The random variable $X$ has a probability distribution $P(X)$ of the following form,where $k$ is some number:
$P(X) = \begin{cases} k, & \text{if } x=0 \\ 2k, & \text{if } x=1 \\ 3k, & \text{if } x=2 \\ 0, & \text{otherwise} \end{cases}$
Find $P(X < 2)$,$P(X \leq 2)$,and $P(X \geq 2)$.
$P(X) = \begin{cases} k, & \text{if } x=0 \\ 2k, & \text{if } x=1 \\ 3k, & \text{if } x=2 \\ 0, & \text{otherwise} \end{cases}$
Find $P(X < 2)$,$P(X \leq 2)$,and $P(X \geq 2)$.
For any probability distribution,the sum of all probabilities must be $1$.
$\sum P(X) = k + 2k + 3k = 6k = 1 \implies k = \frac{1}{6}$.
$P(X<2) = P(X=0) + P(X=1) = k + 2k = 3k = 3 \times \frac{1}{6} = \frac{1}{2}$.
$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = k + 2k + 3k = 6k = 6 \times \frac{1}{6} = 1$.
$P(X \geq 2) = P(X=2) + P(X>2) = 3k + 0 = 3k = 3 \times \frac{1}{6} = \frac{1}{2}$.
$\sum P(X) = k + 2k + 3k = 6k = 1 \implies k = \frac{1}{6}$.
$P(X<2) = P(X=0) + P(X=1) = k + 2k = 3k = 3 \times \frac{1}{6} = \frac{1}{2}$.
$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = k + 2k + 3k = 6k = 6 \times \frac{1}{6} = 1$.
$P(X \geq 2) = P(X=2) + P(X>2) = 3k + 0 = 3k = 3 \times \frac{1}{6} = \frac{1}{2}$.
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The random variable $X$ has the following probability distribution:
| $X$ | $8$ | $12$ | $16$ | $20$ | $24$ |
|---|---|---|---|---|---|
| $P(X)$ | $K$ | $\frac{1}{6}$ | $\frac{3}{8}$ | $2K$ | $\frac{1}{12}$ |
Then the value of $K$ is:
| $X$ | $8$ | $12$ | $16$ | $20$ | $24$ |
|---|---|---|---|---|---|
| $P(X)$ | $K$ | $\frac{1}{6}$ | $\frac{3}{8}$ | $2K$ | $\frac{1}{12}$ |
Then the value of $K$ is:
For the probability distribution of $X$ given below,the variance of $X$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $P(X = x)$ $0.2$ $0.3$ $0.1$ $0.15$ $0.25$
| $X = x$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|---|
| $P(X = x)$ | $0.2$ | $0.3$ | $0.1$ | $0.15$ | $0.25$ |
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