If `m and sigma^(2)` are the mean and variance of the random variable X, whose distribution is given by:
`{:(X=x:,0,1,2,3),(P(X=x):,(1)/(3),(1)/(2),0,(1)/(6)):}`

Options

A. `m=sigma^(2)=2`
B. `m=1,sigma^(2)=2`
C. `m=sigma^(2)=1`
D. `m=2,sigma^(2)=1`

Answer

To solve the problem, we need to find the mean (m) and variance (σ²) of the random variable X given its probability distribution. The distribution is as follows:

- P(X = 0) = 1/3
- P(X = 1) = 1/2
- P(X = 2) = 0
- P(X = 3) = 1/6

### Step 1: Calculate the Mean (m)

The mean (expected value) of a discrete random variable is calculated using the formula:

\[ m = E(X) = \sum (x_i \cdot P(X = x_i)) \]

Substituting the values from the distribution:

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

Calculating each term:

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

Now, summing these values:

\[ m = 0 + \frac{1}{2} + 0 + \frac{1}{2} = 1 \]

Thus, the mean \(m = 1\).

### Step 2: Calculate the Variance (σ²)

To find the variance, we first need to calculate \(E(X^2)\):

\[ E(X^2) = \sum (x_i^2 \cdot P(X = x_i)) \]

Substituting the values:

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

Calculating each term:

- \(0^2 \cdot \frac{1}{3} = 0\)
- \(1^2 \cdot \frac{1}{2} = \frac{1}{2}\)
- \(2^2 \cdot 0 = 0\)
- \(3^2 \cdot \frac{1}{6} = \frac{9}{6} = \frac{3}{2}\)

Now, summing these values:

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

Now, we can calculate the variance using the formula:

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

Substituting the values we found:

\[ \sigma^2 = 2 - (1)^2 = 2 - 1 = 1 \]

Thus, the variance \(σ² = 1\).

### Final Result

The mean \(m\) is 1 and the variance \(σ²\) is also 1.

### Summary

- Mean \(m = 1\)
- Variance \(σ² = 1\)