When an unfair dice is thrown,the probability | vedclass

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(D) Given $P(X=k) = k^2 P$ for $k = 1, 2, 3, 4, 5, 6$. Since the sum of all probabilities must be $1$,we have: $\sum_{k=1}^6 P(X=k) = 1$ $P(1^2) + P(2

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$\frac{441}{91}$

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(D) Given $P(X=k) = k^2 P$ for $k = 1, 2, 3, 4, 5, 6$. Since the sum of all probabilities must be $1$,we have: $\sum_{k=1}^6 P(X=k) = 1$ $P(1^2) + P(2^2) + P(3^2) + P(4^2) + P(5^2) + P(6^2) = 1$ $P(1 + 4 + 9 + 16 + 25 + 36) = 1$ $91P = 1 \Rightarrow P = \frac{1}{91}$. The mean of $X$ is given by $E(X) = \sum_{k=1}^6 k \cdot P(X=k)$. $E(X) = \sum_{k=1}^6 k \cdot (k^2 P) = P \sum_{k=1}^6 k^3$. $E(X) = \frac{1}{91} (1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3)$. $E(X) = \frac{1}{91} (1 + 8 + 27 + 64 + 125 + 216) = \frac{441}{91}$.

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$0.2$	$0.3$	$0.15$	$0.25$	$0.1$

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

$x = x_{i}$	$0$	$1$	$2$
$P_{i}$	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

When an unfair dice is thrown,the probability of getting a number $k$ on it is $P(X=k)=k^2 P$,where $k=1, 2, 3, 4, 5, 6$ and $X$ is the random variable denoting a number on the dice,then the mean of $X$ is

Similar Questions

If the probability distribution function of a random variable $X$ is given as follows:
$X=x_i$ $-2$ $-1$ $0$ $1$ $2$
$P(X=x_i)$ $0.2$ $0.3$ $0.15$ $0.25$ $0.1$

Then $F(0)$ is equal to:

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:

For the probability distribution given by

$x = x_{i}$ $0$ $1$ $2$

$P_{i}$ $\frac{25}{36}$ $\frac{5}{18}$ $\frac{1}{36}$

the standard deviation $(\sigma)$ is

If $x$ is a random variable with $PMF$ as follows: $P(X = x) = \begin{cases} \frac{5}{16}, & x = 0, 1 \\ \frac{kx}{48}, & x = 2 \\ \frac{1}{4}, & x = 3 \end{cases}$ then find $E(x)$.

The probability distribution of a random variable $X$ is given below.

$X = x$ $0$ $1$ $2$ $3$

$P(X = x)$ $\frac{1}{10}$ $\frac{2}{10}$ $\frac{3}{10}$ $\frac{4}{10}$

Then the variance of $X$ is

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