Answer :

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

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

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

$\Rightarrow 60+x+y=72$

$\Rightarrow x+y=12\qquad\ldots(1)$

Variance $=9.25=\dfrac{1}{n} \sum _{i=1}^{8}(x_i-\bar{{}x})^{2}$

$9.25=\dfrac{1}{8}\big[(-3)^{2}+(-2)^{2}+(1)^{2}+(3)^{2}+(3)^{2}+(4)^{2}+x^{2}+y^{2}-2 \times 9(x+y)+2 \times(9)^{2}\big]$

$9.25=\dfrac{1}{8}\big[9+4+1+9+9+16+x^{2}+y^{2}-18(12)+162\big]\qquad$ $\big[$ Using $(1)\big]$

$9.25=\dfrac{1}{8}\big[48+x^{2}+y^{2}-216+162\big]$

$9.25=\dfrac{1}{8}\big[x^{2}+y^{2}-6\big]$

$\Rightarrow x^{2}+y^{2}=80\qquad\ldots(2)$

From $(1),$ we obtain

$x^{2}+y^{2}+2 x y=144 \qquad\ldots(3)$

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

$2 x y=64\qquad\ldots(4)$

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

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

$\Rightarrow (x-y)^2=16$

$\Rightarrow x- y=\pm 4\qquad\ldots(5)$

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

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

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

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

Answer :

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

The observations are $2, 4, 10, 12, 14,$ $x, y$.

Mean, $\bar{{}x}=\dfrac{2+4+10+12+14+x+y}{7}=8$

$\Rightarrow 56=42+x+y$

$\Rightarrow x+y=14\qquad\ldots(1)$

Variance $=16=\dfrac{1}{n} \sum _{i=1}^{7}(x_i-\bar{{}x})^{2}$

$16=\dfrac{1}{7}\big[(-6)^{2}+(-4)^{2}+(2)^{2}+(4)^{2}+(6)^{2}+x^{2}+y^{2}-2 \times 8(x+y)+2 \times(8)^{2}\big]$

$16=\dfrac{1}{7}\big[36+16+4+16+36+x^{2}+y^{2}-16(14)+2(64)\big]\qquad$ $…\big[$ Using $(1)\big]$

$16=\dfrac{1}{7}\big[108+x^{2}+y^{2}-224+128\big]$

$16=\dfrac{1}{7}\big[12+x^{2}+y^{2}\big]$

$\Rightarrow x^{2}+y^{2}=112-12=100$

$\Rightarrow x^{2}+y^{2}=100\qquad\ldots(2)$

From $(1),$ we obtain

$x^{2}+y^{2}+2 x y=196 \qquad\ldots(3)$

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

$2 x y=196$ $- 100$

$\Rightarrow 2 x y=96 \qquad\ldots (4)$

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

$x^{2}+y^{2} - 2 x y=100 - 96$

$\Rightarrow(x - y)^{2}=4$

$\Rightarrow x - y=\pm 2 \qquad\ldots (5)$

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

$x=8$ and $y=6$ when $x- y=2$

$x=6$ and $y=8$ when $x- y=- {2}$

Thus, the remaining observations are $6$ and $8 .$

Answer :

Let the observations be $x_1, x_2, x_3, x_4, x_5$, and $x_6$

It is given that mean is $8$ and standard deviation is $4 .$

Mean, $\bar{{}x}=\dfrac{x_1+x_2+x_3+x_4+x_5+x_6}{6}=8$

If each observation is multiplied by $3$ and the resulting observations are $y_i$, then

$y_i=3 x_i {\text{, i.e. }} x_i=\dfrac{1}{3} y_i$, for $i=1$ to $6$

$\therefore \ \ $ New mean, $\bar{{}y}=\dfrac{y_1+y_2+y_3+y_4+y_5+y_6}{6}$

$ \begin{aligned} & \hspace{2.1cm}=\dfrac{3(x_1+x_2+x_3+x_4+x_5+x_6)}{6} \\ &\hspace{2.1cm} =3 \times 8 \\ &\hspace{1.9cm}\bar{y} =24\qquad \ldots(1) \end{aligned} $

Standard deviation, $\sigma=\sqrt{\dfrac{1}{n} \sum _{i=1}^{6}(x_i-\bar{{}x})^{2}}$

