Question

The probability distribution of a random variable \(X\) is given below. Then, the mean of \(X\) is 

\(X=x_i\)	\(0\)	\(1\)	\(2\)
\(P(X=x_i)\)	\(3k^2\)	\(2k^2\)	\(k^2+k\)

• A. \(\dfrac79\)
• B. \(\dfrac59\)
• C. \(\dfrac{10}{9}\)
• D. \(\dfrac{13}{9}\)

Hint:

For a discrete random variable, \[ \boxed{E(X)=\sum x_iP(X=x_i).} \] Before finding the mean, always verify that \[ \boxed{\sum P(X=x_i)=1.} \]

Answer 1

The Correct Option is C

Step 1: Use the fact that the total probability is \(1\). We have \[ 3k^2+2k^2+(k^2+k)=1. \] Therefore, \[ 6k^2+k-1=0. \] Factoring, \[ (3k-1)(2k+1)=0. \] Since probabilities must be non-negative, \[ \boxed{k=\frac13.} \]

Step 2: Find the probabilities. Substituting \(k=\dfrac13\), \[ P(X=0) = 3\left(\frac13\right)^2 = \frac13, \] \[ P(X=1) = 2\left(\frac13\right)^2 = \frac29, \] \[ P(X=2) = \left(\frac13\right)^2+\frac13 = \frac19+\frac39 = \frac49. \]

Step 3: Calculate the mean. The mean of a random variable is \[ E(X) = \sum x_iP(X=x_i). \] Hence, \[ E(X) = 0\left(\frac13\right) + 1\left(\frac29\right) + 2\left(\frac49\right) = \frac29+\frac89 = \frac{10}{9}. \] Therefore, \[ \boxed{E(X)=\frac{10}{9}}. \] Hence, the correct option is \(\boxed{(C)}\).