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:
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:
\(P(X=x)=\begin{cases} 3kx & \text{for } x=1,2,3\\ 5k(x+2) & \text{for } x=4,5 \\ 0& \text{otherwise}\end{cases}\)
The mean of the distribution is:
Updated On: Jun 14, 2024
- \(\frac{92}{23}\)
- \(\frac{337}{83}\)
- \(\frac{65}{34}\)
- \(\frac{10}{83}\)
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The Correct Option is B
Solution and Explanation
The correct option is (B): \(\frac{337}{83}\)
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