If the mean and the variance of $6,4, a, 8, b, 12,10, 13$ are $9$ and $9.25$ respectively, then $a+b+a b$ is equal to :
- [JEE MAIN 2025]
- A$105$
- B$103$
- C$100$
- D$106$
Similar Questions
If the variance of observations ${x_1},\,{x_2},\,......{x_n}$ is ${\sigma ^2}$, then the variance of $a{x_1},\,a{x_2}.......,\,a{x_n}$, $\alpha \ne 0$ is
If the variance of observations ${x_1},\,{x_2},\,......{x_n}$ is ${\sigma ^2}$, then the variance of $a{x_1},\,a{x_2}.......,\,a{x_n}$, $\alpha \ne 0$ is
Consider $10$ observation $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10}$. such that $\sum_{i=1}^{10}\left(x_i-\alpha\right)=2$ and $\sum_{i=1}^{10}\left(x_i-\beta\right)^2=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The $\frac{\beta}{\alpha}$ is equal to :
Consider $10$ observation $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10}$. such that $\sum_{i=1}^{10}\left(x_i-\alpha\right)=2$ and $\sum_{i=1}^{10}\left(x_i-\beta\right)^2=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The $\frac{\beta}{\alpha}$ is equal to :
- [JEE MAIN 2024]
If for a distribution $\Sigma(x-5)=3, \Sigma(x-5)^{2}=43$ and the total number of item is $18,$ find the mean and standard deviation.
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If the variance of the frequency distribution is $3$ then $\alpha$ is ......
$X_i$
$2$
$3$
$4$
$5$
$6$
$7$
$8$
Frequency $f_i$
$3$
$6$
$16$
$\alpha$
$9$
$5$
$6$
If the variance of the frequency distribution is $3$ then $\alpha$ is ......
| $X_i$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
| Frequency $f_i$ | $3$ | $6$ | $16$ | $\alpha$ | $9$ | $5$ | $6$ |
- [JEE MAIN 2023]
Find the standard deviation of the first n natural numbers.
Find the standard deviation of the first n natural numbers.