$\therefore \ \ (4)^{2}=\dfrac{1}{6} \sum _{j=1}^{6}(x_i-\bar{{}x})^{2} $

$\sum _{i=1}^{6}(x_i-\bar{{}x})^{2}=96 \qquad \ldots{(2)}$

From $(1)$ and $(2),$ it can be observed that,

$ \begin{aligned} & \bar{{}y}=3 \bar{{}x} \\ \\ & \bar{{}x}=\dfrac{1}{3} \bar{{}y} \end{aligned} $

Substituting the values of $x_i$ and $\bar{x}^{-}$ in $(2),$ we obtain

$ \begin{aligned} & \sum _{i=1}^{6}\left(\dfrac{1}{3} y_i-\dfrac{1}{3} \bar{{}y}\right)^{2}=96 \\ \\ & \Rightarrow \sum _{i=1}^{6}(y_i-\bar{{}y})^{2}=864 \end{aligned} $

Therefore, variance of new observations $=\left(\dfrac{1}{6} \times 864\right)=144$

Hence, the standard deviation of new observations is $\sqrt{144}=12$

Answer :

The given $n$ observations are $x_1, x_2 - x_n$.

Mean $=\bar{{}x}$

Variance $= σ^{2}$

$\therefore \ \ \sigma^{2}=\dfrac{1}{n} \sum _{i=1}^{n} y_i(x_i-\bar{{}x})^{2}\qquad \ldots(1)$

If each observation is multiplied by $a$ and the new observations are $y_i$, then

$ \begin{aligned} & y_i=a x_i \text{ i.e., } x_i=\dfrac{1}{a} y_i \\ \\ & \therefore \ \ \bar{{}y}=\dfrac{1}{n} \sum _{i=1}^{n} y_i=\dfrac{1}{n} \sum _{i=1}^{n} a x_i=\dfrac{a}{n} \sum _{i=1}^{n} x_i=a \bar{{}x} \quad\left(\bar{{}x}=\dfrac{1}{n} \sum _{i=1}^{n} x_i\right) \end{aligned} $

Therefore, mean of the observations, $a x_1, a x_2 … a x_n$, is $a \bar{{}x}$.

Substituting the values of $x$ and $\bar{{}x}$ in $(1),$ we obtain

$ \begin{aligned} & \sigma^{2}=\dfrac{1}{n} \sum _{i=1}^{n}\left(\dfrac{1}{a} y_i-\dfrac{1}{a} \bar{{}y}\right)^{2} \\ & \Rightarrow a^{2} \sigma^{2}=\dfrac{1}{n} \sum _{i=1}^{n}\left(y_i-\bar{{}y}\right)^{2} \end{aligned} $

Thus, the variance of the observations, $a x_1, a x_2 … ax_n$, is $a^{2} σ^{2}$.

Answer :

(i): Number of observations $(n)=20$

Incorrect mean $=10$

Incorrect standard deviation $=2$

$\bar{{}x}=\dfrac{1}{n} \sum _{i=1}^{20} x_i$

$10=\dfrac{1}{20} \sum _{i=1}^{20} x_i$

$\Rightarrow \sum _{i=1}^{20} x_i=200$

That is, incorrect sum of observations $=200$

Correct sum of observations $=200 - 8 = 192$

$\therefore \ \ $ Correct mean $=\dfrac{\text{ Correct sum }}{19}=\dfrac{192}{19}=10.1$

Standard deviation $\sigma=\sqrt{\dfrac{1}{n} \sum _{i=1}^{n} x_i^{2}-\dfrac{1}{n^{2}}\big(\sum _{i=1}^{n} x_i\big)^{2}}$

$\Rightarrow \hspace{2.4cm}=\sqrt{\dfrac{1}{n} \sum _{i=1}^{n} x_i^{2}-(\bar{{}x})^{2}}$

$\Rightarrow\hspace{2cm}2=\sqrt{\dfrac{1}{20} \text{ Incorrect } \sum _{i=1}^{n} x_i^{2}-(10)^{2}}$

$\Rightarrow\hspace{2cm} 4=\dfrac{1}{20} \text{ Incorrect } \sum _{i=1}^{n} x_i^{2}-100$

$\Rightarrow \ \ $ Incorrect $\sum _{i=1}^{n} x_i^{2}=2080$

$\therefore \ \ $ Correct $\sum _{j=1}^{n} x_i^{2}=\text{ Incorrect } \sum _{i=1}^{n} x_i^{2}-(8)^{2}$

$\hspace{2.8cm}=2080-64=2016$

$\therefore \ \ $ Correct standard deviation $=\sqrt{\dfrac{\text{ Correct } \sum x_i^{2}}{n}-(\text{ Correct mean })^{2}}$

