^5}{5} \right]_0^1 = 1 \].Since the integral is 1, \( f(x) \) is a valid PDF.
03

- Find the Expected Value (Mean)

The expected value \( E(X) \) of a continuous random variable with PDF \( f(x) \) is defined as:\[ E(X) = \int_{-\infty}^{\infty} x f(x) \, dx \].Substitute the given \( f(x) \):\[ E(X) = \int_{0}^{1} x \cdot 5x^4 \, dx \].Compute:\[ E(X) = \int_{0}^{1} 5x^5 \, dx = 5 \left[ \frac{x^6}{6} \right]_0^1 = \frac{5}{6} \].
04

- Calculate the Second Moment

To calculate the variance, first find the second moment \( E(X^2) \):\[ E(X^2) = \int_{-\infty}^{\infty} x^2 f(x) \, dx \].Substitute the given \( f(x) \):\[ E(X^2) = \int_{0}^{1} x^2 \cdot 5x^4 \, dx \].Compute:\[ E(X^2) = \int_{0}^{1} 5x^6 \, dx = 5 \left[ \frac{x^7}{7} \right]_0^1 = \frac{5}{7} \].
05

- Calculate the Variance

Use the variance formula\[ \text{Var}(X) = E(X^2) - (E(X))^2 \].Substitute \( E(X) = \frac{5}{6} \) and \( E(X^2) = \frac{5}{7} \):\[ \text{Var}(X) = \frac{5}{7} - \left( \frac{5}{6} \right)^2 = \frac{5}{7} - \frac{25}{36} \].Find a common denominator to subtract the fractions:\[ \text{Var}(X) = \frac{180}{252} - \frac{175}{252} = \frac{5}{252} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Density Function
A probability density function (PDF) gives the probability that a random variable takes a value within a particular range. In this exercise, the given PDF is \[ f(x) = 5x^4 \, \ 0 \leq x \leq 1 \]. This tells us that the probability is distributed according to this function over the interval \[ [0, 1] \]. The area under the curve of the PDF over its entire range must equal 1. We verify it by calculating the integral of \ f(x) \ from 0 to 1: \[ \int_{0}^{1} 5x^4 \ dx = 1 \]. Since the area is 1, \ f(x) \ is a valid PDF.
Expected Value
The expected value, or mean, of a continuous random variable is a measure of its central tendency. It indicates where the average value of the variable lies when considering the PDF. Mathematically, the expected value is given by: \[ E(X) = \int_{-\infty}^{\infty} x f(x)\ dx \]. For the given PDF, the expected value is calculated as: \[ E(X) = \int_{0}^{1} x \cdot 5x^4 \ dx = \int_{0}^{1} 5x^5 \ dx = \ \frac{5}{6} \]. This means that on average, the value of \ X \ lies closer to \ \frac{5}{6} \ within the range of 0 to 1.
Variance Calculation
Variance is a measure of how much the values of a random variable deviate from the expected value. It indicates the spread or dispersion of the values. To find the variance, we first calculate the second moment, \ E(X^2) \,: \[ E(X^2) = \int_{0}^{1} x^2 \cdot 5x^4 \ dx = \int_{0}^{1} 5x^6 \ dx = \ \frac{5}{7} \]. Using the formula for variance, we subtract the square of the expected value from the second moment: \[ \text{Var}(X) = E(X^2) - (E(X))^2 \]. Substituting the values, we get: \[ \text{Var}(X) = \ \frac{5}{7} - \left( \frac{5}{6} \right)^2 = \ \frac{5}{7} - \ \frac{25}{36} = \ \frac{5}{252} \]. Thus, the variance of \ X \ is \ \frac{5}{252} \.
Second Moment
The second moment of a random variable is the expected value of the square of the variable, \ E(X^2) \,. It is used to measure the spread of the variable around its mean. For the given PDF, we calculate the second moment as follows: \[ E(X^2) = \int_{0}^{1} x^2 \cdot 5x^4 \ dx = \int_{0}^{1} 5x^6 \ dx = \ \frac{5}{7} \]. This value is crucial for calculating the variance and understanding the distribution of \ X \. A higher second moment indicates that values of \ X \ are typically further from the mean.

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Most popular questions from this chapter

Diameter of a Bolt A certain type of bolt must fit through a 20 -millimeter test hole or else it is discarded. If the diameters of the bolts are normally distributed, with \(\mu=18.2\) millimeters and \(\sigma=.8\) millimeters, what percentage of the bolts will be discarded?

If \(X\) is a random variable with density function \(f(x)\) on \(A \leq x \leq B\), the median of \(X\) is that number \(M\) such that $$ \int_{A}^{M} f(x) d x=\frac{1}{2} $$ In other words, \(\operatorname{Pr}(X \leq M)=\frac{1}{2}\). Find the median of the random variable whose density function is \(f(x)=2(x-1), 1 \leq x \leq 2\).

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Find the expected value and variance for each random variable whose probability density function is given. When computing the variance, use formula (5). \(f(x)=\frac{3 \sqrt{x}}{16}, 0 \leq x \leq 4\)
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