If a random variable $X$ has a probability mass function $P(x) = \begin{cases} kx^2, & \text{for } x = 1, 2, 3, 4 \\ 0, & \text{otherwise} \end{cases}$, then the mean of $X$ is ...
- A$3.33$
- B$3.25$
- C$3.00$
- D$2.50$
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If $X$ is a random variable such that $P(X=-2)=P(X=-1)=P(X=2)=P(X=1)=\frac{1}{6}$ and $P(X=0)=\frac{1}{3}$,then the mean of $X$ is
The cumulative distribution function $F(X)$ of a discrete random variable $X$ is given by the following table:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $F(X=x)$ $0.2$ $0.37$ $0.48$ $0.62$ $0.85$ $1$
Then $P[X=4] + P[X=5] = $
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
|---|---|---|---|---|---|---|
| $F(X=x)$ | $0.2$ | $0.37$ | $0.48$ | $0.62$ | $0.85$ | $1$ |
Then $P[X=4] + P[X=5] = $
The p.m.f of a random variable $X$ is given by $P(X) = \frac{2x}{n(n+1)}$ for $x = 1, 2, 3, \ldots, n$,and $0$ otherwise. Then $E(X) = $
If the probability distribution of a random variable $X$ is given by the following table,then $F(0) =$ . . . . . .
$x_i$ $-2$ $-1$ $0$ $1$ $2$ $P(X = x_i)$ $0.2$ $0.5$ $0.15$ $0.25$ $0.1$
| $x_i$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
| $P(X = x_i)$ | $0.2$ | $0.5$ | $0.15$ | $0.25$ | $0.1$ |
DifficultMHT CET 2016
View Solution$A$ player tosses $2$ fair coins. He wins $Rs. 5$ if $2$ heads appear,$Rs. 2$ if $1$ head appears,and $Rs. 1$ if no head appears. Then,the variance of his winning amount is
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