Statistics

Here number of terms are even therefore

median $=\frac{500k+501k}{2}=\frac{1001\text{k}}{2}={\text{X}}_{\text{M}}$

mean deviation about median $=\frac{\sum \left|{\text{X}}_{\text{i}}-{\text{X}}_{\text{M}}\right|}{\text{n}}$

$=\frac{2\left(\frac{\text{k}}{2}+\frac{3\text{k}}{2}+\frac{5\text{k}}{2}+\dots 500\text{ terms }\right)}{1000}$

$$\begin{gathered} =\frac{2\times {\displaystyle \frac{k}{2}}\left(1+3+5+.....500 \mathrm{terms}\right)}{1000} \\ =\frac{2\times {\displaystyle \frac{k}{2}}\left({\displaystyle \frac{500}{2}}\left\{2+\left(500-1\right)\times 2\right\}\right)}{1000} \end{gathered}$$

\(=\frac{2 \cdot \frac{k}{2}(500)^2}{1000}=\frac{500 k}{2}=500(\) given \()\)

\(\therefore k=2\)

Hence, \(k^2=4\)

Given: Mean

$$\begin{gathered} \mu =\frac{-18+x+y+2+9+16}{8}=\frac{7}{2} \\ \Rightarrow \frac{x+y+9}{8}=\frac{7}{2} \\ \Rightarrow x+y+9=28 \\ \Rightarrow x+y=19 ....\left(i\right) \end{gathered}$$

And Variance
$$\begin{gathered} {\sigma }^2=\frac{\sum x_i^2}{8}-{\left(\mu \right)}^2=\frac{293}{4} \\ \Rightarrow \frac{{10}^2+7^2+1^2+x^2+y^2+2^2+9^2+{16}^2}{8}-{\left(\frac{7}{2}\right)}^2=\frac{293}{4} \end{gathered}$$
$\Rightarrow \frac{293}{4}+\frac{49}{4}=\frac{x^2+y^2+491}{8}$

$\Rightarrow \frac{171}{2}=\frac{x^2+y^2+491}{8}$

$\Rightarrow x^2+y^2=193 ...\left(ii\right)$

After solving equation \((i)\) and \((ii)\), We get:

\(x=7\) and \(y=12\) or \(x=12\) and \(y=7\)

Hence numbers are \(7,12,20,5\)

Mean$=\frac{20+12+7+5}{4}=\frac{44}{4}=11$

Let, in first case, \(10\) numbers are \(x_1, x_2, \ldots \ldots x_9, \alpha\)

Then, $\frac{{\displaystyle \sum _{i=1}^9}{x}_i+\alpha }{10}=10$

$\Rightarrow \alpha +\sum _{i=1}^9{x}_i=100\Rightarrow \sum _{i=1}^9{x}_i=100-\alpha$

And ${\sigma }^2=\left(\frac{{\displaystyle \sum x_i^2+{\alpha }^2}}{n}\right)-{\left(\frac{{\displaystyle \sum x_i+\alpha }}{n}\right)}^2$

$\Rightarrow 2=\left(\frac{{\displaystyle \sum x_i^2+{\alpha }^2}}{10}\right)-{\left(10\right)}^2$

$\Rightarrow \frac{{\displaystyle \sum x_i^2+{\alpha }^2}}{n}=102$

$\Rightarrow x_1^2+x_2^2+\dots \dots x_9^2+{\alpha }^2=1020$

$\Rightarrow \sum x_i^2=1020-{\alpha }^2$

In second case, let number are

$x_1,x_2,\dots \dots x_9,\beta$

Now \(\frac{\sum_{i=1}^9 x_i+\beta}{10}=\frac{100-\alpha+\beta}{10}\)

\(\Rightarrow 10.1=\frac{100-\alpha+\beta}{10}\)

\(\Rightarrow \alpha-\beta=-1\)

Now, ${\sigma }^2=\left(\frac{{\displaystyle \sum x_i^2+{\beta }^2}}{n}\right)-{\left(\frac{{\displaystyle \sum x_i+\beta }}{n}\right)}^2$

$\Rightarrow \frac{{\displaystyle \sum x_i^2+{\beta }^2}}{10}-{\left(10.1\right)}^2=1.99$

$$\begin{gathered} \Rightarrow 1.99=\frac{1020-{\alpha }^2+{\beta }^2}{10}-102.01 \\ \Rightarrow 1020-{\alpha }^2+{\beta }^2=1040 \end{gathered}$$

$\Rightarrow {\beta }^2-{\alpha }^2=20$

\(\Rightarrow \alpha^2-\beta^2=-20\)

\(\Rightarrow (\alpha-\beta)(\alpha+\beta)=-20 \)

\(\Rightarrow (-1)(\alpha+\beta)=-20\)

