The probability distribution of a random vari | vedclass

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(A) The mean of the random variable $X$ is given by $E(X) = \sum p_i x_i = \frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \dots + \frac{n}{n}$.$E(X) = \fra

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$\frac{n^2-1}{12}$

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(A) The mean of the random variable $X$ is given by $E(X) = \sum p_i x_i = \frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \dots + \frac{n}{n}$.$E(X) = \frac{1}{n} (1 + 2 + 3 + \dots + n) = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$.The expected value of $X^2$ is given by $E(X^2) = \sum p_i x_i^2 = \frac{1^2}{n} + \frac{2^2}{n} + \frac{3^2}{n} + \dots + \frac{n^2}{n}$.$E(X^2) = \frac{1}{n} (1^2 + 2^2 + 3^2 + \dots + n^2) = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$.The variance of $X$ is given by $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$.$\operatorname{Var}(X) = \frac{(n+1)(2n+1)}{6} - \left( \frac{n+1}{2} \right)^2$.$\operatorname{Var}(X) = \frac{2n^2 + 3n + 1}{6} - \frac{n^2 + 2n + 1}{4}$.Taking the least common multiple as $12$:$\operatorname{Var}(X) = \frac{2(2n^2 + 3n + 1) - 3(n^2 + 2n + 1)}{12} = \frac{4n^2 + 6n + 2 - 3n^2 - 6n - 3}{12} = \frac{n^2 - 1}{12}$.

$X = x$	$1$	$2$	$3$	$\dots$	$n$
$P(X = x)$	$\frac{1}{n}$	$\frac{1}{n}$	$\frac{1}{n}$	$\dots$	$\frac{1}{n}$

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

The probability distribution of a random variable $X$ is given by the following table:
$X = x$ $1$ $2$ $3$ $\dots$ $n$
$P(X = x)$ $\frac{1}{n}$ $\frac{1}{n}$ $\frac{1}{n}$ $\dots$ $\frac{1}{n}$

Then $\operatorname{Var}(X) = $

Similar Questions

Consider the probability distribution
$\begin{array}{|r|c|c|c|c|c|} \hline X=x & 1 & 2 & 3 & 4 & 5 \\ \hline P(X=x) & K & 2K & K^2 & 2K & 5K^2 \\ \hline \end{array}$
Then the value of $P(X > 2)$ is

For the following probability distribution, the standard deviation of the random variable $X$ is:
$X: 0, 1, 2$
$P(X): 0.3, 0.4, 0.3$

The variance of the following probability distribution is:
$x$ $0$ $1$ $2$
$P(X)$ $\frac{9}{16}$ $\frac{3}{8}$ $\frac{1}{16}$

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

$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$

Determine $k$.

If $X$ is a Poisson variate with mean $2$,then $P\left(X>\frac{3}{2}\right)=$

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The probability distribution of a random variable $X$ is given by the following table:$X = x$$1$$2$$3$$\dots$$n$$P(X = x)$$\frac{1}{n}$$\frac{1}{n}$$\frac{1}{n}$$\dots$$\frac{1}{n}$Then $\operatorname{Var}(X) = $