Problem 23 - Calculus for Biology and Medicine

Question

Let \(X\) be uniformly distributed on the set $$ S=\\{1,2,3, \ldots, 10\\} $$ That is, $$ P(X=k)=\frac{1}{10}, \quad k \in S $$ (a) Find \(E(X)\). (b) Find \(\operatorname{var}(X)\).

Step-by-Step Solution

Answer

(a) 5.5, (b) 8.25

Step 1: Understanding Uniform Distribution

A uniform distribution over a finite set implies that each member of the set has an equal probability of occurrence. In this case, we have 10 elements, each with probability \( P(X=k) = \frac{1}{10} \).

Step 2: Calculate the Expected Value E(X)

For a discrete uniform distribution over a set \( S = \{ 1, 2, 3, \ldots, n \} \), the expected value is calculated by the formula: \[ E(X) = \frac{n+1}{2} \].Here, \( n = 10 \), so\[ E(X) = \frac{10+1}{2} = \frac{11}{2} = 5.5 \].

Step 3: Calculate the Variance Var(X)

The variance of a discrete uniform distribution on the set \( S = \{ 1, 2, 3, \ldots, n \} \) is given by the formula: \[ \operatorname{Var}(X) = \frac{(n^2 - 1)}{12} \].For this problem, \( n = 10 \), so\[ \operatorname{Var}(X) = \frac{10^2 - 1}{12} = \frac{100 - 1}{12} = \frac{99}{12} = 8.25 \].

Key Concepts

Expected Value

The expected value of a random variable is like the average value you would expect if you could repeat the random process indefinitely. It's a key concept in probability that gives us a measure of the center of the distribution of the random variable. For a discrete uniform distribution, where each outcome is equally likely, calculating the expected value is straightforward. If we have a set of numbers from 1 to 10, like in our problem, the expected value is computed using the formula:

• \[ E(X) = \frac{n+1}{2} \]

For our set \( S = \{1, 2, 3, \ldots, 10\} \), \( n = 10 \). So, plug in the numbers:

• \[ E(X) = \frac{10+1}{2} = 5.5 \]

This means that, on average, if you pick a number randomly from this set over and over, you'd expect to get 5.5. This may not make sense immediately because 5.5 isn't in the set, but remember: expected value isn't always a value from the set itself; it's a theoretical mean.

Variance

Variance provides a measure of how spread out the values of a random variable are. It tells us how much individual values are likely to differ from the expected value. In the context of a discrete uniform distribution, variance helps us understand how much variation there is from the mean when each outcome is equally probable.To calculate variance for a uniform distribution where the values range from 1 to \( n \), use this formula:

• \[ \operatorname{Var}(X) = \frac{(n^2 - 1)}{12} \]

For our set \( S \), with \( n = 10 \), the calculation is:

• \[ \operatorname{Var}(X) = \frac{10^2 - 1}{12} = \frac{99}{12} = 8.25 \]

The variance of 8.25 means that there is a moderate amount of spread around the mean (which was 5.5, as calculated earlier). The higher the variance, the more spread out the numbers are. For this distribution, a variance of 8.25 indicates some diversity in the values, but not too much, considering the relatively small range.

Discrete Uniform Distribution

A discrete uniform distribution is a simple yet powerful model in probability. With this distribution, every outcome in your set is equally likely. It's perfect for modeling situations like dice rolls or drawing from a deck of cards, where each possibility occurs with the same frequency.In our example, the variable \( X \) was distributed uniformly over the set \( S = \{1, 2, 3, \ldots, 10\} \). Each number between 1 and 10 is equally probable, which means the probability \( P(X = k) \) is \( \frac{1}{10} \) for each \( k \) in the set.This feature makes calculations straightforward:

• The expected value formula \( E(X) = \frac{n+1}{2} \) leverages the symmetry of the distribution.
• The variance formula \( \operatorname{Var}(X) = \frac{(n^2 - 1)}{12} \) reflects the uniform spacing of values in the set.

Discrete uniform distributions are instrumental in demonstrating fundamental statistical concepts, thanks to their simplicity and symmetry. They are ideal learning tools for gaining a deeper understanding of how probability models describe random processes.