Question

If a random variable \( x \) has the probability distribution 

then \( P(3<x \leq 6) \) is equal to

• A. 0.22
• B. 0.33
• C. 0.34
• D. 0.64

Hint:

Always use the normalization condition \( \sum P(x) = 1 \) first to find unknown constants in a probability distribution before calculating specific probabilities.

Answer 1

The Correct Option is D

Since \( x \) is a random variable, the sum of all probabilities must be equal to 1.
Step 1: Use the normalization condition.
\[ 0 + 2k + k + 3k + 2k^2 + 2k + (k^2 + k) + 7k^2 = 1 \] \[ 9k + 10k^2 = 1 \] \[ 10k^2 + 9k - 1 = 0 \] Solving this quadratic equation, we get: \[ k = \frac{-9 + 11}{20} = \frac{1}{10} \] (The negative value is rejected since probability cannot be negative.)
Step 2: Identify the values satisfying \( 3<x \leq 6 \).
The values of \( x \) satisfying \( 3<x \leq 6 \) are: \[ x = 4,\;5,\;6 \]
Step 3: Calculate the required probability.
\[ P(3<x \leq 6) = P(4) + P(5) + P(6) \] \[ = 2k^2 + 2k + (k^2 + k) \] Substituting \( k = \frac{1}{10} \), \[ = 2\left(\frac{1}{10}\right)^2 + 2\left(\frac{1}{10}\right) + \left(\frac{1}{10}\right)^2 + \frac{1}{10} \] \[ = 0.02 + 0.20 + 0.01 + 0.10 = 0.64 \] Final Answer: \[ \boxed{0.64} \]