Hence, \(\alpha+\beta=20\)

\(\sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}+2\right)^2=180 \)
\( \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^2+4 \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}+\sum_{\mathrm{i}=1}^{10} 4=180 \)
\( \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^2+4 \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=180-40=140 \ldots\ (1)\)
\( \text { Also } \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-1\right)^2=90 \)
\( \sum_{i=1}^{10} x_i^2-2 \sum_{i=1}^{10} x_i+\sum_{i=1}^{10} 1=90 \)
\( \sum_{i=1}^{10} x_i^2-2 \sum_{i=1}^{10} x_i=90-10=80 \ldots\ (2)\)
From (1) and (2) we get:
\( \sum_{i=1}^{10} x_i^2=100 \text { and } \sum_{i=1}^{10} x_i=10 \)
\( \sigma^2=\frac{\sum_{i=1}^{10} x_i^2}{N}-\left(\frac{\sum_{i=1}^{10} x_i}{N}\right)^2\)
\( =\frac{100}{10}-\left(\frac{10}{10}\right)^2\)
\( \sigma^2=10-1=9 \)
\( \Rightarrow \sigma=3\)

Mean \(\bar{x}=\frac{\sum x_i}{n}\)

\(8=\frac{2+4+\alpha+8+\beta+12+14}{7}\)

\(56=40+\alpha+\beta\)

\(\alpha+\beta=16\)

and \(\sigma^2=\frac{\sum x_i^2}{n}-(\bar{x})^2\)

\(16=\frac{4+16+\alpha^2+64+\beta^2+144+196}{7}-64\)

\(560=424+\alpha^2+\beta^2\)

\(\alpha^2+\beta^2=136\)

Since \((\alpha+\beta)^2=\alpha^2+\beta^2+2 \alpha \beta\)

\(256=136+2 \alpha \beta\)

\(\alpha \beta=60\)

\(\alpha\) and \(\beta\) are the roots of the quadratic equation \(t^2-(\alpha+\beta) t+\alpha \beta=0\)

\(t^2-16 t+60=0\)

\((t-6)(t-10)=0\)

\(\alpha=6\) and \(\beta=10\)

Now, \(3 \alpha+2=20\) and \(2 \beta+1=21\)

Sum of roots \(=20+21=41\)

Product of roots \(=420\)

Required equation is

\(x^2-41 x+420=0\)

$\begin{array}{|ccccccc|}\hline x & c & 2c & 3c & 4c & 5c & 6c \\ f & 2& 1& 1& 1& 1& 1\\ \hline\end{array}$

Mean $\overline{x}=\frac{(2+2+3+4+5+6)C}{7}=\frac{22C}{7}$

variance ${\sigma }^2=\frac{c^2(2+2^2+3^2+4^2+5^2+6^2)}{7}-{\left(\frac{22c}{7}\right)}^2$

$=\frac{92c^2}{7}-c^2\times \frac{484}{49}$

$=\frac{(644-484)c^2}{49}=\frac{160c^2}{49}$

$160=\frac{160\times c^2}{49}\Rightarrow c=7$

$\begin{array}{|ccccccc|}\hline x & c & 2c & 3c & 4c & 5c & 6c \\ f & 2& 1& 1& 1& 1& 1\\ \hline\end{array}$

Mean $\overline{x}=\frac{(2+2+3+4+5+6)C}{7}=\frac{22C}{7}$

variance ${\sigma }^2=\frac{c^2(2+2^2+3^2+4^2+5^2+6^2)}{7}-{\left(\frac{22c}{7}\right)}^2$

$=\frac{92c^2}{7}-c^2\times \frac{484}{49}$

$=\frac{(644-484)c^2}{49}=\frac{160c^2}{49}$

$160=\frac{160\times c^2}{49}\Rightarrow c=7$

New mean = 44 – 4 × 5.5 = 44 – 22 = 22

Given,
Number of terms \(=10\)
Mean \(=5.5\)

\(\sum_{i=1}^{10} x_i^2=371\)

\(\begin{aligned} & \mu=\frac{\sum_{i=1}^{10} \mathrm{x}_{\mathrm{i}}}{n} \\ & 5.5=\frac{\sum_{i=1}^{10} \mathrm{x}_{\mathrm{i}}}{10} \\ & \sum_{i=1}^{10} \mathrm{x}_{\mathrm{i}}=55\end{aligned}\)

Corrected \(\sum^{10} x_i=55-(4+5)+(6+8)=60\)

Corrected Mean \(\mu=\frac{60}{10}=6\)

Corrected \(\sum_{i=1}^{10} x_i^2=371-\left(4^2+5^2\right)+\left(6^2+8^2\right)=430\)

Now, corrected variance \(=\frac{1}{n} \text { Corrected } \sum_{i=1}^{10} x_i^2-(\text { Corrected Mean } \mu)^2\)

