Question

If \( X \) is a random variable with the probability mass function (p.m.f.) as follows: \[ P(X = x) = \begin{cases} \frac{5}{16}, & x = 0, \\ \frac{kx}{48}, & x = 1, \\ \frac{1}{4}, & x = 2, \\ \frac{1}{4}, & x = 3, \end{cases} \] then find \( E(X) \):

Hint:

To calculate the expected value \( E(X) \), ensure that the total probability sums to 1 and substitute each \( x \cdot P(X = x) \) term carefully into the summation formula.

Answer 1

The expected value \( E(X) \) is given by: \[ E(X) = \sum_x x \cdot P(X = x). \] 
Step 1: Verify the total probability
The total probability must sum to 1: \[ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = \frac{5}{16} + \frac{k}{48} + \frac{1}{4} + \frac{1}{4}. \] 
Substitute \( \frac{1}{4} = \frac{12}{48} \): \[ \frac{5}{16} + \frac{k}{48} + \frac{12}{48} + \frac{12}{48} = 1. \] 
Convert \( \frac{5}{16} \) to a denominator of 48: \[ \frac{5}{16} = \frac{15}{48}. \] 
\[ \frac{15}{48} + \frac{k}{48} + \frac{12}{48} + \frac{12}{48} = 1. \] 
Simplify: \[ \frac{15 + k + 12 + 12}{48} = 1. \] 
\[ \frac{39 + k}{48} = 1 \quad \implies \quad 39 + k = 48 \quad \implies \quad k = 9. \] 
Step 2: Find \( P(X = 1) \)
Substitute \( k = 9 \): \[ P(X = 1) = \frac{k \cdot 1}{48} = \frac{9}{48}. \] 
Step 3: Calculate \( E(X) \)
Substitute the probabilities into the formula for \( E(X) \): \[ E(X) = 0 \cdot \frac{5}{16} + 1 \cdot \frac{9}{48} + 2 \cdot \frac{12}{48} + 3 \cdot \frac{12}{48}. \] 
Simplify: \[ E(X) = 0 + \frac{9}{48} + \frac{24}{48} + \frac{36}{48}. \] 
\[ E(X) = \frac{9 + 24 + 36}{48} = \frac{69}{48}. \] 
\[ E(X) = 1.4375. \] 
Final Answer: \[ \boxed{1.4375} \]