If the probability distribution function of a random variable $X$ is given as follows:
$X=x_i$ $-2$ $-1$ $0$ $1$ $2$ $P(X=x_i)$ $0.2$ $0.3$ $0.15$ $0.25$ $0.1$
Then $F(0)$ is equal to:
| $X=x_i$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|---|
| $P(X=x_i)$ | $0.2$ | $0.3$ | $0.15$ | $0.25$ | $0.1$ |
Then $F(0)$ is equal to:
- A$P(X > 0)$
- B$1 - P(X > 0)$
- C$1 - P(X < 0)$
- D$P(X < 0)$
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Which of the following functions is not a probability density function $(p.d.f.)$ of a continuous random variable $X$?
$A$ random variable $X$ has the following probability distribution:
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|
| $P(X)$ | $k^2$ | $2k$ | $k$ | $2k$ | $5k^2$ |
Then $P(X \geq 2)$ is equal to:
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|
| $P(X)$ | $k^2$ | $2k$ | $k$ | $2k$ | $5k^2$ |
Then $P(X \geq 2)$ is equal to:
$A$ random variable $X$ has the following probability distribution:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $P(X = x)$ $0.15$ $0.23$ $0.10$ $0.12$ $0.20$ $0.08$ $0.07$ $0.05$
For the event $E = \{ X \text{ is a prime number} \}$,$F = \{ X < 4 \}$,then $P(E \cup F)$ is
| $X = x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(X = x)$ | $0.15$ | $0.23$ | $0.10$ | $0.12$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the event $E = \{ X \text{ is a prime number} \}$,$F = \{ X < 4 \}$,then $P(E \cup F)$ is
For the following probability distribution,find the expected value of $X$:
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $P(x)$ $k$ $0.3$ $0.15$ $0.15$ $0.1$ $2k$
| $x$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|---|
| $P(x)$ | $k$ | $0.3$ | $0.15$ | $0.15$ | $0.1$ | $2k$ |
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