\(=\frac{1}{10} \times 430-(6)^2=43-36=7\)

$$\begin{gathered} {\overline{x}}_1=10, {\omega }_1=7, {\overline{x}}_2=5, {\omega }_2=3 \\ \overline{x}=\frac{{\overline{x}}_1{\omega }_1+{\overline{x}}_2{\omega }_2}{{\omega }_1+{\omega }_2}=\frac{10\times 7+5\times 3}{7+3}=\frac{85}{10}=8.5 \end{gathered}$$

$\frac{1+3+a+7+b}{5}=5 \implies 11 +a+b=25 \implies a+b=14$

$\frac{1}{5}\sum (x_i -5)^2 =10 \implies \sum (x_i -5)^2=50$

$74 + (a^2 +b^2) -10(a+b) =50$

Substitute $a+b=14$

$a^2 +b^2 = 116$

$a=10, b=4$.

Mean of $y_n = \frac{2+5+13+11+9}{5}=8$

Variance $= \frac{1}{5}\sum y_n^2 - (\bar{y})^2$

$\sum y_n^2 = 2^2 +5^2 +13^2 +11^2 +9^2 =400$

Variance $=\frac{400}{5}-8^2=80-64=16$

Given \(a_1+a_2+\cdots+a_{10}=50\)

\(\left(a_1+a_2+\ldots+a_{10}\right)^2=2500\)

\(\sum_{i=1}^{10} a_i^2+2 \sum_{\forall k<j} a_k \cdot a_j=2500\)

\(\sum_{i=1}^{10} a_i^2=2500-2(1100)=300\)

Standard Deviation \(=\sqrt{\frac{\sum a_k^2}{10}-\left(\frac{\sum a_k}{10}\right)^2} \)
Standard Deviation \(=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^2}=\sqrt{5}\)

Given \(a_1+a_2+\cdots+a_{10}=50\)

\(\left(a_1+a_2+\ldots+a_{10}\right)^2=2500\)

\(\sum_{i=1}^{10} a_i^2+2 \sum_{\forall k<j} a_k \cdot a_j=2500\)

\(\sum_{i=1}^{10} a_i^2=2500-2(1100)=300\)

Standard Deviation \(=\sqrt{\frac{\sum a_k^2}{10}-\left(\frac{\sum a_k}{10}\right)^2} \)
Standard Deviation \(=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^2}=\sqrt{5}\)

$$\begin{gathered} \text{Given, lower limit }\ell \text{ =12} \\ \text{frequency f = 12} \\ \text{frequency less than 12 F= 18} \\ \text{class height = 18-12=6} \\ \text{median =14} \\ \text{Now, } \\ \text{ medain =ℓ+}\left(\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{\mathrm{f}}\right)\times \mathrm{h} \\ 14 = 12+\left(\frac{{\displaystyle \frac{\mathrm{N}}{2}-18}}{12}\right)\times 6 \\ \frac{14-12}{6}=\left(\frac{{\displaystyle \frac{\mathrm{N}}{2}-18}}{12}\right) \\ 4=\frac{\mathrm{N}}{2}-18 \\ \mathrm{N}=44 \end{gathered}$$

$$\begin{gathered} {\sigma }^2=\frac{\sum x^2}{n}-{\left(\frac{\sum x}{n}\right)}^2 \\ \Rightarrow {\sigma }^2=\frac{2^2+3^2+5^2+8^2+{12}^2}{5}-{\left(\frac{2+3+5+8+12}{5}\right)}^2 \\ \Rightarrow {\sigma }^2=13.2 \end{gathered}$$

Data 2, 3, 5, 8, 12
Median = 5
Mean deviation from the median

$$\begin{gathered} M=\frac{\sum \left|x- \mathrm{Median}\right|}{n}=\frac{3+2+0+3+7}{5} \\ M=\frac{15}{5}=3 \\ {\sigma }^2-\mathrm{M}=10.2 \end{gathered}$$

Let the mean of the \(20\) observations be \(\bar{x}\) and the standard deviation be \(\sigma\).

