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 $12$:$\operatorname{Var}(X) = \frac{2(2n^2 + 3n + 1) - 3(n^2 + 2n + 1)}{12} = \frac{4n^2 + 6n + 2 - 3n^2 - 6n - 3}{12} = \frac{n^2 - 1}{12}$.

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

Number of smart phones $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
Probability $(P(x))$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

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 a game of tossing a six-sided fair die. If the face that comes up is $k$, the player wins Rs. $36/k$ and loses Rs. $2k$, where $k \in \{1, 2, 3, 4, 5, 6\}$. What is the expected winning amount in this game in Rs.?

The following table shows the probability distribution of smart phones sold in a shop per day:
Number of smart phones $(x)$ $0$ $1$ $2$ $3$ $4$ $5$
Probability $(P(x))$ $k$ $0.3$ $0.15$ $0.15$ $0.1$ $2k$

Then $E(x) = ?$

In a game,a man wins $₹ 40$ if he gets $5$ or $6$ on a throw of a fair die and loses $₹ 20$ for getting any other number on the die. If he decides to throw the die either until he gets a $5$ or $6$ or to a maximum of $3$ throws,then his expected gain/loss (in rupees) is:

$A$ person draws a card from a pack of $52$ cards and replaces it. After shuffling,he draws a card again. If he repeats this process,what is the probability that he draws a red card for the first time on the third draw?

Two dice are thrown simultaneously. If $X$ denotes the number of sixes,find the expectation of $X$.

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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) = $