Question

For a discrete random variable X, whose probability distribution is defined as :
\(P(x)=\begin{cases} 2k(x+1) ;& x = 0,1 \\ 3kx; &  x=2 \\ k(5-x) & x=3 \end{cases}\)
The value of mean will be

• A. \(\frac{6}{7}\)
• B. \(\frac{15}{7}\)
• C. \(\frac{12}{7}\)
• D. \(\frac{11}{7}\)

Answer 1

The Correct Option is D

To find the mean of the given discrete random variable \(X\) with its probability distribution, we must first determine the value of \(k\) by ensuring the sum of probabilities equals 1.
The probability distribution is:
\(P(x)=\begin{cases} 2k(x+1) & x = 0,1 \\ 3kx & x=2 \\ k(5-x) & x=3 \end{cases}\)
Calculating each probability:
\(P(0)=2k(0+1)=2k\)
\(P(1)=2k(1+1)=4k\)
\(P(2)=3k \cdot 2=6k\)
\(P(3)=k(5-3)=2k\)
The sum of these probabilities must equal 1:
\(2k+4k+6k+2k=14k=1\)
Solving for \(k\):
\(k=\frac{1}{14}\)
With \(k\) known, we can compute probabilities:
\(P(0)=2k=\frac{2}{14}=\frac{1}{7}\)
\(P(1)=4k=\frac{4}{14}=\frac{2}{7}\)
\(P(2)=6k=\frac{6}{14}=\frac{3}{7}\)
\(P(3)=2k=\frac{2}{14}=\frac{1}{7}\)
Now calculate the mean (\(E[X]\)):
\(E[X]=\sum xP(x)=0\cdot\frac{1}{7}+1\cdot\frac{2}{7}+2\cdot\frac{3}{7}+3\cdot\frac{1}{7}\)
\(E[X]=0+\frac{2}{7}+\frac{6}{7}+\frac{3}{7}\)
\(E[X]=\frac{2+6+3}{7}=\frac{11}{7}\)
Thus, the mean of the distribution is \(\frac{11}{7}\).