Question

A continuous random variable \(X\) has a uniform distribution if it has a probability density function of the form $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{b-a} & \text { if } a

Short answer

(a) Integral equals 1, (b) Mean: \( \frac{a+b}{2} \), Variance: \( \frac{(b-a)^2}{12} \), (c) Probability: 0.2.

Step-by-step solution

Verify Probability Density Function Normalization

To ensure the function is a legitimate probability density function, we need to confirm that the entire area under the curve equals 1. For a uniform distribution, this means checking that \( \int_{a}^{b} \frac{1}{b-a} \, dx = 1 \). Calculating, we find:\[\int_{a}^{b} \frac{1}{b-a} \, dx = \frac{1}{b-a} \cdot (b-a) = 1.\]Thus, the probability density function is normalized.

Calculate the Mean

The mean \( \mu \) of a uniform distribution \( X \sim U(a, b) \) is the midpoint of the interval \([a, b]\). It is given by:\[\mu = \frac{a + b}{2}.\]

Calculate the Variance

The variance \( \sigma^2 \) of a uniform distribution \( X \sim U(a, b) \) is calculated using the formula:\[\sigma^2 = \frac{(b-a)^2}{12}.\]

Calculate the Probability for Specific Values of a and b

Given that \( a = 0 \) and \( b = 10 \), the probability that \( X < 2 \) is the area under the density function from \( 0 \) to \( 2 \).The probability is calculated as:\[P(X < 2) = \int_{0}^{2} \frac{1}{10} \, dx = \frac{1}{10} \cdot 2 = \frac{2}{10} = 0.2.\]

Key concepts

Probability Density Function

The Probability Density Function (PDF) is a fundamental concept in probability and statistics, responsible for describing the likelihood of a continuous random variable falling within a certain range of values. For a uniform distribution, which is one of the simplest forms of probability distributions, the PDF is particularly straightforward. The uniform distribution over an interval \(a, b\) is represented by the following PDF: \[ f(x)=\begin{cases} \frac{1}{b-a} & \text{if } a < x < b \ 0 & \text{otherwise} \end{cases} \]
This means that the probability of the variable taking on any value within this interval is evenly distributed, as indicated by the constant height. In order for a PDF to be valid, it must be normalized. This requires that the total area under the curve (PDF) equals 1. For our uniform distribution, this can be verified by integrating the PDF over the interval \(a, b\), confirming that the area is indeed 1. Thus, we can conclude that the function describes a legitimate probability distribution.

Mean of Uniform Distribution

The mean of a uniform distribution is often referred to as the average or expected value. For a uniform distribution, this average is intuitively the midpoint of the interval \(a, b\).
The formula for the mean \(\mu\) is given by: \[ \mu = \frac{a + b}{2} \] This formula expresses the point exactly halfway between \(a\) and \(b\). This reflects the symmetry of the uniform distribution, where each point between \(a\) and \(b\) is equally likely. Understanding the mean is crucial as it provides insights into the central tendency of the distribution. In practical applications, it helps us locate the "center" of all possible outcomes, representing where outcomes are likely to cluster.

Variance of Uniform Distribution

Variance is a measure of how much variability there is in a distribution. It helps us understand the spread or dispersion of the data points around the mean. For a uniform distribution, this dispersion is fixed across the interval. The formula to determine variance \(\sigma^2\) for a uniform distribution is: \[ \sigma^2 = \frac{(b-a)^2}{12} \] This expression gives us the average of the squared deviations from the mean. The factor of 12 in the denominator arises from the arithmetic calculations performed over the continuous and evenly distributed data points between \(a\) and \(b\). An interesting observation is that as the range \(b-a\) increases, so does the variance, indicating that the outcomes are more spread out.

Probability Calculation

Probability calculations in the context of a uniform distribution involve determining the likelihood of the variable falling within a particular subinterval. Since each point in the interval \(a, b\) has the same probability density, calculating probabilities simplifies to computing areas under the constant PDF. For example, if we need to find the probability that a random variable \(X < c\) within \(a, b\), we calculate:
\[ P(X < c) = \int_a^c \frac{1}{b-a} \, dx = \frac{c-a}{b-a} \]
In scenarios like our exercise, where \(a = 0\) and \(b = 10\), to find the probability that the variable is less than 2, we evaluate:
\[ P(X < 2) = \int_0^2 \frac{1}{10} \, dx = 0.2 \] This calculation shows how probabilities are obtained in a uniform setting, highlighting the simplicity and practicality of the uniform distribution for modeling equal-likelihood outcomes across a defined range.