Mean and variance of eight observations are 9 and 9.25, respectively. If six of observations are 6,7

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13.Statistics

hard

The mean and variance of eight observations are $9$ and $9.25,$ respectively. If six of the observations are $6,7,10,12,12$ and $13,$ find the remaining two observations.

$4,8$

Solution

Let the remaining two observations be $x$ and $y$.

Therefore, the observations are $6,7,10,12,12,13, x, y$

Mean, $\bar{x}=\frac{6+7+10+12+12+13+x+y}{8}=9$

$\Rightarrow 60+x+y=72$

$\Rightarrow x+y=12$ ………..$(1)$

Variance $ = 9.25 = \frac{1}{n}\sum\limits_{i = 1}^8 {{{\left( {{x_i} – \bar x} \right)}^2}} $

$9.25=\frac{1}{8}[(-3)^{2}+(-2)^{2}+(1)^{2}+(3)^{2}+(4)^{2}$

$+x^{2}+y^{2}-2 \times 9(x+y)+2 \times(9)^{2}]$

$9.25=\frac{1}{8}\left[9+4+1+9+9+16+x^{2}+y^{2}-18(12)+162\right]$ ……..[ using $(1)$ ]

$9.25=\frac{1}{8}\left[48+x^{2}+y^{2}-216+162\right]$

$9.25=\frac{1}{8}\left[x^{2}+y^{2}-6\right]$

$\Rightarrow x^{2}+y^{2}=80$ ………$(2)$

From $(1),$ we obtain

$x^{2}+y^{2}+2 x y=144$ ……..$(3)$

From $(2)$ and $(3),$ we obtain

$2 x y=64$ ……….$(4)$

Subtracting $(4)$ from $(2),$ we obtain

$x^{2}+y^{2}-2 x y=80-64=16$

$\Rightarrow x-y=\pm 4 $ ………..$(5)$

Therefore, from $(1)$ and $(5),$ we obtain

$x=8$ and $y=4,$ when $x-y=4$

$x=4$ and $y=8,$ when $x-y=-4$

Thus, the remaining observations are $4$ and $8$

Standard 11

Mathematics

A data consists of $n$ observations

${x_1},{x_2},……,{x_n}.$ If $\sum\limits_{i – 1}^n {{{({x_i} + 1)}^2}} = 9n$ and $\sum\limits_{i – 1}^n {{{({x_i} – 1)}^2}} = 5n,$ then the standard deviation of this data is

hard

(JEE MAIN-2019)

The variance of $20$ observation is $5$ . If each observation is multiplied by $2$ , then the new variance of the resulting observations, is

medium

The mean and standard deviation of $15$ observations were found to be $12$ and $3$ respectively. On rechecking it was found that an observation was read as $10$ in place of $12$ . If $\mu$ and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15\left(\mu+\mu^2+\sigma^2\right)$ is equal to$……………….$

hard

(JEE MAIN-2024)

Let the observations $\mathrm{x}_{\mathrm{i}}(1 \leq \mathrm{i} \leq 10)$ satisfy the equations, $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)=10$ and $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)^{2}=40$ If $\mu$ and $\lambda$ are the mean and the variance of the observations, $\mathrm{x}_{1}-3, \mathrm{x}_{2}-3, \ldots ., \mathrm{x}_{10}-3,$ then the ordered pair $(\mu, \lambda)$ is equal to :

hard

(JEE MAIN-2020)

Calculate mean, variance and standard deviation for the following distribution.

Classes $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$

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hard

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${f_i}$	$3$	$7$	$12$	$15$	$8$	$3$	$2$

The mean and variance of eight observations are $9$ and $9.25,$ respectively. If six of the observations are $6,7,10,12,12$ and $13,$ find the remaining two observations.

Solution

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