The p.m.f. of a random variable $X$ is $P(x) = \begin{cases} \frac{2x}{n(n+1)}, & x = 1, 2, 3, \ldots, n \\ 0, & \text{otherwise} \end{cases}$,then $E(X)$ is
- A$\frac{n+1}{6}$
- B$\frac{2n+1}{6}$
- C$\frac{n+1}{3}$
- D$\frac{2n+1}{3}$
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If $X$ is a random variable with the following probability distribution:
$X=x$ $-3$ $6$ $9$ $P(X=x)$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{1}{3}$
Then the variance of $X$ is:
| $X=x$ | $-3$ | $6$ | $9$ |
| $P(X=x)$ | $\frac{1}{6}$ | $\frac{1}{2}$ | $\frac{1}{3}$ |
Then the variance of $X$ is:
Let $X$ be a random variable with probability distribution function,$P(X=x)=K\left(\frac{2}{5}\right)^x, x=1, 2, 3, \ldots$. Then,the value of $K$ is
If the probability mass function (p.m.f.) of a random variable $X$ is $P(X=x) = \frac{1}{10}$ for $x = 1, 2, 3, \ldots, 10$,and $0$ otherwise,then $\operatorname{Var}(X)$ is equal to:
The probability distribution of $x$ is given by the following table:
$x$ $0$ $1$ $2$ $3$ $P(x)$ $0.2$ $k$ $k$ $2k$
Find the value of $k$.
| $x$ | $0$ | $1$ | $2$ | $3$ |
| $P(x)$ | $0.2$ | $k$ | $k$ | $2k$ |
Find the value of $k$.
DifficultKCET 2015
View Solution$A$ random variable $X$ takes values $0, 1, 2, 3, \dots$ with probabilities $P(X=x) = k(x+1)\left(\frac{1}{2}\right)^x$. If $k$ is a constant,then $P(X=1) = $
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