A random variable X takes the value 1,2,3 and 4 such that
`2P(X=1)=3P(X=2)=P(X=3)=5P(X=4)`. If `sigma^(2)` is the variance and `mu` is the mean of X then `sigma^(2)+mu^(2)=`

Options

A. `421/61`
B. `570/61`
C. `149/61`
D. `3480/3721`

Answer

To solve the problem step by step, we will first express the probabilities in terms of a single variable and then find the mean and variance of the random variable \(X\).

### Step 1: Define the probabilities
Given the relationships:
\[ 2P(X=1) = 3P(X=2) = P(X=3) = 5P(X=4) \]
Let \(P(X=1) = 2k\), \(P(X=2) = 3k\), \(P(X=3) = k\), and \(P(X=4) = 5k\).

### Step 2: Set up the equation for total probability
The total probability must equal 1:
\[ P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1 \]
Substituting the probabilities we defined:
\[ 2k + 3k + k + 5k = 1 \]
This simplifies to:
\[ 11k = 1 \]
Thus, we find:
\[ k = \frac{1}{11} \]

### Step 3: Calculate individual probabilities
Now we can calculate the individual probabilities:
\[ P(X=1) = 2k = \frac{2}{11}, \quad P(X=2) = 3k = \frac{3}{11}, \quad P(X=3) = k = \frac{1}{11}, \quad P(X=4) = 5k = \frac{5}{11} \]

### Step 4: Calculate the mean (\(\mu\))
The mean \(\mu\) is given by:
\[ \mu = E(X) = \sum x_i P(X=x_i) = 1 \cdot P(X=1) + 2 \cdot P(X=2) + 3 \cdot P(X=3) + 4 \cdot P(X=4) \]
Substituting the probabilities:
\[ \mu = 1 \cdot \frac{2}{11} + 2 \cdot \frac{3}{11} + 3 \cdot \frac{1}{11} + 4 \cdot \frac{5}{11} \]
Calculating each term:
\[ \mu = \frac{2}{11} + \frac{6}{11} + \frac{3}{11} + \frac{20}{11} = \frac{31}{11} \]

### Step 5: Calculate \(E(X^2)\)
Next, we calculate \(E(X^2)\):
\[ E(X^2) = \sum x_i^2 P(X=x_i) = 1^2 \cdot P(X=1) + 2^2 \cdot P(X=2) + 3^2 \cdot P(X=3) + 4^2 \cdot P(X=4) \]
Substituting the probabilities:
\[ E(X^2) = 1^2 \cdot \frac{2}{11} + 2^2 \cdot \frac{3}{11} + 3^2 \cdot \frac{1}{11} + 4^2 \cdot \frac{5}{11} \]
Calculating each term:
\[ E(X^2) = \frac{2}{11} + \frac{12}{11} + \frac{9}{11} + \frac{80}{11} = \frac{103}{11} \]

### Step 6: Calculate the variance (\(\sigma^2\))
The variance \(\sigma^2\) is given by:
\[ \sigma^2 = E(X^2) - \mu^2 \]
Calculating \(\mu^2\):
\[ \mu^2 = \left(\frac{31}{11}\right)^2 = \frac{961}{121} \]
Now substituting into the variance formula:
\[ \sigma^2 = \frac{103}{11} - \frac{961}{121} \]
To perform this subtraction, we convert \(\frac{103}{11}\) to a fraction with a denominator of 121:
\[ \frac{103}{11} = \frac{103 \cdot 11}{121} = \frac{1133}{121} \]
Thus,
\[ \sigma^2 = \frac{1133}{121} - \frac{961}{121} = \frac{172}{121} \]

### Step 7: Calculate \(\sigma^2 + \mu^2\)
Now we can find \(\sigma^2 + \mu^2\):
\[ \sigma^2 + \mu^2 = \frac{172}{121} + \frac{961}{121} = \frac{1133}{121} \]

### Final Answer
Thus, the final answer is:
\[ \sigma^2 + \mu^2 = \frac{1133}{121} \]