The p.d.f. of a random variable $x$ is given by $f(x) = \frac{1}{4a}$ for $0 < x < 4a$ $(a > 0)$ and $f(x) = 0$ otherwise. If $P(x < \frac{3a}{2}) = k P(x > \frac{5a}{2})$,then $k = . . . . . .$
- A$1$
- B$\frac{1}{2}$
- C$\frac{1}{8}$
- D$\frac{3}{2}$
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If the function $f$ defined by $f(x) = \begin{cases} K(x-x^2) & \text{if } 0 < x < 1 \\ 0 & \text{otherwise} \end{cases}$ is the probability density function (p.d.f.) of a random variable $X$,then the value of $P(X < \frac{1}{2})$ is
Let a pair of dice be thrown and the random variable $X$ be the sum of the numbers that appear on the two dice. Find the mean or expectation of $X$.
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View SolutionIf a random variable $X$ has the following probability distribution values,then $P(X \geq 6)$ has the value:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$
| $X$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ |
| $P(X)$ | $0$ | $k$ | $2k$ | $2k$ | $3k$ | $k^2$ | $2k^2$ | $7k^2 + k$ |
If $3$ is the variance of a Poisson distribution,then $P(1 < x < 4) = $
If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$,whose distribution is given by:
$X=x$ $0$ $1$ $2$ $3$ $P(X=x)$ $\frac{1}{3}$ $\frac{1}{2}$ $0$ $\frac{1}{6}$
Then:
| $X=x$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|
| $P(X=x)$ | $\frac{1}{3}$ | $\frac{1}{2}$ | $0$ | $\frac{1}{6}$ |
Then:
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