If a cubical die is thrown,then the mean and variance of the random variable $X$,representing the number on the face that shows up,are respectively:
- A$\frac{2}{7}, \frac{12}{35}$
- B$\frac{7}{2}, \frac{12}{35}$
- C$\frac{1}{7}, \frac{1}{12}$
- D$\frac{7}{2}, \frac{35}{12}$
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Similar Questions
If the mean of the following probability distribution of a random variable $X$ is $\frac{46}{9}$,then the variance of the distribution is:
$X$ $0$ $2$ $4$ $6$ $8$ $P(X)$ $a$ $2a$ $a+b$ $2b$ $3b$
| $X$ | $0$ | $2$ | $4$ | $6$ | $8$ |
| $P(X)$ | $a$ | $2a$ | $a+b$ | $2b$ | $3b$ |
DifficultJEE MAIN 2024
View SolutionLet a sample space be $S = \{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?
Outcome Probability $\omega_{1}$ $\frac{1}{12}$ $\omega_{2}$ $\frac{1}{12}$ $\omega_{3}$ $\frac{1}{6}$ $\omega_{4}$ $\frac{1}{6}$ $\omega_{5}$ $\frac{1}{6}$ $\omega_{6}$ $\frac{3}{2}$
| Outcome | Probability |
| $\omega_{1}$ | $\frac{1}{12}$ |
| $\omega_{2}$ | $\frac{1}{12}$ |
| $\omega_{3}$ | $\frac{1}{6}$ |
| $\omega_{4}$ | $\frac{1}{6}$ |
| $\omega_{5}$ | $\frac{1}{6}$ |
| $\omega_{6}$ | $\frac{3}{2}$ |
Easy
View SolutionIf $m$ and $\sigma^2$ are the mean and variance of the random variable $X$,whose distribution is given by:
$X=x$ $0$ $1$ $2$ $3$ $P(X=x)$ $\frac{1}{3}$ $\frac{1}{2}$ $0$ $\frac{1}{6}$
Then:
| $X=x$ | $0$ | $1$ | $2$ | $3$ |
|---|---|---|---|---|
| $P(X=x)$ | $\frac{1}{3}$ | $\frac{1}{2}$ | $0$ | $\frac{1}{6}$ |
Then:
For the given probability distribution,find $E(X^2)$.$X$ $1$ $2$ $3$ $4$ $P(X)$ $\frac{1}{10}$ $\frac{1}{5}$ $\frac{3}{10}$ $\frac{2}{5}$
| $X$ | $1$ | $2$ | $3$ | $4$ |
| $P(X)$ | $\frac{1}{10}$ | $\frac{1}{5}$ | $\frac{3}{10}$ | $\frac{2}{5}$ |
If the probability density function (p.d.f.) of a continuous random variable $X$ is given by $f(x) = \begin{cases} k(9 + 8x - x^2), & \text{for } -1 \leq x \leq 4 \\ 0, & \text{otherwise} \end{cases}$, then the value of $k$ is:
DifficultMHT CET 2026
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