\(\sum_{i=1}^{20}\left(x_i+5\right)^2=2500\)\(\ldots(1)\)

\(\sum_{i=1}^{20}\left(x_i-5\right)^2=100\)\(\ldots(2)\)

\((1)-(2)\)

\(\Rightarrow 20 \sum x_i=2400 \Rightarrow \sum x_i=120\)

mean, \(\bar{x}=6\)

\((1)+(2)\)

\(\Rightarrow 2 \sum x_{i}^2+40\left(5^2\right)=2600\)

\(\sum x_i^2=800\)

\(\sigma^2=\frac{\sum x_{i}^2}{\mathrm{n}}-(\bar{x})^2=\frac{800}{20}-36=4\)

\(\bar{x}: \sigma=6: 2=3: 1\)

Given, lower limit \(\ell=12\)
frequency \(\mathrm{f}=12\)
frequency less than \(12 \mathrm{~F}=18\)
class height \(=18-12=6\)
median \(=14\)
Now,

Median

$$\begin{gathered} M=\ell +\left(\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{\mathrm{f}}\right)\times \mathrm{h} \\ 14 = 12+\left(\frac{{\displaystyle \frac{\mathrm{N}}{2}-18}}{12}\right)\times 6 \\ \frac{14-12}{6}=\left(\frac{{\displaystyle \frac{\mathrm{N}}{2}-18}}{12}\right) \\ 4=\frac{\mathrm{N}}{2}-18 \\ \mathrm{N}=44 \end{gathered}$$

\(\begin{aligned}
& \sum x=55-4-5+6+8=60 \\
& \sum x^2=328-16-25+36+64=387 \\
& \bar{x}=6 \\
& \sigma^2=\frac{387}{10}-6^2=38.7-36=2.7
\end{aligned}\)

$$\begin{gathered} \text{Consider} \sum _{i=1}^{10}\left(x_i-2\right)=30 \\ \Rightarrow \sum _{i=1}^{10}{x}_i-2\times 10=30 \\ \Rightarrow \sum _{i=1}^{10}{x}_i=50 \\ \text{Variance} =\frac{4}{5} \\ \Rightarrow \frac{4}{5}=\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{10}-{\left(\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{10}\right)}^2 \\ \Rightarrow \frac{4}{5}=\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{10}-5^2 \\ \Rightarrow \Sigma {\mathrm{x}}_{\mathrm{i}}^2=258 \\ \text{Next,} \sum _{i=1}^{10}{\left(x_i-\beta \right)}^2=98 \\ \underset{i=1}{\overset{10}{\Rightarrow \sum }}\left(x_i^2-2\beta ·x_i+{\beta }^2\right)=98 \\ \Rightarrow 258-2\beta \left(50\right)+10{\beta }^2=98 \\ \Rightarrow \left(\beta -8\right)\left(\beta -2\right)=0 \\ \text{As} \beta >2, \text{we get }\beta =8 \end{gathered}$$

$$\begin{gathered} \text{Consider, } \\ 2\left({\mathrm{x}}_1-1\right)+4\beta ,2\left({\mathrm{x}}_2-1\right)+4\beta , \\ \dots .2\left({\mathrm{x}}_{10}-1\right)+4\beta \\ \text{Put} \beta =8 \\ \text{we get }2{\mathrm{x}}_1+30,2{\mathrm{x}}_2+30,\dots ..2{\mathrm{x}}_{10}+30 \\ \mu =2\left(5\right)+30=40 \\ {\sigma }^2=2^2\left(\frac{4}{5}\right)=\frac{16}{5} \\ \frac{\mathrm{\beta }\mu }{{\sigma }^2}=\frac{8\times 40}{16/5}=100 \end{gathered}$$

Mean \(=\frac{\sum_{r=0}^n r(r-1)^n C_r}{\sum_{r=0}^n{ }^n C_r}\)

Using \(\sum_{r=0}^n r(r-1){ }^n C_r=n(n-1) 2^{n-2}\) and \(\sum_{r=0}^n{ }^n C_r=2^n\):

Mean \(=\frac{n(n-1) 2^{n-2}}{2^n}= \frac{n(n-1)}{4}\)

\(\frac{n(n-1)}{4}=60\)

\(n(n-1)=240\)

\(n=16\)

Total frequency \(=2^{16}=65536\)

Median lies at the \(32768^{\text {th }}\) and \(32769^{\text {th }}\) observations

Cumulative frequency up to \(r=7\) is \(\sum_{r=0}^7{ }^{16} C_r=26333\), and cumulative frequency up to \(r=8\) is \(\sum_{r=0}^8{ }^{16} C_r=39203\)

So both middle observations correspond to \(r=8\)

Median \(=8(8-1)=56\)

2,3,3,4,5,7,a,b are the observations

$$\begin{array}{l} \text{Mean} = 4 = \frac{2 + 6 + 4 + 12 + a + b}{8} \Rightarrow \boxed{a + b = 8} \\ \text{Variance} = 2 = \frac{4 + 18 + 16 + 25 + 49 + a^2 + b^2}{8} - 4^2 = 2 \\ a^2 + b^2 = 32 \\ a = b = 4 \\ \end{array}$$

Mean deviation about mode

$$\begin{array}{l}=\frac{1}{8}(\mathrm{\mid }2-4\mathrm{\mid }+\mathrm{\mid }3-4\mathrm{\mid }+\mathrm{\mid }3-4\mathrm{\mid }+\mathrm{\mid }5-4\mathrm{\mid }+\mathrm{\mid }7-4\mathrm{\mid })\\ =\frac{1}{8}(2+1+1+1+3)=1\end{array}$$

\(\begin{aligned}
&\begin{aligned}
& l=12, f=18 \quad f=12 \\
& h=6 \\
& M=14, \quad \Rightarrow 12+\frac{n / 2-18}{12} \times 6 =14\\
& 4=n / 2-18 \Rightarrow \quad n=44
\end{aligned}
\end{aligned}\)

Let remaining two observasions are \(a\) and \(b\).

