If the variance of $10$ natural numbers $1,1,1, \ldots ., 1, k$ is less than $10 ,$ then the maximum possible value of $k$ is ...... .
$12$
$11$
$14$
$21$
Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance
Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .
Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.
If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, …., (x_{50} - 4)^2$ is
The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$
Height | Weight | |
Mean | $162.6\,cm$ | $52.36\,kg$ |
Variance | $127.69\,c{m^2}$ | $23.1361\,k{g^2}$ |
Can we say that the weights show greater variation than the heights?
If the mean and variance of the data $65,68,58,44$, $48,45,60, \alpha, \beta, 60$ where $\alpha>\beta$ are $56$ and $66.2$ respectively, then $\alpha^2+\beta^2$ is equal to
The mean and variance of $7$ observations are $8$ and $16,$ respectively. If five observations are $2, 4, 10,12,14,$ then the absolute difference of the remaining two observations is