*88*

**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. \( E\left((X – 3)^2\right) \approx 11.67 \)