Question

The probability distribution of a discrete random variable X is defined as:
\(P(X=x)=\begin{cases} 3kx & \text{for } x=1,2,3\\  5k(x+2) & \text{for } x=4,5 \\ 0& \text{otherwise}\end{cases}\)
The mean of the distribution is:

• A. \(\frac{92}{23}\)
• B. \(\frac{337}{83}\)
• C. \(\frac{65}{34}\)
• D. \(\frac{10}{83}\)

Answer 1

The Correct Option is B

To find the mean of the probability distribution of the discrete random variable \(X\), we start by determining the probability mass function and then calculating the mean using the formula for expected value. The probability distribution is given by:
\(P(X=x)=\begin{cases}3kx & \text{for } x=1,2,3\\5k(x+2) & \text{for } x=4,5\\0 & \text{otherwise}\end{cases}\)
First, ensure the probabilities sum to 1:\(3k(1)+3k(2)+3k(3)+5k(4+2)+5k(5+2)=1\)
Calculating each term:

1. \(3k(1)=3k\)
2. \(3k(2)=6k\)
3. \(3k(3)=9k\)
4. \(5k(6)=30k\)
5. \(5k(7)=35k\)

Sum:\(3k+6k+9k+30k+35k=83k\)
Set this equal to 1:\(83k=1\Rightarrow k=\frac{1}{83}\)
Now, calculate the mean \(E(X)\):
Expected value formula:\(E(X)=\sum x_iP(X=x_i)\)

1. \(x=1\), \(P(X=1)=3k=3\left(\frac{1}{83}\right)=\frac{3}{83}\), contribution to \(E(X)\) is \(\frac{3}{83}\times1=\frac{3}{83}\)
2. \(x=2\), \(P(X=2)=6k=\frac{6}{83}\), contribution to \(E(X)\) is \(\frac{12}{83}\)
3. \(x=3\), \(P(X=3)=9k=\frac{9}{83}\), contribution to \(E(X)\) is \(\frac{27}{83}\)
4. \(x=4\), \(P(X=4)=30k=\frac{30}{83}\), contribution to \(E(X)\) is \(\frac{120}{83}\)
5. \(x=5\), \(P(X=5)=35k=\frac{35}{83}\), contribution to \(E(X)\) is \(\frac{175}{83}\)

Sum the contributions:
\(\frac{3}{83}+\frac{12}{83}+\frac{27}{83}+\frac{120}{83}+\frac{175}{83}=\frac{337}{83}\)
Thus, the mean of the distribution is \(\frac{337}{83}\).