$\hspace{3.9cm}=\sqrt{\dfrac{2016}{19}-(10.1)^{2}} $

$\hspace{3.9cm}=\sqrt{106.1-102.01} =\sqrt{4.09} =2.02$

(ii): When $8$ is replaced by $12 ,$

Incorrect sum of observations $=200$

$\therefore \ \ $ Correct sum of observations $=200 - 8+12=204$

$\therefore \ \ $ Correct mean $=\dfrac{\text{ Correct sum }}{20}$

$\hspace{2.3cm}=\dfrac{204}{20}=10.2$

Standard deviation $\sigma=\sqrt{\dfrac{1}{n} \sum _{i=1}^{n} x_i{ }^{2}-\dfrac{1}{n^{2}}\big(\sum _{i=1}^{n} x_i\big)^{2}}$

$\Rightarrow \hspace{2.4cm} =\sqrt{\dfrac{1}{n} \sum _{i=1}^{n} x_i{ }^{2}-(\bar{{}x})^{2}}$

$\Rightarrow\hspace{2cm} 2=\sqrt{\dfrac{1}{20} \text{ Incorrect } \sum _{i=1}^{n} x_i^{2}-(10)^{2}}$

$\Rightarrow \hspace{1.9cm}4=\dfrac{1}{20}$ Incorrect $\sum _{i=1}^{n} x_i^{2}-100$

$\Rightarrow \ \ $ Incorrect $\sum _{i=1}^{n} x_i^{2}=2080$

$\therefore \ \ \ \ $ Correct $\sum _{i=1}^{n} x_i^{2} \ =$ Incorrect $\sum _{i=1}^{n} x_i^{2}-(8)^{2}+(12)^{2}$

$\hspace{3.1cm} =2080-64+144 =2160$

$\therefore \ \ $ Correct standard deviation $=\sqrt{\dfrac{\text{ Correct } \sum x_i{ }^{2}}{n}-(\text{ Correct mean })^{2}}$

$ \begin{aligned} & \hspace{3.8cm} =\sqrt{\dfrac{2160}{20}-(10.2)^{2}} \\ & \hspace{3.8cm}=\sqrt{108-104.04} \\ & \hspace{3.8cm}=\sqrt{3.96} \\ & \hspace{3.8cm}=1.98 \end{aligned} $

Answer :

Given, number of observation $\mathrm{n}=100$

Incorrect mean $(\bar{x})=20$

We know that

$ \begin{aligned} & \bar{x}=\dfrac{\sum x_i}{n} \\ \\ & \Rightarrow 20=\dfrac{1}{100} \sum_{i=1}^{100} x_i \\ \\ & \Rightarrow \sum_{i=1}^{100} x_i=20 \times 100=2000 \end{aligned} $

So, the incorrect sum of observations $=2000$.

Given incorrect observations are $21,21,18$ and these has to be omitted.

So, correct sum of observations $=2000-21-21-18=2000-60=1940$

Correct mean $=\dfrac{\text { Correct sum }}{100-3}=\dfrac{1940}{97}=20$

Given, incorrect standard deviation $\sigma=3$

$ \begin{aligned} & \sigma=\sqrt{\dfrac{\sum_{i=1}^n x_i^2}{n}-\left(\dfrac{\sum_{i=1}^n x_i}{n}\right)^2} \\ \\ & \sigma^2=\dfrac{\sum_{i=1}^n x_i^2}{n}-(\bar{x})^2 \\ \\ & \Rightarrow 3^2=\dfrac{\sum x_i^2}{100}-(20)^2 \\ \\ & \Rightarrow \text { Incorrect } \sum x_i^2=100(9+400)=40900 \end{aligned} $

$ \begin{aligned} \text { Correct } \sum_{i=1}^n x_i^2 & =\text { Incorrect } \sum_{i=1}^n x_i^2-(21)^2-(21)^2-(18)^2 \\ \\ & =40900-441-441-324 \\ \\ & =39694 \end{aligned} $

$ \begin{aligned} \therefore \text { Correct standard deviation } & =\sqrt{\dfrac{\text { correct } \sum x_i^2}{n}-(\text { Correct mean })^2} \\ \\ & =\sqrt{\dfrac{39694}{97}-(20)^2} \\ \\ & =\sqrt{409.216-400} \\ \\ & =\sqrt{9.216} \\ \\ & =3.036 \end{aligned} $