Then, Mean$=\frac{2+3+5+10+11+13+15+21+a+b}{10}=9$

$\Rightarrow \frac{80+a+b}{10}=9\Rightarrow a+b=10$ …\((i)\)

And variance $=\frac{\Sigma x_i^2}{10}-{\left(\frac{\Sigma x_i}{10}\right)}^2=34.2$

$\Rightarrow \frac{2^2+3^2+5^2+{10}^2+{11}^2+{13}^2+{15}^2+{21}^2+a^2+b^2}{10}-{\left(9\right)}^2=34.2$

$\Rightarrow 1094+a^2+b^2-810=342$

$\Rightarrow a^2+b^2=58$ …\((ii)\)

from \((i)\) and \((ii)\), we get

$a=7, b=3$

or $a=3, b=7$

Numbers are $2,3,3,5,7,10,11,13,15,21$ (even number of terms)

Median $=\frac{7+10}{2}=8.5$

Now, mean deviation about the median:

$\text{M}.\text{D}=\frac{6.5+5.5+5.5+3.5+1.5+1.5+2.5+4.5+6.5+12.5}{10}$

$=\frac{50}{5}=5$

\(\begin{aligned}
& \sum x=55-4-5+6+8=60 \\
& \sum x^2=328-16-25+36+64=387 \\
& \bar{x}=6 \\
& \sigma^2=\frac{387}{10}-6^2=38.7-36=2.7
\end{aligned}\)

Let remaining two observations are \(a\) and \(b\).

Then, Mean$=\frac{2+3+5+10+11+13+15+21+a+b}{10}=9$

$\Rightarrow \frac{80+a+b}{10}=9\Rightarrow a+b=10$ …\((i)\)

And variance $=\frac{\Sigma x_i^2}{10}-{\left(\frac{\Sigma x_i}{10}\right)}^2=34.2$

$\Rightarrow \frac{2^2+3^2+5^2+{10}^2+{11}^2+{13}^2+{15}^2+{21}^2+a^2+b^2}{10}-{\left(9\right)}^2=34.2$

$\Rightarrow 1094+a^2+b^2-810=342$

$\Rightarrow a^2+b^2=58$ …\((ii)\)

from \((i)\) and \((ii)\), we get

$a=7, b=3$

or $a=3, b=7$

Numbers are $2,3,3,5,7,10,11,13,15,21$ (even number of terms)

Median $=\frac{7+10}{2}=8.5$

Now, mean deviation about the median:

$\text{M}.\text{D}=\frac{6.5+5.5+5.5+3.5+1.5+1.5+2.5+4.5+6.5+12.5}{10}$

$=\frac{50}{5}=5$

$$\begin{gathered} \text{Consider} \sum _{i=1}^{10}\left(x_i-2\right)=30 \\ \Rightarrow \sum _{i=1}^{10}{x}_i-2\times 10=30 \\ \Rightarrow \sum _{i=1}^{10}{x}_i=50 \\ \text{Variance} =\frac{4}{5} \\ \Rightarrow \frac{4}{5}=\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{10}-{\left(\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{10}\right)}^2 \\ \Rightarrow \frac{4}{5}=\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{10}-5^2 \\ \Rightarrow \Sigma {\mathrm{x}}_{\mathrm{i}}^2=258 \\ \text{Next,} \sum _{i=1}^{10}{\left(x_i-\beta \right)}^2=98 \\ \underset{i=1}{\overset{10}{\Rightarrow \sum }}\left(x_i^2-2\beta ·x_i+{\beta }^2\right)=98 \\ \Rightarrow 258-2\beta \left(50\right)+10{\beta }^2=98 \\ \Rightarrow \left(\beta -8\right)\left(\beta -2\right)=0 \\ \text{As} \beta >2, \text{we get }\beta =8 \end{gathered}$$

$$\begin{gathered} \text{Consider, } \\ 2\left({\mathrm{x}}_1-1\right)+4\beta ,2\left({\mathrm{x}}_2-1\right)+4\beta ,\dots .2\left({\mathrm{x}}_{10}-1\right)+4\beta \\ \text{Put} \beta =8 \\ \text{we get }2{\mathrm{x}}_1+30,2{\mathrm{x}}_2+30,\dots ..2{\mathrm{x}}_{10}+30 \\ \mu =2\left(5\right)+30=40 \\ {\sigma }^2=2^2\left(\frac{4}{5}\right)=\frac{16}{5} \\ \frac{\mathrm{\beta }\mu }{{\sigma }^2}=\frac{8\times 40}{16/5}=100 \end{gathered}$$

