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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $1$

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(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 $11$.

$x_i$	$92$	$93$	$97$	$98$	$102$	$104$	$109$
$f_i$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

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 ...... .

Similar Questions

Find the mean and variance for the following data:

$x_i$ $92$ $93$ $97$ $98$ $102$ $104$ $109$

$f_i$ $3$ $2$ $3$ $2$ $6$ $3$ $3$

The variance of the first $50$ even natural numbers is

The mean and standard deviation of $100$ items are $50$ and $4$, respectively. Then the sum of the squares of all the items is:

The variance of the observations $8, 12, 13, 15, 22$ is:

The $S.D.$ of $15$ items is $6$. If each item is decreased or increased by $1$,then the standard deviation will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module