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: $$\begin{array}{rc} \hline \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) \\ \hline-3 & 0.2 \\ -1 & 0.3 \\ 1.5 & 0.4 \\ 2 & 0.1 \\ \hline \end{array}$$ Find the mean, the variance, and the standard deviation of \(X\).

Step-by-Step Solution

Answer

Mean: -0.1, Variance: 3.39, Standard Deviation: 1.84.

Step 1: Calculating the Mean

The mean of a discrete random variable \( X \) is given by the formula: \( \mu = \sum x_i P(X=x_i) \). Using the table provided, we calculate as follows: - 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 \(2 \times 0.1 = 0.2\) Summing these values gives the mean: \( \mu = -0.6 + (-0.3) + 0.6 + 0.2 = -0.1 \).

Step 2: Calculating Variance

The variance of a discrete random variable \( X \) is determined by \( \sigma^2 = \sum (x_i - \mu)^2 P(X=x_i) \). Now substituting the values: - For \( x = -3 \), \((-3 - (-0.1))^2 \times 0.2 = (2.9)^2 \times 0.2 = 8.41 \times 0.2 = 1.682\) - For \( x = -1 \), \((-1 - (-0.1))^2 \times 0.3 = (0.9)^2 \times 0.3 = 0.81 \times 0.3 = 0.243\) - For \( x = 1.5 \), \((1.5 - (-0.1))^2 \times 0.4 = (1.6)^2 \times 0.4 = 2.56 \times 0.4 = 1.024\) - For \( x = 2 \), \((2 - (-0.1))^2 \times 0.1 = (2.1)^2 \times 0.1 = 4.41 \times 0.1 = 0.441\) Adding these values, the variance is \( \sigma^2 = 1.682 + 0.243 + 1.024 + 0.441 = 3.39 \).

Step 3: Calculating Standard Deviation

The standard deviation is the square root of the variance, \( \sigma \). Using the variance calculated, \( \sigma = \sqrt{3.39} \approx 1.84 \).

Key Concepts

Discrete Random Variable

A discrete random variable is a type of random variable that can take on a countable number of distinct values. These values can be finite or countably infinite. In the context of our probability mass function (PMF) table, the discrete random variable, denoted by \( X \), can take values \(-3, -1, 1.5,\) and \(2\).
Each of these values has an associated probability, as shown in the PMF table.
The probabilities are important because they sum up to 1, reflecting the total certainty of all possible outcomes happening.
Understanding a discrete random variable involves knowing both the possible values it can take and the likelihood or probability of each value.

• Discrete variables are used when outcomes are distinct and separate.
• Unlike continuous variables which could take any value in a range, discrete variables "jump" from one value to the next.
• The key feature of discrete variables is that they are countable!

Discreet random variables are fundamental in probability because they form the basis of many statistical calculations, such as mean, variance, and standard deviation, which we'll explore next.

Mean Calculation

The mean of a discrete random variable, also known as the expected value, is a measure of the central tendency of a probability distribution. Think of it as the long-term average result you would expect if you repeated the experiment many times.
To find the mean, you multiply each possible value of the random variable by its probability and then sum up all these products. Mathematically, it’s expressed as \[ \mu = \sum x_i P(X=x_i) \]For the given exercise, the calculations for each value \( x \) are:

• \( x = -3 \): Multiply by \( 0.2 \), giving \( -0.6 \)
• \( x = -1 \): Multiply by \( 0.3 \), giving \( -0.3 \)
• \( x = 1.5 \): Multiply by \( 0.4 \), resulting in \( 0.6 \)
• \( x = 2 \): Multiply by \( 0.1 \), resulting in \( 0.2 \)

By adding these products together, the mean \( \mu \) is calculated as \( -0.6 + (-0.3) + 0.6 + 0.2 = -0.1 \).
Thus, the mean of this distribution is \(-0.1\).

• The mean gives us an idea of where the "center" of the distribution is located.
• This calculation helps predict the average outcome over many trials.

Variance and Standard Deviation

Variance and standard deviation are measures that tell us about the spread of our data values relative to the mean. They show us how much the values of a random variable differ from the mean.
The variance \( \sigma^2 \) is calculated using the formula:\[ \sigma^2 = \sum (x_i - \mu)^2 P(X=x_i) \]This formula calculates the expectation of the squared deviation of each value from the mean. For our table:

• \( x = -3 \): The squared deviation \((2.9)^2 \times 0.2 = 1.682\)
• \( x = -1 \): The squared deviation \((0.9)^2 \times 0.3 = 0.243\)
• \( x = 1.5 \): The squared deviation \((1.6)^2 \times 0.4 = 1.024\)
• \( x = 2 \): The squared deviation \((2.1)^2 \times 0.1 = 0.441\)

Summing these values gives us the variance: \( \sigma^2 = 1.682 + 0.243 + 1.024 + 0.441 = 3.39 \).
The standard deviation \( \sigma \) is simply the square root of the variance: \( \sigma = \sqrt{3.39} \approx 1.84 \).

• Variance measures the average squared deviations from the mean, indicating how spread out the values are.
• Standard deviation, being the square root of variance, provides a more interpretable measure of spread by bringing it back to the units of the original data.
• Higher values in standard deviation indicate a wider spread around the mean, while lower values indicate a tighter clustering of data around the mean.