Observations: 2, 3, 3, 4, 5, 7, a, b
Mean = 4 and Standard Deviation = 2.
Mode = 3.
Using mean: (2+3+3+4+5+7+a+b)/8 = 4 ⇒ a+b = 8.
Using variance: [(2−4)²+(3−4)²+(3−4)²+(4−4)²+(5−4)²+(7−4)²+(a−4)²+(b−4)²]/8 = 4.
This gives (a−4)²+(b−4)² = 16.
Since a+b=8, let b=8−a. Then 2(a−4)²=16 ⇒ (a−4)²=8 ⇒ a,b = 4±2√2.
Mean deviation about mode (3) = [1+0+0+1+2+4+(1+2√2)+(1+2√2)]/8 = (10+4√2)/8 = (5+2√2)/4.
Answer: Option (A).

Let the data be:
$6, 4, a, 8, b, 12, 10, 13$

There are $n = 8$ observations, with mean $\bar{x} = 9$ and variance $\sigma^2 = 9.25$.

$$\frac{6 + 4 + a + 8 + b + 12 + 10 + 13}{8} = 9$$

Sum of known numbers: $6+4+8+12+10+13 = 53$.

$$\frac{53 + a + b}{8} = 9 \Rightarrow 53 + a + b = 72 \Rightarrow a + b = 19.$$

$$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$$

$$9.25 = \frac{\sum x_i^2}{8} - 9^2 = \frac{\sum x_i^2}{8} - 81$$ $$\frac{\sum x_i^2}{8} = 90.25 \Rightarrow \sum x_i^2 = 722.$$

$$529 + a^2 + b^2 = 722 \Rightarrow a^2 + b^2 = 193.$$

$$(a + b)^2 = a^2 + b^2 + 2ab \Rightarrow 19^2 = 193 + 2ab$$

$$361 = 193 + 2ab \Rightarrow 2ab = 168 \Rightarrow ab = 84.$$

$$a+b+ab=19+84=103.$$

$$\begin{gathered} \text{Given,} \\ \text{Number of terms =10} \\ \text{Mean = 5.5} \\ \sum _{i=1}^{10}{x}_i^2=371 \\ \text{We know, } \\ \mu =\frac{\sum _{i=1}^{10}{\mathrm{x}}_{\mathrm{i}}}{n} \\ 5.5=\frac{\sum _{i=1}^{10}{\mathrm{x}}_{\mathrm{i}}}{10} \\ \sum _{i=1}^{10}{\mathrm{x}}_{\mathrm{i}}=55 \\ \mathrm{Corrected} \sum _{\mathrm{i}=1}^{10}{\mathrm{x}}_{\mathrm{i}}=55-(4+5)+(6+8)=60 \\ \mathrm{Corrected} \mathrm{Mean} \mathrm{\mu }=\frac{\mathrm{corrected} \sum _{\mathrm{i}=1}^{10}{\mathrm{x}}_{\mathrm{i}}}{10} \\ \mathrm{Corrected} \mathrm{Mean} \mathrm{\mu }=\frac{60}{10}=6 \\ \mathrm{Now}, \sum _{\mathrm{i}=1}^{10}{\mathrm{x}}_{\mathrm{i}}^2=371 \\ \mathrm{Corrected} \sum _{\mathrm{i}=1}^{10}{\mathrm{x}}_{\mathrm{i}}^2=371-\left(4^2+5^2\right)+\left(6^2+8^2\right)=430 \\ \mathrm{Now}, \mathrm{corrected} \mathrm{variance} = \\ \frac{1}{\mathrm{n}}\mathrm{Corrected} \sum _{\mathrm{i}=1}^{10}{\mathrm{x}}_{\mathrm{i}}^2-{\left(\mathrm{Corrected} \mathrm{Mean} \mathrm{\mu }\right)}^2 \\ =\frac{1}{10}\times 430-{\left(6\right)}^2=43-36=7 \end{gathered}$$

\(\frac{\sum x_i}{100}=40 \)
\(\Rightarrow \sum x_i=4000 \)
\(\sum x_i=4000-50+40=3990 \)
\(\mu=\frac{3990}{10}=39.9 \)
\((5.1)^2=\frac{\sum x_i^2}{100}-(40)^2 \)
\(\sigma=\sqrt{\frac{\left\{(5.1)^2+1600\right\} \times 100-(50)^2+(40)^2}{100}-(39.90)^2}=5 \)
\(10(\mu+\sigma)=10(39.90+5)=449\)

