Question

If the probability distribution is given by, \[ \begin{array}{|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 \\ \hline p(n) & 8a-1 \, /30 & 4a-1 \, /30 & 2a+1 \, /30 & b \\ \hline \end{array} \] If it is given that \( \sigma^2 + \mu^2 = 2 \), where \( \sigma \) is the standard deviation and \( \mu \) is the mean of the distribution, then \( \frac{a}{b} \) is:

• A. \( \frac{22}{71} \)
• B. \( \frac{110}{71} \)
• C. \( \frac{220}{71} \)
• D. \( \frac{1110}{71} \)

Hint:

In probability distribution problems, make sure that the sum of all probabilities equals 1 and use the relationships for mean and variance to solve for unknowns.

Answer 1

The Correct Option is D

Step 1: Understand the given probability distribution.
The sum of all probabilities must be equal to 1, so we write the following equation based on the probability distribution: \[ \frac{8a-1}{30} + \frac{4a-1}{30} + \frac{2a+1}{30} + b = 1 \]
Step 2: Solve for \( b \).
Simplifying the equation: \[ \frac{8a-1 + 4a-1 + 2a+1}{30} + b = 1 \] \[ \frac{14a - 1}{30} + b = 1 \] \[ b = 1 - \frac{14a - 1}{30} = \frac{30 - (14a - 1)}{30} = \frac{31 - 14a}{30} \]
Step 3: Use the formula for the mean \( \mu \).
The mean \( \mu \) is given by: \[ \mu = \sum x \cdot p(x) = 0 \cdot \frac{8a-1}{30} + 1 \cdot \frac{4a-1}{30} + 2 \cdot \frac{2a+1}{30} + 3 \cdot b \] \[ \mu = \frac{(4a-1) + 2(2a+1) + 3b}{30} \] Substitute \( b = \frac{31 - 14a}{30} \): \[ \mu = \frac{(4a-1) + 2(2a+1) + 3\left(\frac{31 - 14a}{30}\right)}{30} \] Simplifying this equation will give the mean \( \mu \).
Step 4: Use the variance equation.
The variance \( \sigma^2 \) is given by: \[ \sigma^2 = \sum (x - \mu)^2 \cdot p(x) \] We use the formula for variance and the equation \( \sigma^2 + \mu^2 = 2 \) to solve for \( a \) and \( b \).
Step 5: Solve for \( \frac{a}{b} \).
After solving the system of equations using the provided conditions, we find that the value of \( \frac{a}{b} \) is \( \frac{1110}{71} \).