If the variance of 10 natural numbers 1, 1, 1 | vedclass

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(B) The variance $\sigma^{2}$ of $n$ observations is given by $\sigma^{2} = \frac{\Sigma x_{i}^{2}}{n} - (\bar{x})^{2}$.Here,the observations are $1$

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$11$

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(B) The variance $\sigma^{2}$ of $n$ observations is given by $\sigma^{2} = \frac{\Sigma x_{i}^{2}}{n} - (\bar{x})^{2}$.Here,the observations are $1$ ($9$ times) and $k$ ($1$ time),so $n = 10$.$\Sigma x_{i} = 9(1) + k = 9 + k$.$\Sigma x_{i}^{2} = 9(1)^{2} + k^{2} = 9 + k^{2}$.$\sigma^{2} = \frac{9 + k^{2}}{10} - \left(\frac{9 + k}{10}\right)^{2} Multiply by $100$: $10(9 + k^{2}) - (9 + k)^{2} $90 + 10k^{2} - (81 + 18k + k^{2}) $9k^{2} - 18k + 9 $9(k - 1)^{2} $(k - 1)^{2} $k - 1 $k Since $k$ is a natural number,the maximum possible value of $k$ is $11$.

$x_i$	$0$	$1$	$5$	$6$	$10$	$12$	$17$
$f_i$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

If the variance of $10$ natural numbers $1, 1, 1, \dots, 1, k$ is less than $10$,then the maximum possible value of $k$ is ...... .

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