A random variable X takes values 1,2,3 and 4 with probabilities `(1)/(6),(1)/(3),(1)/(3),(1)/(6)` respectively, then its mean and variance is equal to

Options

A. `(5)/(2),(11)/(12)`
B. `(5)/(2),(11)/(16)`
C. `(5)/(3),(11)/(16)`
D. `(5)/(3),(11)/(12)`

Answer

To find the mean and variance of the random variable \( X \) that takes values 1, 2, 3, and 4 with corresponding probabilities \( \frac{1}{6}, \frac{1}{3}, \frac{1}{3}, \frac{1}{6} \), we can follow these steps:

### Step 1: Define the random variable and its probabilities
Let \( X = \{1, 2, 3, 4\} \) and the corresponding probabilities be:
- \( P(X=1) = \frac{1}{6} \)
- \( P(X=2) = \frac{1}{3} \)
- \( P(X=3) = \frac{1}{3} \)
- \( P(X=4) = \frac{1}{6} \)

### Step 2: Calculate the mean (expected value) of \( X \)
The mean \( \mu \) of a random variable is calculated using the formula:
\[ \mu = E(X) = \sum_{i=1}^{n} x_i P(X=x_i) \]
Substituting the values:
\[ \mu = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{3} + 3 \cdot \frac{1}{3} + 4 \cdot \frac{1}{6} \]

### Step 3: Simplify the mean calculation
Calculating each term:
- \( 1 \cdot \frac{1}{6} = \frac{1}{6} \)
- \( 2 \cdot \frac{1}{3} = \frac{2}{3} = \frac{4}{6} \)
- \( 3 \cdot \frac{1}{3} = 1 = \frac{6}{6} \)
- \( 4 \cdot \frac{1}{6} = \frac{4}{6} \)

Now, summing these:
\[ \mu = \frac{1}{6} + \frac{4}{6} + \frac{6}{6} + \frac{4}{6} = \frac{1 + 4 + 6 + 4}{6} = \frac{15}{6} = \frac{5}{2} \]

### Step 4: Calculate the variance of \( X \)
The variance \( \sigma^2 \) is calculated using the formula:
\[ \sigma^2 = E(X^2) - (E(X))^2 \]
First, we need to calculate \( E(X^2) \):
\[ E(X^2) = \sum_{i=1}^{n} x_i^2 P(X=x_i) \]
Calculating \( E(X^2) \):
\[ E(X^2) = 1^2 \cdot \frac{1}{6} + 2^2 \cdot \frac{1}{3} + 3^2 \cdot \frac{1}{3} + 4^2 \cdot \frac{1}{6} \]
Calculating each term:
- \( 1^2 \cdot \frac{1}{6} = \frac{1}{6} \)
- \( 2^2 \cdot \frac{1}{3} = 4 \cdot \frac{1}{3} = \frac{4}{3} = \frac{8}{6} \)
- \( 3^2 \cdot \frac{1}{3} = 9 \cdot \frac{1}{3} = 3 = \frac{18}{6} \)
- \( 4^2 \cdot \frac{1}{6} = 16 \cdot \frac{1}{6} = \frac{16}{6} \)

Now, summing these:
\[ E(X^2) = \frac{1}{6} + \frac{8}{6} + \frac{18}{6} + \frac{16}{6} = \frac{1 + 8 + 18 + 16}{6} = \frac{43}{6} \]

### Step 5: Calculate the variance
Now, substituting back into the variance formula:
\[ \sigma^2 = E(X^2) - (E(X))^2 = \frac{43}{6} - \left(\frac{5}{2}\right)^2 \]
Calculating \( \left(\frac{5}{2}\right)^2 = \frac{25}{4} \):
\[ \sigma^2 = \frac{43}{6} - \frac{25}{4} \]

### Step 6: Find a common denominator and simplify
The common denominator for 6 and 4 is 12:
\[ \sigma^2 = \frac{43 \cdot 2}{12} - \frac{25 \cdot 3}{12} = \frac{86}{12} - \frac{75}{12} = \frac{11}{12} \]

### Final Answer
Thus, the mean and variance of the random variable \( X \) are:
- Mean \( \mu = \frac{5}{2} \)
- Variance \( \sigma^2 = \frac{11}{12} \)