(C) For a probability distribution,the sum of all probabilities must be equal to $1$.Therefore,$0.1 + k + 0.2 + 2k + 0.3 + k = 1$.Combining the terms,

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(C) For a probability distribution,the sum of all probabilities must be equal to $1$.Therefore,$0.1 + k + 0.2 + 2k + 0.3 + k = 1$.Combining the terms,we get $0.6 + 4k = 1$.$4k = 0.4$,which implies $k = 0.1$.The mean of $X$,denoted by $E(X)$,is given by $\sum x_i P(x_i)$.$E(X) = (-2 \times 0.1) + (-1 \times 0.1) + (0 \times 0.2) + (1 \times 0.2) + (2 \times 0.3) + (3 \times 0.1)$.$E(X) = -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3$.$E(X) = 0.8$.

Class 12 Mathematics Probability Probability distribution

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The following table represents the probability distribution of a random variable $X$ for some $k \in Q$. Find the mean of $X$.
$\begin{array}{|c|c|c|c|c|c|c|} \hline X=x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(X=x) & 0.1 & k & 0.2 & 2k & 0.3 & k \\ \hline \end{array}$

AP EAMCET 2020 All AP EAMCET

A
$1.7$
B
$1.8$
C
$0.8$
D
$0.7$

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The following table represents the probability distribution of a random variable $X$ for some $k \in Q$. Find the mean of $X$.$\begin{array}{|c|c|c|c|c|c|c|} \hline X=x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(X=x) & 0.1 & k & 0.2 & 2k & 0.3 & k \\ \hline \end{array}$

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The following table represents the probability distribution of a random variable $X$ for some $k \in Q$. Find the mean of $X$.
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