The following is the p.d.f. of a continuous random variable $X$: $f(x) = \begin{cases} \frac{x}{8} & , \text{if } 0 < x < 4 \\ 0 & , \text{otherwise} \end{cases}$
Then $F(0.5)$,$F(1.7)$,and $F(5)$ are respectively:
Then $F(0.5)$,$F(1.7)$,and $F(5)$ are respectively:
- A$\frac{1}{64}, 1, 0.18$
- B$0.0156, 0.18, 1$
- C$0.18, 0.0156, 1$
- D$1, 0.0156, 0.18$
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Similar Questions
If $X$ is a Poisson variable such that $3 P(X=4)=\frac{1}{2} P(X=2)+P(X=0)$,then the mean of $X$ is
If $X$ is a random variable with cumulative distribution function $F(x)$ and its probability distribution is given by the following table:
$X = x$ $-1.5$ $-0.5$ $0.5$ $1.5$ $2.5$ $P(X = x)$ $0.05$ $0.2$ $0.15$ $0.25$ $0.35$
Then,find the value of $F(1.5) - F(-0.5)$.
| $X = x$ | $-1.5$ | $-0.5$ | $0.5$ | $1.5$ | $2.5$ |
|---|---|---|---|---|---|
| $P(X = x)$ | $0.05$ | $0.2$ | $0.15$ | $0.25$ | $0.35$ |
Then,find the value of $F(1.5) - F(-0.5)$.
For the random variable $X$ with the probability distribution given by the table:
$X = x$ $0$ $1$ $2$ $3$ $P(X = x)$ $K$ $K + \frac{1}{7}$ $2K$ $\frac{2}{5}$
The mean of $X$ is:
| $X = x$ | $0$ | $1$ | $2$ | $3$ |
| $P(X = x)$ | $K$ | $K + \frac{1}{7}$ | $2K$ | $\frac{2}{5}$ |
The mean of $X$ is:
Suppose three coins are tossed simultaneously. If $X$ denotes the number of heads,then the probability distribution of $X$ is:
The variance of a Poisson variate $X$ is $2$. Then $P(X \geq 3) = $
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