If the probability mass function (p.m.f.) of a random variable $X$ is given by the table below,then find the value of $k$.
$X$ $1$ $2$ $3$ $4$ $5$ $P(X = x)$ $k$ $\frac{k}{3}$ $\frac{k}{4}$ $\frac{k}{2}$ $\frac{k}{2}$
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $P(X = x)$ | $k$ | $\frac{k}{3}$ | $\frac{k}{4}$ | $\frac{k}{2}$ | $\frac{k}{2}$ |
- A$\frac{15}{31}$
- B$\frac{1}{12}$
- C$\frac{11}{12}$
- D$\frac{12}{31}$
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The cumulative distribution function of a discrete random variable $X$ is given by the following table:
$X = x$ $-4$ $-2$ $0$ $2$ $4$ $6$ $8$ $10$ $F(X = x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$
Then,calculate $\frac{P(X \leqslant 0)}{P(X > 0)}$.
| $X = x$ | $-4$ | $-2$ | $0$ | $2$ | $4$ | $6$ | $8$ | $10$ |
|---|---|---|---|---|---|---|---|---|
| $F(X = x)$ | $0.1$ | $0.3$ | $0.5$ | $0.65$ | $0.75$ | $0.85$ | $0.90$ | $1$ |
Then,calculate $\frac{P(X \leqslant 0)}{P(X > 0)}$.
Suppose that two cards are drawn at random from a deck of cards. Let $X$ be the number of aces obtained. Then the value of $E(X)$ is
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View SolutionIf 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:
If a random variable $X$ has the p.d.f. $f(x) = \begin{cases} \frac{k}{x^2+1} & , \text{if } 0 < x < \infty \\ 0 & , \text{otherwise} \end{cases}$,then the c.d.f. of $X$ is:
The probability distribution of a discrete random variable $X$ is given below. If $E(X^2) = \Sigma x^2 P(X=x)$, then $6 E(X^2) - \operatorname{Var}(X) =$
$X=x$ $-1$ $0$ $1$ $2$ $P(X=x)$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$
| $X=x$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|
| $P(X=x)$ | $\frac{1}{3}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{3}$ |
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