Question

The probability distribution of a discrete random variable (r.v.) \(X\) is given below:

\(X=x\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(P(X=x)\)	\(k\)	\(2k\)	\(4k\)	\(2k\)	\(k\)

Then the value of \(P[X\leq E(X)]\) is:

• A. \(\frac{7}{10}\)
• B. \(\frac{1}{10}\)
• C. \(\frac{3}{10}\)
• D. \(\frac{9}{10}\)

Hint:

Use the total probability equal to 1 to find k, compute the expected value, and add the probabilities up to that value.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The probabilities of a distribution add up to 1, and \(E(X)=\sum xP(X=x)\).

Step 2: Find k
\(k+2k+4k+2k+k=10k=1\), so \(k=\dfrac1{10}\).

Step 3: Find E(X)
\(E(X)=0\cdot k+1\cdot2k+2\cdot4k+3\cdot2k+4\cdot k=20k=2\).

Step 4: Calculation
\(P(X\leq2)=k+2k+4k=7k=\dfrac7{10}\).

Final Answer:
The probability P[X less than or equal to E(X)] is 7 over 10. \[ \boxed{\text{(A) }\dfrac{7}{10}} \]