Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

A PDF for a continuous random variable \(X\) is given. Use the PDF to find (a) \(P(X \geq 2),(b) E(X)\), and \((c)\) the \(\mathrm{CDF}\). $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{40}, & \text { if }-20 \leq x \leq 20 \\ 0, & \text { otherwise } \end{array}\right. $$

Step-by-Step Solution

Answer

(a) \(P(X \geq 2) = \frac{9}{20}\), (b) \(E(X) = 0\), (c) CDF: \(F(x) = 0\) for \(x < -20\), \(\frac{x+20}{40}\) for \(-20 \leq x < 20\), \(1\) for \(x \geq 20\).

Step 1: Understanding the Probability Density Function (PDF)

The given PDF is a uniform distribution over the interval \(-20 \leq x \leq 20\) with a constant density of \(\frac{1}{40}\).This means the random variable \(X\) can take any value within this interval with equal probability, and outside this interval, the probability is zero.

Step 2: Calculating the Probability \(P(X \geq 2)\)

For \(X \geq 2\), we need the probability over the interval \([2, 20]\). Since the distribution is uniform, we calculate the length of this interval, and multiply by the constant density: \\[P(X \geq 2) = (20 - 2) \times \frac{1}{40} = \frac{18}{40} = \frac{9}{20}.\]

Step 3: Calculating the Expectation \(E(X)\)

For a uniform distribution, the expectation \(E(X)\) is the midpoint of the interval. Here, the interval is \([-20, 20]\), so the midpoint is:\[E(X) = \frac{-20 + 20}{2} = 0.\]

Step 4: Constructing the Cumulative Distribution Function (CDF)

The CDF \(F(x)\) is the integral of the PDF from the lower limit of the distribution to \(x\). For a uniform distribution on \([-20, 20]\):- For \(x < -20\), \(F(x) = 0\).- For \(-20 \leq x < 20\), \\[F(x) = \int_{-20}^{x} \frac{1}{40} \, dt = \frac{x + 20}{40}.\]- For \(x \geq 20\), \(F(x) = 1\).

Key Concepts

Probability Density Function (PDF)

A probability density function (PDF) describes the likelihood of a continuous random variable to take on a particular value. Unlike discrete random variables, which have a probability mass function, continuous random variables like the one provided here have PDFs.
The key aspect of a PDF is that it must integrate over its entire range to 1, as a total probability is always 100%. In our specific example, the PDF is defined as \[ f(x) = \begin{cases} \frac{1}{40}, & \text{if } -20 \le x \le 20 \0, & \text{otherwise.}\end{cases}\]
This represents a uniform distribution over the interval from -20 to 20. The PDF is constant within its interval, which means each value between -20 and 20 is equally probable. Beyond these bounds, the probability is zero. This uniformity is significant as it simplifies many computations, such as finding probabilities over specific sub-intervals.

Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) is a function used to determine the probability that a continuous random variable takes on a value less than or equal to a given point. It is essentially the cumulative area under the PDF from the lower bound up to that point.
In our case, the cumulative distribution function can be written as:

• For \(x < -20\), \(F(x) = 0\) – since no value less than -20 is part of the distribution.
• For \(-20 \le x \le 20\), \(F(x) = \int_{-20}^{x} \frac{1}{40} \, dt = \frac{x + 20}{40}\)
• For \(x > 20\), \(F(x) = 1\) – indicating that the entire distribution is covered.

The CDF increases linearly within the interval because the distribution is uniform and displays no preference for individual values. This function allows for the determination of probabilities that a variable will fall into various ranges.

Uniform Distribution

Uniform distribution is a type of continuous probability distribution where all outcomes are equally likely within a given interval. This simplicity is one of the distinct characteristics of a uniform distribution.
For the PDF we have, the interval is \([-20, 20] \), with the constant value of \( \frac{1}{40}\) as the density. This translates to equal probability for each value within this range. With a uniform distribution, every sub-interval of equal length has the same probability.
Uniform distributions are quite helpful in scenarios involving simple random samples or when modeling signals with constant amplitude within a specified range. Understanding this distribution enables easier computations like expectations and cumulative probabilities, which are achievable with simple arithmetic due to the even spread of values across the interval.

Expectation (E(X))

The expectation, or expected value, of a continuous random variable is a measure of the central tendency, often referred to as the 'mean'. It provides an idea of where the center of the distribution lies.
For a uniform distribution, the expectation is the middle point of the interval since all points within the interval are equally likely. Mathematically, it is the average of the interval's bounds. For our uniform distribution over \([-20, 20] )\), the expectation is:\[E(X) = \frac{-20 + 20}{2} = 0.\]
This result signifies that, on average, the random variable is expected to take a value of zero, even though it could be anywhere from -20 to 20. Understanding the expected value is crucial, especially in fields like economics and engineering, where it helps in predicting outcomes and making decisions based on statistical data.