For the following cumulative distribution function $F(x)$ of a random variable $X$,find $P(3 < X \leq 5)$.
$x$ $1$ $2$ $3$ $4$ $5$ $6$ $F(x)$ $0.2$ $0.37$ $0.48$ $0.62$ $0.85$ $1$
| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
|---|---|---|---|---|---|---|
| $F(x)$ | $0.2$ | $0.37$ | $0.48$ | $0.62$ | $0.85$ | $1$ |
- A$0.37$
- B$0.48$
- C$0.62$
- D$0.85$
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State whether the following table represents a probability distribution of a random variable $X$. Give reasons for your answer.
$X$ $0$ $1$ $2$ $P(X)$ $0.4$ $0.4$ $0.2$
| $X$ | $0$ | $1$ | $2$ |
| $P(X)$ | $0.4$ | $0.4$ | $0.2$ |
Easy
View SolutionIf $P(X=2)=0.3, P(X=3)=0.4, P(X=4)=0.3$,then the variance of random variable $X$ is (in $.6$)
The value of the constant $c$,so that $P(x)=c\left(\frac{2}{3}\right)^{x}$,$x=1,2,3, \ldots$ is the probability distribution function of a discrete random variable $X$ is
$A$ random variable $X$ has the following probability distribution:
$X=x$ $1$ $2$ $3$ $4$ $P(X=x)$ $0.1$ $0.2$ $0.3$ $0.4$
The mean and standard deviation of $X$ are respectively:
| $X=x$ | $1$ | $2$ | $3$ | $4$ |
| $P(X=x)$ | $0.1$ | $0.2$ | $0.3$ | $0.4$ |
The mean and standard deviation of $X$ are respectively:
$A$ random variable $X$ takes the values $0, 1$ and $2$. If $P(X=1)=P(X=2)$ and $P(X=0)=0.4$,then the mean of the random variable $X$ is
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