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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State which of the following is not the probability distribution of a random variable. Give reasons for your answer.
$Z$ $3$ $2$ $1$ $0$ $-1$ $P(Z)$ $0.3$ $0.2$ $0.4$ $0.1$ $0.05$
| $Z$ | $3$ | $2$ | $1$ | $0$ | $-1$ |
| $P(Z)$ | $0.3$ | $0.2$ | $0.4$ | $0.1$ | $0.05$ |
Easy
View SolutionFor the following probability distribution,find the $Var(X)$.
$X$ $-2$ $-1$ $0$ $1$ $2$ $3$ $P(X)$ $0.1$ $0.2$ $0.2$ $0.3$ $0.15$ $0.05$
(Given : $(0.25)^2 = 0.0625$,$(0.35)^2 = 0.1225$,$(0.45)^2 = 0.2025$)
| $X$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ |
| $P(X)$ | $0.1$ | $0.2$ | $0.2$ | $0.3$ | $0.15$ | $0.05$ |
(Given : $(0.25)^2 = 0.0625$,$(0.35)^2 = 0.1225$,$(0.45)^2 = 0.2025$)
If the probability that an individual will suffer a bad reaction from an injection is $0.001$, then the probability that out of $2000$ individuals, exactly $3$ individuals suffer a bad reaction is
The variance of a Poisson variate $X$ is $2$. Then $P(X \geq 3) = $
In a hospital, on an average if there are $35$ births in a week, then the probability that there will be less than $3$ births in a day is:
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