Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose a random variable \(Z\) has CDF $$ F(z)= \begin{cases}0, & \text { if } z<0 \\ z^{2} / 9, & \text { if } 0 \leq z \leq 3 \\ 1, & \text { if } z>3\end{cases} $$ Find each of the following: (a) \(P(Z>1)\) (b) \(P(1

Step-by-Step Solution

Answer

(a) \(\frac{8}{9}\), (b) \(\frac{1}{3}\), (c) \(f(z) = \frac{2z}{9}\), (d) 2

Step 1: Calculate P(Z > 1)

To find the probability that a random variable Z is greater than 1, we can use the cumulative distribution function (CDF). It is given by:\[ P(Z > 1) = 1 - P(Z \leq 1) \]Where the CDF at 1 is:\[ F(1) = \left( \frac{1^2}{9} \right) = \frac{1}{9} \]Thus,\[ P(Z > 1) = 1 - \frac{1}{9} = \frac{8}{9} \]

Step 2: Calculate P(1 < Z < 2)

To calculate the probability that the random variable Z is between 1 and 2, use the CDF:\[ P(1 < Z < 2) = F(2) - F(1) \]Calculate each:\[ F(2) = \left( \frac{2^2}{9} \right) = \frac{4}{9} \]\[ F(1) = \frac{1}{9} \]Now subtract:\[ P(1 < Z < 2) = \frac{4}{9} - \frac{1}{9} = \frac{3}{9} = \frac{1}{3} \]

Step 3: Determine the PDF of Z

The PDF (probability density function) is found by differentiating the CDF over its respective interval. For \(0 \leq z \leq 3\), we have:\[ f(z) = \frac{d}{dz} \left( \frac{z^2}{9} \right) = \frac{2z}{9} \]

Step 4: Find E(Z)

The expected value \(E(Z)\) can be found using the PDF. We integrate over the range where the PDF is non-zero:\[ E(Z) = \int_{0}^{3} z \cdot \frac{2z}{9} \, dz = \frac{2}{9} \int_{0}^{3} z^2 \, dz \]Now calculate the integral:\[ E(Z) = \frac{2}{9} \left[ \frac{z^3}{3} \right]_{0}^{3} = \frac{2}{9} \cdot \left( \frac{27}{3} - 0 \right) = \frac{2}{9} \cdot 9 = 2 \]

Key Concepts

Probability Density Function (PDF)

The Probability Density Function, or PDF, is a crucial concept in understanding how continuous random variables behave. Unlike its cousin for discrete random variables, the probability mass function, the PDF helps describe the probability of a random variable taking on a particular value within a continuous range.
To derive a PDF from a given CDF, one must differentiate the CDF in its range of applicability.Consider a continuous random variable Z whose cumulative distribution function (CDF) is defined piecewise. To find its PDF, we differentiate the CDF on the interval where it's non-zero. In the case of the exercise:

• The CDF of Z is given by \( F(z) = \frac{z^2}{9} \) for \( 0 \leq z \leq 3 \).

So, the PDF over this range is computed by:
\[ f(z) = \frac{d}{dz} \left( \frac{z^2}{9} \right) = \frac{2z}{9} \]
Once we obtain the PDF, it serves as the foundation for calculating probabilities of continuous outcomes over intervals, rather than specific values.

Expected Value (Expectation)

The expected value or expectation of a random variable gives us the average or mean of the outcomes if the experiment related to the random variable were repeated an infinite number of times. It's akin to the center of mass in geometry.For a continuous random variable, like our Z from the exercise, the expected value is calculated using its PDF by integrating the product of the variable and its density function over the distribution's support:

• Here, we need to compute \( E(Z) = \int_{0}^{3} z \cdot \frac{2z}{9} \, dz \).

The integral simplifies the process of finding the average outcome over the entire range. The computed expectation from the exercise is \( E(Z) = 2 \). This means that, on average, the central tendency or expected outcome of Z is 2, considering its probability distribution.

Random Variable Probability

Understanding random variable probabilities involves knowing the chance that a variable falls within a specified range. This doesn't mean calculating the likelihood of a specific point but rather an interval. The cumulative distribution function (CDF) is used extensively here.To find the probability that Z lies in a particular interval, we rely on the CDF. For example, in our step-by-step exercise, we needed:

• \( P(Z > 1) = 1 - F(1) \) where F(1) was found using the CDF formula \( F(z) = \frac{z^2}{9} \).
• Similarly, \( P(1 < Z < 2) \) involved calculating the spread between two points, i.e., \( F(2) - F(1) \).

Through these computations, we realize that the CDF provides holistic insight into the behavior and likelihoods associated with the random variable. This foundational knowledge aids in making informed decisions and predictions based on probabilistic data.