Answer :

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

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

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

$\Rightarrow 60+x+y=72$

$\Rightarrow x+y=12$

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

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

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

[Using (1)]

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

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

$\Rightarrow x^{2}+y^{2}=80$

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

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

$\Rightarrow x- y=-(5)$

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

$\Rightarrow 56=42+x+y$

$\Rightarrow x+y=14$

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

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

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

…[Using (1)]

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

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

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

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

From (1), we obtain

$x^{2}+y^{2}+2 x y=196 -(3)$

From (2) and (3), we obtain

$2 x y=196$ - 100

$\Rightarrow 2 x y=96 - (4)$

Subtracting (4) from (2), we obtain

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

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

$\Rightarrow x - y=- (5)$

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

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

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

$$ \begin{aligned} & =\dfrac{3(x_1+x_2+x_3+x_4+x_5+x_6)}{6} \\ & =3 \times 8 \\ & =24 \end{aligned} $$

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

$$ \begin{align*} \therefore & (4)^{2}=\dfrac{1}{6} \sum _{j=1}^{6}(x_j-\bar{{}x})^{2} \\ & \sum _{i=1}^{6}(x_i-\bar{{}x})^{2}=96 \tag{2} \end{align*} $$

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

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

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

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 $= σ^{2}$

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

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_j=\dfrac{1}{n} \sum _{i=1}^{n} a x_i=\dfrac{a}{n} \sum _{i=1}^{n} x_i=a \bar{{}x} \quad(\bar{{}x}=\dfrac{1}{n} \sum _{i=1}^{n} x_i) \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}(\dfrac{1}{a} y_i-\dfrac{1}{a} \bar{{}y})^{2} \\ & \Rightarrow a^{2} \sigma^{2}=\dfrac{1}{n} \sum _{i=1}^{n}(y_i-\bar{{}y})^{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 $\sigma=\sqrt{\dfrac{1}{n} \sum _{i=1}^{n} x_i^{2}-\dfrac{1}{n^{2}}(\sum _{i=1}^{n} x_i)^{2}}=\sqrt{\dfrac{1}{n} \sum _{i=1}^{n} x_i^{2}-(\bar{{}x})^{2}}$

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

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

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

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

$ =2080-64 $

$ =2016 $

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

$ \begin{aligned} & =\sqrt{\dfrac{2016}{19}-(10.1)^{2}} \\ & =\sqrt{106.1-102.01} \\ & =\sqrt{4.09} \\ & =2.02 \end{aligned} $

(ii) When 8 is replaced by 12 ,

Incorrect sum of observations $=200$

$\therefore$ Correct sum of observations $=200$ - $8+12=204$ $\therefore$ Correct mean $=\dfrac{\text{ Correct sum }}{20}=\dfrac{204}{20}=10.2$

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

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

$\Rightarrow 4=\dfrac{1}{20}$ 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 $\sum _{i=1}^{n} x_i^{2}-(8)^{2}+(12)^{2}$

$ \begin{aligned} & =2080-64+144 \\ & =2160 \end{aligned} $

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

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

Answer :

Number of observations $(n)=100$

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

Incorrect standard deviation $(\sigma)=3$

$\Rightarrow 20=\dfrac{1}{100} \sum _{i=1}^{100} x_i$

$\Rightarrow \sum _{i=1}^{100} x_i=20 \times 100=2000$

$\therefore$ Incorrect sum of observations $=2000$

$\Rightarrow$ Correct sum of observations $=2000$ - 21 - 21 - $18=2000$ - $60=1940$ $\therefore$ Correct mean $=\dfrac{\text{ Correct sum }}{100-3}=\dfrac{1940}{97}=20$

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

$\Rightarrow 3=\sqrt{\dfrac{1}{100} \times \text{ Incorrect } \sum x_i^{2}-(20)^{2}}$

$\Rightarrow$ Incorrect $\sum x_i^{2}=100(9+400)=40900$

Correct $\sum _{i=1}^{n} x_i^{2}=Incorrect \sum _{i=1}^{n} x_i^{2}-(21)^{2}-(21)^{2}-(18)^{2}$

$ =40900-441-441-324 $

$ =39694 $

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

$ \begin{aligned} & =\sqrt{\dfrac{39694}{97}-(20)^{2}} \\ & =\sqrt{409.216-400} \\ & =\sqrt{9.216} \\ & =3.036 \end{aligned} $