If the probability function of a random variable $X$ is given by $P(X=j) = \frac{1}{2^j}$ for $j = 1, 2, 3, \ldots, \infty$,then the variance of $X$ is:
- A$2$
- B$3$
- C$4$
- D$5$
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Suppose that a random variable $X$ follows a Poisson distribution. If $P(X=1) = P(X=2)$,then $P(X=5)$ is equal to:
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
| $X = x$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|
| $P(X = x)$ | $\frac{1}{10}$ | $\frac{2}{10}$ | $\frac{3}{10}$ | $\frac{4}{10}$ |
Then the variance of $X$ is
DifficultAP EAMCET 2011
View SolutionIf a discrete random variable $X$ has the probability distribution $P(X=x) = k \frac{2^{2x+1}}{(2x+1)!}$ for $x = 0, 1, 2, \ldots, \infty$,then $k =$
The cumulative distribution function (c.d.f.) $F(x)$ of a discrete random variable $X$ is given by the following table:
$X$ $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$ $F(X)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Then,find $P[X=3]$.
| $X$ | $-3$ | $-1$ | $0$ | $1$ | $3$ | $5$ | $7$ | $9$ |
|---|---|---|---|---|---|---|---|---|
| $F(X)$ | $0.1$ | $0.3$ | $0.5$ | $0.65$ | $0.75$ | $0.85$ | $0.90$ | $1$ |
Then,find $P[X=3]$.
The probability distribution of a random variable $X$ is given below. Then,the standard deviation of $X$ is
$X=x_i$ $2$ $3$ $5$ $7$ $12$ $P(X=x_i)$ $3k$ $k$ $k$ $2k$ $k$
| $X=x_i$ | $2$ | $3$ | $5$ | $7$ | $12$ |
|---|---|---|---|---|---|
| $P(X=x_i)$ | $3k$ | $k$ | $k$ | $2k$ | $k$ |
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