$\frac{x_1+x_2+x_3+\dots +x_n}{n}=\overline{x}$

$\therefore$ New mean is ${\overline{x}}_B^{\text{'}}=\frac{{\displaystyle \sum _{r=1}^n\left(x_i+2r\right)}}{n}$

$=\overline{x}+\frac{2n(n+1)}{2n}$

$\overline{x}\text{'}\Rightarrow \overline{x}+(n+1)$

\(\begin{aligned}
&\begin{aligned}
& l=12, f=18 \quad f=12 \\
& h=6 \\
& M=14, \quad \Rightarrow 12+\frac{n / 2-18}{12} \times 6 =14\\
& 4=n / 2-18 \Rightarrow \quad n=44
\end{aligned}
\end{aligned}\)

Given: \(y=ax+b\)

\(\Sigma y_i=a \Sigma x_i+\Sigma b\)

\(=a \times(1+2+\ldots \ldots+19)+19 b\)

$\Rightarrow \frac{{\displaystyle \sum y_i}}{19}=\frac{a\times 19\times 20}{2\times 19}+b$

\(\Rightarrow 30=10 a+b \ldots(i)\)

Variance of $X=\frac{{\displaystyle \sum x_i^2}}{19}-{\left(\frac{{\displaystyle \sum x_i}}{19}\right)}^2$

$=\frac{19\times 20\times 39}{19\times 6}-{\left(10\right)}^2=30$

Variance of \(Y=a^2(\) variance of \(X)\)

\(\Rightarrow 750=a^2 \times 30\)

\(\Rightarrow a^2=25 \Rightarrow a= \pm 5\)

if \(a=5 \Rightarrow b=30-50=-20\) ...from \((i)\)

if \(a=-5 \Rightarrow b=30+50=80\) ...from \((i)\)

Sum of values of \(b=80-20=60\)

\( \frac{\sum_{i=1}^n x_i}{n}=8 \)
\( \Rightarrow \frac{48+x_n}{n}=8 \Rightarrow x_n=8 n-48\)
\( \frac{\sum x_1^2}{n}-(\bar{x})^2=16\)
\( \Rightarrow \frac{496+x_n^2}{n}-64=16\)
\( \Rightarrow 496+x_n^2=80 n \)
\( \Rightarrow 496+(8 n-48)^2=80 n \)
\( \Rightarrow31+4(n-6)^2=5 n\)
\( \Rightarrow 4 n^2-48 n+175=5 n\)
\( \Rightarrow 4 n^2-53 n+175=0\)

\(\Rightarrow(4 n-25)(n-7)=0\)

\(n=7, \frac{25}{4}\)

but \(n\) must be a natural number, so \(n=7\)

For set \(1\)

\(\bar{x}=1, \sigma_1^2=13\)

For set \(2\)

\(\bar{y}=2, \sigma_2^2=1\)

Combined variance \(=\frac{\mathrm{n}_1 \sigma_1^2+\mathrm{n}_2 \sigma_2^2}{\mathrm{n}_1+\mathrm{n}_2}+\frac{\mathrm{n}_1 \mathrm{n}_2}{\left(\mathrm{n}_1+\mathrm{n}_2\right)^2}(\overline{\mathrm{x}}-\overline{\mathrm{y}})^2\)

\(=\frac{4 \times 13+6 \times 1}{10}+\frac{4 \times 6}{10^2}(2-1)^2\)

\(=\frac{58}{10}+\frac{24}{100}=\frac{580+24}{100}=\frac{604}{100}=6.04\)

$\frac{1+3+a+7+b}{5}=5⟹11+a+b=25⟹a+b=14$

$\frac{1}{5}\sum (x_i-5{)}^2=10⟹\sum (x_i-5{)}^2=50$

$74+(a^2+b^2)-10(a+b)=50$

Substitute $a+b=14$

$a^2 +b^2 = 116$

$a=10, b=4$.

Let $y_n=n+x_n$, therefore

Mean of $y_n = \frac{2+5+13+11+9}{5}=8$

Variance $= \frac{1}{5}\sum y_n^2 - (\bar{y})^2$

$\sum y_n^2 = 2^2 +5^2 +13^2 +11^2 +9^2 =400$

Variance $=\frac{400}{5}-8^2=80-64=16$

$$\begin{gathered} {\overline{x}}_1=10, {\omega }_1=7, {\overline{x}}_2=5, {\omega }_2=3 \\ \overline{x}=\frac{{\overline{x}}_1{\omega }_1+{\overline{x}}_2{\omega }_2}{{\omega }_1+{\omega }_2}=\frac{10\times 7+5\times 3}{7+3}=\frac{85}{10}=8.5 \end{gathered}$$

