A random variable $X$ has the following probability distribution : $\begin{array}{|c|c|c|c|c|} \hline X=x & 1 & 2 & 3 & 4 \ \hline P(X=x) & 0.1 & 0.2 & 0.3 & 0.4 \ \hline \end{array}$ The mean and standard deviation of $X$ are respectively.

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Aicharya · Accepted Answer

The mean $E(X)$ and standard deviation $\sigma$ are calculated from the given probability distribution using standard formulas.

$E(X) = \sum x_i P(X=x_i) = (1)(0.1) + (2)(0.2) + (3)(0.3) + (4)(0.4) = 3.0$

$E(X^2) = \sum x_i^2 P(X=x_i) = (1)(0.1) + (4)(0.2) + (9)(0.3) + (16)(0.4) = 10.0$

The variance is $\text{Var}(X) = E(X^2) - [E(X)]^2 = 10.0 - 9.0 = 1.0$,
giving standard deviation $\sigma = \sqrt{1.0} = 1$.

The values $E(X)=3$ and $\sigma=1$ correspond to option B.

Final answer: $\boxed{\text{B}}$

A random variable $X$ has the following probability distribution : $\begin{array}{|c|c|c|c|c|} \hline X=x &…

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