Problem 22 - Calculus for Biology and Medicine

Question

Suppose that the probability mass function of a discrete random variable \(X\) is given by the following table: $$\begin{array}{rc} \hline \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) \\ \hline-1 & 0.1 \\ -0.5 & 0.2 \\ 0.1 & 0.1 \\ 0.5 & 0.25 \\ 1 & 0.35 \\ \hline \end{array}$$ Find the mean, the variance, and the standard deviation of \(X\).

Step-by-Step Solution

Answer

Mean: 0.275, Variance: 0.487875, Standard Deviation: 0.698.

Step 1: Calculate the Mean

To find the mean (expected value) of the random variable \(X\), multiply each value \(x_i\) by its probability \(P(X=x_i)\) and sum the results: \[ E(X) = \sum_{i} x_i \cdot P(X=x_i) = (-1)(0.1) + (-0.5)(0.2) + (0.1)(0.1) + (0.5)(0.25) + (1)(0.35) = -0.1 - 0.1 + 0.01 + 0.125 + 0.35 = 0.275. \] Thus, the mean of \(X\) is 0.275.

Step 2: Compute the Variance

To compute the variance, use the formula \( \text{Var}(X) = E(X^2) - (E(X))^2 \). First, find \(E(X^2)\): \[ E(X^2) = \sum_{i} x_i^2 \cdot P(X=x_i) = (-1)^2(0.1) + (-0.5)^2(0.2) + (0.1)^2(0.1) + (0.5)^2(0.25) + (1)^2(0.35) = 0.1 + 0.05 + 0.001 + 0.0625 + 0.35 = 0.5635. \] Then, calculate the variance: \[ \text{Var}(X) = 0.5635 - (0.275)^2 = 0.5635 - 0.075625 = 0.487875. \] Thus, the variance of \(X\) is 0.487875.

Step 3: Determine the Standard Deviation

The standard deviation is the square root of the variance. Calculate it: \[ \sigma(X) = \sqrt{0.487875} \approx 0.698. \] Therefore, the standard deviation of \(X\) is approximately 0.698.

Key Concepts

Mean of a Discrete Random Variable

The mean, often referred to as the expected value, of a discrete random variable is a measure of its central tendency. It represents the average outcome you would expect if the experiment described by the random variable is repeated many times. Calculating the mean involves two main steps.

• First, multiply each possible value of the random variable by its corresponding probability. This weighted multiplication ensures that more probable outcomes contribute more significantly to the mean.
• Next, sum up all these weighted values.

For the given example, you calculate the expected value by applying these steps to each outcome of the variable.
The formula for the mean is: \[ E(X) = \sum{x_i \cdot P(X=x_i)} \]where \(x_i\) are the possible values the random variable can take, and \(P(X=x_i)\) is the probability of \(X\) being \(x_i\). By performing this calculation, as shown in the solution, we determine the mean to be 0.275.

Variance of a Discrete Random Variable

Variance provides insight into the spread or variability of a random variable's possible values. It is a measure of how much the outcomes deviate from the mean value on average. Calculating the variance consists of a few clear steps.

• Start by calculating the expected value of the square of the random variable, written as \(E(X^2)\). This involves multiplying the square of each outcome by its respective probability and then summing these values.
• Subtract the square of the mean, previously calculated, from this sum.

Mathematically, this is described by:\[ \text{Var}(X) = E(X^2) - (E(X))^2 \]In the exercise solution, it was determined that the variance amounts to 0.487875. A higher variance indicates that the outcomes are more spread across the potential values.

Standard Deviation of a Discrete Random Variable

The standard deviation is a related concept to variance, providing a more interpretable measure of spread by bringing variance back to the original units of the variable. Standard deviation is particularly useful because, unlike variance, it is in the same units as the random variable itself, making it easier to conceptualize.
To find the standard deviation, simply compute the square root of the variance:
\[ \sigma(X) = \sqrt{\text{Var}(X)} \]This operation reverses the squaring that occurs when calculating variance, allowing for a more direct understanding of data spread. In the exercise, the standard deviation was calculated to be approximately 0.698. With the standard deviation, it becomes clear how tightly the data is clustered around the mean, with smaller values indicating less variability, and larger values indicating more.