(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 $X$.
$\begin{array}{|c|c|c|c|c|c|c|} \hline X=x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(X=x) & 0.1 & k & 0.2 & 2k & 0.3 & k \\ \hline \end{array}$

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$1.7$
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$1.8$
C
$0.8$
D
$0.7$

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$A$ random variable $X$ has the following probability distribution:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X = x)$ $k$ $3k$ $5k$ $7k$ $8k$ $k$

Then $P(2 \leq X < 5) = $

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In a hospital, on an average if there are $35$ births in a week, then the probability that there will be less than $3$ births in a day is:

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$A$ random variable $X$ has the following probability distribution:

$X=x_i$ $-2$ $-1$ $0$ $1$ $2$

$P(X=x_i)$ $1/6$ $k$ $1/4$ $k$ $1/6$

The variance of this random variable is

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The probability distribution of a random variable $X$ is given by:
$X = x_i$ $0$ $1$ $2$ $3$ $4$
$P(X = x_i)$ $0.4$ $0.3$ $0.1$ $0.1$ $0.1$

Then the variance of $X$ is:

$X = x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

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$A$ random variable $X$ has the probability distribution given below. Its variance is:
$X$ $1$ $2$ $3$ $4$ $5$
$P(X=x)$ $K$ $2K$ $3K$ $2K$ $K$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

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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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