Exercise
A discrete random variable \( X \) has the following probability distribution:
\[
\begin{array}{|c|c|c|c|c|}
\hline
x & 2 & 4 & 6 & 10 \\
\hline
p(x) & \frac{1}{6} & \frac{1}{3} & \frac{1}{3} & \frac{1}{6} \\
\hline
\end{array}
\]
Answer the following questions:
1. Calculate the probability \( P(4 \leq X < 10) \).
2. Calculate the mean (expected value) and variance of the random variable \( X \).
3. Calculate \( E\left((X – 3)^2\right) \).
—
Solution
Question 1: Calculate the probability \( P(4 \leq X < 10) \).
We need to find the probability that \( X \) takes values in the range \( 4 \leq X < 10 \).
\[
P(4 \leq X < 10) = P(X = 4) + P(X = 6)
\]
Substituting the given probabilities:
\[
P(4 \leq X < 10) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}
\]
So, \( P(4 \leq X < 10) = \frac{2}{3} \).
Question 2: Calculate the mean (expected value) and variance of the random variable \( X \).
The expected value \( E(X) \) is calculated as:
\[
E(X) = \sum_{i} x_i \cdot p(x_i)
\]
Substituting the values:
\[
E(X) = 2 \cdot \frac{1}{6} + 4 \cdot \frac{1}{3} + 6 \cdot \frac{1}{3} + 10 \cdot \frac{1}{6}
\]
\[
E(X) = \frac{2}{6} + \frac{4}{3} + 2 + \frac{10}{6}
\]
\[
E(X) = \frac{2 + 8 + 12 + 10}{6} = \frac{32}{6} = \frac{16}{3} \approx 5.33
\]
So, the mean \( E(X) = \frac{16}{3} \).
The variance \( \text{Var}(X) \) is calculated as:
\[
\text{Var}(X) = E(X^2) – [E(X)]^2
\]
First, we calculate \( E(X^2) \):
\[
E(X^2) = \sum_{i} x_i^2 \cdot p(x_i)
\]
\[
E(X^2) = 2^2 \cdot \frac{1}{6} + 4^2 \cdot \frac{1}{3} + 6^2 \cdot \frac{1}{3} + 10^2 \cdot \frac{1}{6}
\]
\[
E(X^2) = \frac{4}{6} + \frac{16}{3} + \frac{36}{6} + \frac{100}{6}
\]
\[
E(X^2) = \frac{4 + 32 + 72 + 100}{6} = \frac{208}{6} \approx 34.67
\]
Now, calculate the variance:
\[
\text{Var}(X) = 34.67 – \left(\frac{16}{3}\right)^2 = 34.67 – \frac{256}{9} \approx 6.222
\]
So, the variance \( \text{Var}(X) \approx 6.222 \).
Question 3: Calculate \( E\left((X – 3)^2\right) \).
We calculate this as:
\[
E\left((X – 3)^2\right) = \sum_{i} (x_i – 3)^2 \cdot p(x_i)
\]
Substituting the values:
\[
E\left((X – 3)^2\right) = (2 – 3)^2 \cdot \frac{1}{6} + (4 – 3)^2 \cdot \frac{1}{3} + (6 – 3)^2 \cdot \frac{1}{3} + (10 – 3)^2 \cdot \frac{1}{6}
\]
\[
E\left((X – 3)^2\right) = (-1)^2 \cdot \frac{1}{6} + 1^2 \cdot \frac{1}{3} + 3^2 \cdot \frac{1}{3} + 7^2 \cdot \frac{1}{6}
\]
\[
E\left((X – 3)^2\right) = \frac{1}{6} + \frac{1}{3} + 9 \cdot \frac{1}{3} + \frac{49}{6}
\]
\[
E\left((X – 3)^2\right) = \frac{1 + 2 + 18 + 49}{6} = \frac{70}{6} \approx 11.67
\]
So, \( E\left((X – 3)^2\right) \approx 11.67 \).
—
Summary of Answers
1. \( P(4 \leq X < 10) = \frac{2}{3} \)
2. \( E(X) = \frac{16}{3} \approx 5.33 \) and \( \text{Var}(X) \approx 6.222 \)
3. \( E\left((X – 3)^2\right) \approx 11.67 \)
