Chapter 5: Problem 6
For each probability density function, over the given interval, find \(\mathrm{E}(x), \mathrm{E}\left(x^{2}\right),\) the mean, the variance, and the standard deviation. $$ f(x)=\frac{1}{4} x, \quad[1,3] $$
Short Answer
Expert verified
\( \mathrm{E}(x) = \frac{13}{6}, \mathrm{E}(x^2) = 5, \text{Var}(x) = \frac{11}{36}, \sigma(x) = \frac{\sqrt{11}}{6} \).
Step by step solution
01
Calculate the Expectation E(x)
To find the expectation \( \mathrm{E}(x) \), we use the formula for the expected value of a continuous random variable:\[ \mathrm{E}(x) = \int_{a}^{b} x \, f(x) \, dx \] where \( f(x) = \frac{1}{4}x \) and the interval is \([1,3]\). Thus, \[ \mathrm{E}(x) = \int_{1}^{3} x \, \frac{1}{4}x \, dx = \frac{1}{4} \int_{1}^{3} x^2 \, dx \]. Calculating this integral, we get \[ \mathrm{E}(x) = \frac{1}{4} \left[ \frac{x^3}{3} \right]_{1}^{3} = \frac{1}{4} \left( \frac{27}{3} - \frac{1}{3} \right) = \frac{1}{4} \times \frac{26}{3} = \frac{13}{6} \].
02
Calculate the Expectation E(x²)
For \( \mathrm{E}(x^2) \), we apply the expected value formula again:\[ \mathrm{E}(x^2) = \int_{a}^{b} x^2 \, f(x) \, dx \] with \( f(x) = \frac{1}{4}x \). Hence, \[ \mathrm{E}(x^2) = \int_{1}^{3} x^2 \times \frac{1}{4}x \, dx = \frac{1}{4} \int_{1}^{3} x^3 \, dx \]. Calculating the integral gives us \[ \frac{1}{4} \left[ \frac{x^4}{4} \right]_{1}^{3} \]. Therefore, \[ \mathrm{E}(x^2) = \frac{1}{4} \left( \frac{81}{4} - \frac{1}{4} \right) = \frac{1}{4} \times \frac{80}{4} = 5 \].
03
Calculate the Variance Var(x)
The variance \( \mathrm{Var}(x) \) is defined as \( \mathrm{E}(x^2) - (\mathrm{E}(x))^2 \). We've calculated \( \mathrm{E}(x^2) = 5 \) and \( \mathrm{E}(x) = \frac{13}{6} \). Thus,\[ \mathrm{Var}(x) = 5 - \left( \frac{13}{6} \right)^2 = 5 - \frac{169}{36} \]. Converting \(5\) to a fraction with a denominator of \(36\), we get \[ 5 = \frac{180}{36} \]. So, \[ \mathrm{Var}(x) = \frac{180}{36} - \frac{169}{36} = \frac{11}{36} \].
04
Calculate the Standard Deviation (x)
The standard deviation \( \sigma(x) \) is the square root of the variance. Hence,\[ \sigma(x) = \sqrt{\mathrm{Var}(x)} = \sqrt{\frac{11}{36}} = \frac{\sqrt{11}}{6} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Expected Value
The expected value, often denoted as \( \mathrm{E}(X) \), is a fundamental concept in probability and statistics, representing the average or mean value of a continuous random variable over an interval. Think of it as the center of gravity of the probability distribution—it gives us a notion of the long-term average outcome if we repeat the process many times. To find the expected value of a continuous random variable \( X \), we integrate the product of \( x \) and its probability density function (PDF) \( f(x) \) over the given interval \[ \mathrm{E}(x) = \int_{a}^{b} x \cdot f(x) \, dx \]. Here, our PDF is \( f(x) = \frac{1}{4}x \) and the interval is \([1, 3]\). After computing the integral, we determine that \( \mathrm{E}(x) = \frac{13}{6} \). This value represents the expected or average position \( X \) would take in many repetitions of the process.
Variance and Standard Deviation
Variance measures how much the values of a continuous random variable diverge from the expected value. In simpler terms, it quantifies the spread or dispersion of a set of data points. A high variance indicates that the data points are spread out over a wider range, while a low variance suggests they are clustered closely around the expected value. The variance is calculated as \[ \mathrm{Var}(X) = \mathrm{E}(x^2) - (\mathrm{E}(x))^2 \].
- First, compute \( \mathrm{E}(x^2) \) using the integral \( \int_{a}^{b} x^2 \cdot f(x) \, dx \), which turns out to be 5 in this exercise.
- With \( \mathrm{E}(x) = \frac{13}{6} \), we find that \( (\mathrm{E}(x))^2 = \frac{169}{36} \).
- This gives us a variance \( \mathrm{Var}(x) = \frac{180}{36} - \frac{169}{36} = \frac{11}{36} \).
Continuous Random Variable
A continuous random variable is a variable that can take any value within a given range. Unlike discrete variables, which have specific outcomes, continuous variables can assume a continuum of values. This range could be all numbers between 1 and 10, or any possible temperatures from freezing to boiling, for instance.
Some important characteristics of continuous random variables include:
- They are associated with a probability density function (PDF), which describes the probabilities of the outcomes—however, for continuous variables, it's the area under the PDF curve over an interval that gives a probability, not the curve's height at a single point.
- They often require integration over a given interval to calculate probabilities, expected values, or variances, since we're dealing with infinite possibilities within any finite interval.
- Continuous random variables are not limited to real-life measurements but are used in theoretical contexts to model phenomena where outcomes are too numerous or continuous for discrete probability models.