Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

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

Step 1: Verify the Probability Density Function

To solve part (a), we need to verify that the integral of the probability density function over all possible values equals 1.The function is defined as:\[f(x) = \begin{cases}\frac{1}{b-a}, & \text{if } a < x < b \0, & \text {if } x \leq a \text{ or } x \geq b\end{cases}\]We calculate \(\int_{-\infty}^{\infty} f(x) \, dx\) by splitting it into three ranges:1. For \(x < a\), \(f(x) = 0\).2. For \(a \leq x \leq b\), \(f(x) = \frac{1}{b-a}\).3. For \(x > b\), \(f(x) = 0\).Thus, the integral becomes:\[\int_{-\infty}^{\infty} f(x) \, dx = \int_{a}^{b} \frac{1}{b-a} \, dx = \frac{1}{b-a} \int_{a}^{b} 1 \, dx = \frac{1}{b-a} \times (b-a) = 1.\]Therefore, the integral of the probability density function is 1.

Step 2: Calculate the Mean

To find the mean \( \mu \) of a uniform distribution, we calculate the expected value:\[\mu = E[X] = \int_{a}^{b} x f(x) \, dx = \int_{a}^{b} x \frac{1}{b-a} \, dx.\]\[\mu = \frac{1}{b-a} \int_{a}^{b} x \, dx = \frac{1}{b-a} \left[ \frac{x^2}{2} \right]_{a}^{b} = \frac{1}{b-a} \left( \frac{b^2}{2} - \frac{a^2}{2} \right).\]\[\mu = \frac{1}{2(b-a)} (b^2 - a^2) = \frac{1}{2} (b + a).\]Thus, the mean \( \mu \) is \( \frac{a+b}{2} \).

Step 3: Calculate the Variance

The variance \( \sigma^2 \) is given by the expectation of the squared deviations from the mean:\[\sigma^2 = E[X^2] - (E[X])^2.\]Firstly, calculate \( E[X^2] \):\[E[X^2] = \int_{a}^{b} x^2 \frac{1}{b-a} \, dx = \frac{1}{b-a} \left[ \frac{x^3}{3} \right]_{a}^{b} = \frac{1}{b-a} \left( \frac{b^3}{3} - \frac{a^3}{3} \right).\]Now, substitute the expectation \( E[X] = \frac{a+b}{2} \) into the variance formula:\[\sigma^2 = \frac{1}{b-a} \left( \frac{b^3}{3} - \frac{a^3}{3} \right) - \left(\frac{b+a}{2}\right)^2.\]Simplify to find:\[\sigma^2 = \frac{b^2 + a^2 - 2ab}{12} = \frac{(b-a)^2}{12}.\]

Step 4: Calculate the Probability for a Given Interval

For part (c), the problem is defined with \( a = 0 \) and \( b = 10 \), and asks for the probability that \( X < 2 \). With these bounds, the function becomes:\[f(x) = \frac{1}{10-0} = \frac{1}{10} \text{ for } 0 < x < 10.\]Calculate the probability:\[P(X < 2) = \int_{0}^{2} \frac{1}{10} \, dx = \frac{1}{10} \times (2 - 0) = \frac{2}{10} = 0.2.\]

Step 5: Conclusion

The uniform distribution has been understood and computed for the given conditions. For a uniform distribution on \([a, b]\), the probability density function's integral over the entire range is 1. The mean is \( \frac{a+b}{2} \) and the variance is \( \frac{(b-a)^2}{12} \). With \( a = 0 \) and \( b = 10 \), the probability that \( X < 2 \) is 0.2.

Key Concepts

Continuous Random Variable

A continuous random variable is a type of random variable that can take any numerical value in a given range. Unlike discrete random variables, which can only take specific values, continuous random variables have an infinite number of possible values within a given interval.
A great example of a continuous random variable is the time it takes for a person to run a mile. This time can be any positive real number, meaning it can take any value within a continuous range.
For continuous random variables, we use the probability density function (PDF) to find probabilities of different outcomes. The PDF helps us understand how probabilities are distributed over the continuum of possible outcomes in a specific range.

Probability Density Function

The probability density function (PDF) of a continuous random variable defines the likelihood of the variable to take on a particular value within a given range. Since continuous random variables can assume an infinite number of values, the probability of a specific outcome is zero. Instead, we calculate the probability over an interval.
The PDF for a uniformly distributed continuous random variable is constant within a specified range from \(a\) to \(b\), and zero outside this interval. In mathematical terms, for a uniform distribution, the PDF is defined as:

• \(f(x) = \frac{1}{b-a}\) for \(a < x < b\)
• \(f(x) = 0\) for \(x \leq a\) or \(x \geq b\)

The PDF must satisfy the condition that the total area under the curve is equal to 1, which represents the total probability of all possible outcomes.

Mean and Variance

In probability theory and statistics, the mean and variance are fundamental concepts that describe the distribution of a random variable. For a uniform distribution, the mean and variance give us information about the center and the spread of values, respectively.
The mean \(\mu\) of a uniform distribution, defined on the interval \([a, b]\), is calculated as the average of the two boundaries:\[\mu = \frac{a+b}{2}.\]This formula indicates that the mean is the middle point of the interval.
The variance \(\sigma^2\) measures how much the values of the random variable differ from the mean. For a uniform distribution, the variance is given by:\[\sigma^2 = \frac{(b-a)^2}{12}.\]A smaller variance indicates that the data values are close to the mean, while a larger variance indicates that the data values are more spread out.

Probability Calculation

Probability calculations using a probability density function involve determining the area under the curve of the PDF within a specific interval. For a continuous random variable with a uniform distribution, this is relatively straightforward as the PDF is constant in the given interval.
To find the probability that a random variable \(X\) is less than a certain value within the interval \([a, b]\), for instance, we compute:\[P(X < c) = \int_{a}^{c} \frac{1}{b-a} \, dx = \frac{c-a}{b-a}.\]For practical applications such as determining the probability that \(X\) is less than 2 when \(a = 0\) and \(b = 10\):

• The PDF becomes \(f(x) = \frac{1}{10}\).
• Calculate \(P(X < 2) = \int_{0}^{2} \frac{1}{10} \, dx = \frac{2-0}{10} = 0.2\).

This probability of 0.2 means there's a 20% chance that \(X\) takes a value less than 2.