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

The mean is 0.285; variance is 0.483135; standard deviation is approximately 0.695.

Step 1: Calculate the Mean

The mean of a discrete random variable is calculated using the formula \( \mu = \sum x_i P(X = x_i) \). Here, compute each product and sum them:- \((-1) \times 0.1 = -0.1\)- \((-0.5) \times 0.2 = -0.1\)- \((0.1) \times 0.1 = 0.01\)- \((0.5) \times 0.25 = 0.125\)- \((1) \times 0.35 = 0.35\)Sum these products: \(-0.1 - 0.1 + 0.01 + 0.125 + 0.35 = 0.285\).Thus, the mean is \( \mu = 0.285 \).

Step 2: Calculate the Variance

The variance is calculated using the formula \( \sigma^2 = \sum (x_i - \mu)^2 P(X = x_i) \).For each \(x_i\), compute \((x_i - \mu)^2 \):- \((-1 - 0.285)^2 \times 0.1 = 1.664025 \times 0.1 = 0.1664025\)- \((-0.5 - 0.285)^2 \times 0.2 = 0.616225 \times 0.2 = 0.123245\)- \((0.1 - 0.285)^2 \times 0.1 = 0.034225 \times 0.1 = 0.0034225\)- \((0.5 - 0.285)^2 \times 0.25 = 0.046225 \times 0.25 = 0.01155625\)- \((1 - 0.285)^2 \times 0.35 = 0.510025 \times 0.35 = 0.17850875\)Sum these values: \(0.1664025 + 0.123245 + 0.0034225 + 0.01155625 + 0.17850875 = 0.483135\).Thus, the variance is \( \sigma^2 = 0.483135 \).

Step 3: Calculate the Standard Deviation

The standard deviation is the square root of the variance: \( \sigma = \sqrt{0.483135} \approx 0.695 \).

Key Concepts

Understanding the Mean of a Discrete Random Variable

In probability and statistics, the mean of a discrete random variable is a measure of its central tendency. Imagine you have a set of different outcomes, each with its probability of occurrence. The mean is like the 'average' outcome you might expect if you repeat an experiment many times.
To calculate the mean, you use the formula \[\mu = \sum x_i P(X = x_i),\]where

• \(x_i\) represents each possible outcome,
• \(P(X = x_i)\) represents the probability of each outcome.

What you do is simply multiply each outcome by its probability, and then sum up all these products. This gives you a single number, the mean, which in our specific exercise turned out to be 0.285.
This mean tells us, over a large number of trials, we can expect the average value to hover around 0.285. But remember, each individual trial won't necessarily give you this number - it varies due to the random nature of the variable.

Delving Into Variance Calculation

After understanding the mean, let's talk about variance, another critical measure in statistics. Variance tells us how much the random variable's values deviate from the mean. It's a measure of the data's spread – are the values tightly packed around the mean, or are they spread out? This is important because it gives context to the mean; two different datasets could have the same mean, but wildly different variances.
In our case, calculating variance employs the formula:\[\sigma^2 = \sum (x_i - \mu)^2 P(X = x_i).\]Here’s what we do:

• Subtract the mean \(\mu\) from each outcome \(x_i\),
• Square the result to ensure a positive number (since we're interested in magnitude of variance, not direction),
• Multiply by the probability \(P(X = x_i)\),
• Finally, sum up these values to get the variance.

For our exercise, the variance came out to be 0.483135. A larger variance indicates a more spread out distribution, while a smaller variance indicates values closer to the mean.

Decoding Standard Deviation Calculation

The last stop on our journey through statistics is the standard deviation. This measure, more intuitive for many people, gives insight into the same concept as variance but expressed in the original units of our data instead of squared units.
The formula is straightforward: to obtain the standard deviation, you take the square root of the variance:\[\sigma = \sqrt{\sigma^2}.\]For our example, the variance was 0.483135, so the standard deviation is the square root of this value, approximately 0.695.
Why is this number important? Well, it tells us an average deviation from the mean. Unlike variance, which can be hard to interpret because it's in squared units, the standard deviation is in the same units as your data, making it easier to understand. If the standard deviation is small, most values are close to the mean. If it's large, values are more spread out. Together with the mean and variance, the standard deviation gives you a fuller picture of the random variable's behavior.