MHT CET · Maths · Probability
A random variable \(X\) has the following probability distribution
\( \begin{array}{c|cccc} X = x_i & 1 & 2 & 3 & 4 \\ \hline P(X = x_i) & 0.2 & 0.4 & 0.3 & 0.1 \end{array} \)
The mean and variance of X are respectively
- A 2.3 and 6.1
- B 2.3 and 0.81
- C 2.3 and 0.1
- D 2.3 and 0.9
Answer & Solution
Correct Answer
(B) 2.3 and 0.81
Step-by-step Solution
Detailed explanation
\(E[X] = 1(0.2) + 2(0.4) + 3(0.3) + 4(0.1) = 2.3\) \(E[X^2] = 1^2(0.2) + 2^2(0.4) + 3^2(0.3) + 4^2(0.1) = 6.1\)
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