Problem 21 - Calculus for Biology and Medicine

Question

Suppose that the probability mass function of a discrete random variable \(X\) is given by the following table: $$ \begin{array}{cc} \hline \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) \\ \hline-3 & 0.2 \\ -1 & 0.3 \\ 1.5 & 0.4 \\ 2 & 0.1 \\ \hline \end{array} $$ Find the mean, the variance, and the standard deviation of \(X\).

Step-by-Step Solution

Answer

Mean is -0.1, variance is 3.39, and standard deviation is approximately 1.84.

Step 1: Find the Mean of X

The mean of a discrete random variable, \(X\), can be calculated using the formula \(\mu = \sum x_i P(X = x_i)\). Substitute each value from the table: \(\mu = (-3)(0.2) + (-1)(0.3) + (1.5)(0.4) + (2)(0.1) = -0.6 - 0.3 + 0.6 + 0.2 = -0.1\). Hence, the mean \(\mu = -0.1\).

Step 2: Find the Variance of X

The variance is calculated using \(\sigma^2 = \sum (x_i - \mu)^2 P(X = x_i)\). First, substitute \(\mu = -0.1\) into the formula: variance, \(\sigma^2 = ((-3 + 0.1)^2)(0.2) + ((-1 + 0.1)^2)(0.3) + ((1.5 + 0.1)^2)(0.4) + ((2 + 0.1)^2)(0.1)\). Simplifying, \(= (2.9^2)(0.2) + (0.9^2)(0.3) + (1.6^2)(0.4) + (2.1^2)(0.1)\). Final calculations yield \(= 1.682 + 0.243 + 1.024 + 0.441 = 3.39\). Hence, \(\sigma^2 = 3.39\).

Step 3: Find the Standard Deviation of X

The standard deviation \(\sigma\) is the square root of the variance: \(\sigma = \sqrt{\sigma^2} = \sqrt{3.39} \approx 1.84\). Therefore, the standard deviation of \(X\) is approximately \(1.84\).

Key Concepts

Mean of Random Variable

The mean of a random variable is often referred to as the expected value. It's a way to find the average by considering how likely each outcome is. For a discrete random variable, like in this exercise, we can calculate the mean using a special formula. It's written like this: \( \mu = \sum x_i P(X = x_i) \).

Here's what each part means:

• \( x_i \) is a possible value that the random variable can take.
• \( P(X = x_i) \) is the probability that the random variable equals \( x_i \).

To find \( \mu \), simply multiply each \( x_i \) by its probability, then add all these products together. In our problem:

• For \( x = -3 \), the contribution is \((-3) \times 0.2 = -0.6\).
• For \( x = -1 \), the contribution is \((-1) \times 0.3 = -0.3\).
• For \( x = 1.5 \), the contribution is \((1.5) \times 0.4 = 0.6\).
• For \( x = 2 \), the contribution is \((2) \times 0.1 = 0.2\).

Adding these contributions gives \(-0.6 + (-0.3) + 0.6 + 0.2 = -0.1\). So, the mean \( \mu \) equals \(-0.1\). This is the average expected value when considering the probabilities.

Variance Calculation

Variance tells us about the spread of the random variable's possible values, showing us how much the values differ from the mean on average. The formula for calculating variance \( \sigma^2 \) is:\[ \sigma^2 = \sum (x_i - \mu)^2 P(X = x_i) \]Here's a deeper dive into what that means:

• \( x_i - \mu \): This part measures how far each value is from the mean \( \mu \). We use this for each value.
• \( (x_i - \mu)^2 \): Squaring each difference ensures all differences are positive and it penalizes larger deviations more.
• \( P(X = x_i) \): We multiply by the probability of each value to weight it according to how likely it is.

In our problem:

• For \( x = -3 \), \((2.9)^2 \times 0.2 = 1.682\).
• For \( x = -1 \), \((0.9)^2 \times 0.3 = 0.243\).
• For \( x = 1.5 \), \((1.6)^2 \times 0.4 = 1.024\).
• For \( x = 2 \), \((2.1)^2 \times 0.1 = 0.441\).

Adding these gives us the total variance: \( 1.682 + 0.243 + 1.024 + 0.441 = 3.39\). Thus, \( \sigma^2 = 3.39 \). This shows the extent to which the values of the random variable \(X\) are spread out from the mean.

Standard Deviation Calculation

Standard deviation is another way of looking at how much the values of a random variable are spread out or deviate from the mean. It's basically the square root of the variance. The formula is simple:\[ \sigma = \sqrt{\sigma^2} \]The great thing about standard deviation is that it returns the measure of spread in the same unit as our initial data. This makes it easier to understand and relate back to the variable we are analyzing.

In our specific case, we calculated the variance to be \( \sigma^2 = 3.39 \). The next step is to take the square root of this number:\[ \sigma = \sqrt{3.39} \approx 1.84 \]So, the standard deviation of \(X\) is approximately \(1.84\). This value gives you a sense of how much the data set as a whole typically varies from the mean. It's very useful for understanding the distribution of values.