Let X be a random variable such that `f(X=-2)=f(X=-1)=f(X=2)=f(X=1)=(1)/(6)` and `f(X=0)=(1)/(3)`. Find the mean and variance of X.

Answer

To find the mean and variance of the random variable \( X \) with the given probability mass function, we can follow these steps:

### Step 1: Identify the values and probabilities
We have the following values of \( X \) and their corresponding probabilities:

- \( f(X = -2) = \frac{1}{6} \)
- \( f(X = -1) = \frac{1}{6} \)
- \( f(X = 0) = \frac{1}{3} \)
- \( f(X = 1) = \frac{1}{6} \)
- \( f(X = 2) = \frac{1}{6} \)

### Step 2: Calculate the mean
The mean \( \mu \) of a discrete random variable is calculated using the formula:

\[ \mu = E(X) = \sum (x \cdot f(x)) \]

Substituting the values:

\[ \mu = (-2) \cdot \frac{1}{6} + (-1) \cdot \frac{1}{6} + 0 \cdot \frac{1}{3} + 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} \]

Calculating each term:

\[ \mu = -\frac{2}{6} - \frac{1}{6} + 0 + \frac{1}{6} + \frac{2}{6} \]

Combining the terms:

\[ \mu = -\frac{2}{6} - \frac{1}{6} + \frac{1}{6} + \frac{2}{6} = -\frac{2}{6} = -\frac{1}{3} \]

### Step 3: Calculate the variance
The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = E(X^2) - (E(X))^2 \]

First, we need to calculate \( E(X^2) \):

\[ E(X^2) = \sum (x^2 \cdot f(x)) \]

Substituting the values:

\[ E(X^2) = (-2)^2 \cdot \frac{1}{6} + (-1)^2 \cdot \frac{1}{6} + 0^2 \cdot \frac{1}{3} + 1^2 \cdot \frac{1}{6} + 2^2 \cdot \frac{1}{6} \]

Calculating each term:

\[ E(X^2) = 4 \cdot \frac{1}{6} + 1 \cdot \frac{1}{6} + 0 + 1 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} \]

Combining the terms:

\[ E(X^2) = \frac{4}{6} + \frac{1}{6} + 0 + \frac{1}{6} + \frac{4}{6} = \frac{10}{6} = \frac{5}{3} \]

Now, we can find the variance:

\[ \sigma^2 = E(X^2) - (E(X))^2 = \frac{5}{3} - \left(-\frac{1}{3}\right)^2 \]

Calculating \( \left(-\frac{1}{3}\right)^2 \):

\[ \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \]

Now substituting back into the variance formula:

\[ \sigma^2 = \frac{5}{3} - \frac{1}{9} \]

To subtract these fractions, we need a common denominator. The least common multiple of 3 and 9 is 9:

\[ \sigma^2 = \frac{15}{9} - \frac{1}{9} = \frac{14}{9} \]

### Final Results
Thus, the mean \( \mu \) and variance \( \sigma^2 \) of the random variable \( X \) are:

- Mean \( \mu = 0 \)
- Variance \( \sigma^2 = \frac{14}{9} \)