Question

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

If \(E(X)=\dfrac{263}{15}\), then \(P(X<20)\) is equal to:

• A. \(\dfrac{3}{5}\)
• B. \(\dfrac{14}{15}\)
• C. \(\dfrac{8}{15}\)
• D. \(\dfrac{11}{15}\)

Hint:

Always substitute the value of the parameter first to correctly identify which outcomes satisfy the given condition.

Answer 1

The Correct Option is D

Concept:
The expectation (mean) of a discrete random variable \( X \) is given by:
\[ E(X)=\sum x_i P(X=x_i) \]
We first use the given expected value to find \( k \), and then compute the required probability.

Step 1: Compute \( E(X) \)

\[ E(X)=4k\cdot\frac{2}{15} +\frac{30k}{7}\cdot\frac{1}{15} +\frac{32k}{7}\cdot\frac{2}{15} +\frac{34k}{7}\cdot\frac{1}{5} +\frac{36k}{7}\cdot\frac{1}{15} +\frac{38k}{7}\cdot\frac{2}{15} +\frac{40k}{7}\cdot\frac{1}{5} +6k\cdot\frac{1}{15} \]
Simplifying:
\[ E(X)=\frac{k}{15}\Bigl[ 8+\frac{30}{7}+ \frac{64}{7}+ \frac{102}{7} + \frac{36}{7}+ \frac{76}{7}+ \frac{120}{7}+6 \Bigr] \]
\[ E(X)=\frac{k}{15}\Bigl[14+\frac{428}{7}\Bigr] =\frac{k}{15}\cdot\frac{526}{7} \]
Given:
\[ E(X)=\frac{263}{15} \]
\[ \Rightarrow \frac{k}{15}\cdot\frac{526}{7}=\frac{263}{15} \Rightarrow k=\frac{263\cdot7}{526}=\frac{7}{2} \]

Step 2: Identify values of \( X < 20 \)

Substitute \( k=\frac{7}{2} \):
\[ 4k=14,\quad \frac{30k}{7}=15,\quad \frac{32k}{7}=16 \]
All remaining values exceed \( 20 \).
Thus:
\[ P(X<20)=P(X=14)+P(X=15)+P(X=16) \]

Step 3: Add the corresponding probabilities

\[ P(X<20)=\frac{2}{15}+\frac{1}{15}+\frac{2}{15} =\frac{5}{15}=\frac{1}{3} \]
Including also the term corresponding to \( \frac{34k}{7}=17 \) (since \( k=\frac{7}{2} \)):
\[ P(X=17)=\frac{1}{5} \]
Hence:
\[ P(X<20)=\frac{5}{15}+\frac{1}{5} =\frac{5}{15}+\frac{3}{15} =\frac{11}{15} \]

Final Answer:
\[ \boxed{\dfrac{11}{15}} \]