Question

The probability distribution of a random variable \( X \) is given below:

\( X \)	1	2	4	2k	3k	5k
\( P(X) \)	\( \frac{1}{2} \)	\( \frac{1}{5} \)	\( \frac{3}{25} \)	\( \frac{1}{10} \)	\( \frac{1}{25} \)	\( \frac{1}{25} \)

Answer 1

Expected Value and Probability from Distribution

Given:

Expected value \( E(X) = 2.94 \)

Step 1: Write the expression for \( E(X) \)

\[ E(X) = 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{5} + 4 \cdot \frac{3}{25} + (2k) \cdot \frac{1}{10} + (3k) \cdot \frac{1}{25} + (5k) \cdot \frac{1}{25} \]

Step 2: Simplify constant and variable terms

Constants:

\[ 1 \cdot \frac{1}{2} = \frac{1}{2},\quad 2 \cdot \frac{1}{5} = \frac{2}{5},\quad 4 \cdot \frac{3}{25} = \frac{12}{25} \]

Convert to denominator 50:

\[ \frac{1}{2} = \frac{25}{50},\quad \frac{2}{5} = \frac{20}{50},\quad \frac{12}{25} = \frac{24}{50} \Rightarrow \text{Sum} = \frac{25 + 20 + 24}{50} = \frac{69}{50} \]

Variable terms:

\[ \frac{2k}{10} = \frac{k}{5} = \frac{10k}{50},\quad \frac{3k}{25} + \frac{5k}{25} = \frac{8k}{25} = \frac{16k}{50} \Rightarrow \text{Total variable part} = \frac{10k + 16k}{50} = \frac{26k}{50} \]

Full expression for expected value:

\[ E(X) = \frac{69 + 26k}{50} \]

Step 3: Use given value \( E(X) = 2.94 \)

Equating:

\[ \frac{69 + 26k}{50} = 2.94 \Rightarrow 69 + 26k = 2.94 \cdot 50 = 147 \Rightarrow 26k = 147 - 69 = 78 \Rightarrow k = \frac{78}{26} = \boxed{3} \]

Step 4: Find \( P(X \leq 4) \)

Values of \( X \leq 4 \) are: 1, 2, 4

\[ P(X \leq 4) = \frac{1}{2} + \frac{1}{5} + \frac{3}{25} = \frac{25}{50} + \frac{10}{50} + \frac{6}{50} = \frac{41}{50} \]

✅ Final Answers:

• \( \boxed{k = 3} \)
• \( \boxed{P(X \leq 4) = \frac{41}{50}} \)