Here number of terms are even therefore

median $=\frac{500k+501k}{2}=\frac{1001\text{k}}{2}={\text{X}}_{\text{M}}$

mean deviation about median $=\frac{\sum \left|{\text{X}}_{\text{i}}-{\text{X}}_{\text{M}}\right|}{\text{n}}$

$=\frac{2\left(\frac{\text{k}}{2}+\frac{3\text{k}}{2}+\frac{5\text{k}}{2}+\dots 500\text{ terms }\right)}{1000}$

$$\begin{gathered} =\frac{2\times {\displaystyle \frac{k}{2}}\left(1+3+5+.....500 \mathrm{terms}\right)}{1000} \\ =\frac{2\times {\displaystyle \frac{k}{2}}\left({\displaystyle \frac{500}{2}}\left\{2+\left(500-1\right)\times 2\right\}\right)}{1000} \end{gathered}$$

\(=\frac{2 \cdot \frac{k}{2}(500)^2}{1000}=\frac{500 k}{2}=500(\) given \()\)

\(\therefore k=2\)

Hence, \(k^2=4\)

\( \frac{8+13+17+\mathrm{b}+21+\mathrm{a}+51+67+103}{9}=40 \)
\( \mathrm{a}+\mathrm{b}+280=360 \Rightarrow \mathrm{a}+\mathrm{b}=80 \ldots .(1) \)

For median \(21\), after arranging the \(9\) numbers, \(21\) must be the fifth term.

The numbers less than \(21\) are \(8,13,17\). Hence one of \(a,b\) must also be less than \(21\).

Since \(a>b\), we take \(b<21\) and \(a>21\)

Mean deviation about median \(=\frac{\sum\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{n}} \)
\( \Rightarrow \frac{\sum\left|\mathrm{x}_{\mathrm{i}}-21\right|}{9}=26 \)
\( \Rightarrow \frac{1}{9}(13+8+4+21-\mathrm{b}+0+\mathrm{a}-21+30+46+82)=26 \)
\( \Rightarrow \mathrm{a}-\mathrm{b}=51 \quad \ldots .(2) \)

From (1) and (2)
\( \Rightarrow 2 \mathrm{a}=131\)

Given: \(y=ax+b\)

\(\Sigma y_i=a \Sigma x_i+\Sigma b\)

\(=a \times(1+2+\ldots \ldots+19)+19 b\)

$\Rightarrow \frac{{\displaystyle \sum y_i}}{19}=\frac{a\times 19\times 20}{2\times 19}+b$

\(\Rightarrow 30=10 a+b \ldots(i)\)

Variance of $X=\frac{{\displaystyle \sum x_i^2}}{19}-{\left(\frac{{\displaystyle \sum x_i}}{19}\right)}^2$

$=\frac{19\times 20\times 39}{19\times 6}-{\left(10\right)}^2=30$

Variance of \(Y=a^2(\) variance of \(X)\)

\(\Rightarrow 750=a^2 \times 30\)

\(\Rightarrow a^2=25 \Rightarrow a= \pm 5\)

if \(a=5 \Rightarrow b=30-50=-20\) ...from \((i)\)

if \(a=-5 \Rightarrow b=30+50=80\) ...from \((i)\)

Sum of values of \(b=80-20=60\)

Mean $\left(\overline{x}\right)=10$
$\Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{20}=10$
$\Sigma {\mathrm{x}}_{\mathrm{i}}=10\times 20=200$
If 8 is replaced by 12 , then correct $\Sigma x_i=200-8+12=204$
Correct mean \((\bar{x})=\frac{\sum x_{\mathrm{i}}}{20}\)
$=\frac{204}{20}=10.2$
since standard deviation \(=2\)
therefore variance \(=(\text { S.D. })^2=2^2=4\)
$\Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{20}-{\left(\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{20}\right)}^2=4$
$\Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{20}-(10{)}^2=4$
$\Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{20}=104$
$\Rightarrow \Sigma {\mathrm{x}}_{\mathrm{i}}^2=2080$
Now, replaced '8' by '12'
Then, correct $\Sigma {\mathrm{x}}_{\mathrm{i}}^2=2080-8^2+{12}^2=2160$
$\therefore$ correct Variance

$=\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{20}-{\left(\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{20}\right)}^2$
$=\frac{2160}{20}-(10.2{)}^2$
$=108-104.04$
$=3.96$
Correct standard deviation

$=\sqrt{3.96}$

Mean $\left(\overline{x}\right)=10$
$\Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{20}=10$
$\Sigma {\mathrm{x}}_{\mathrm{i}}=10\times 20=200$
If 8 is replaced by 12 , then correct $\Sigma x_i=200-8+12=204$
Correct mean \((\bar{x})=\frac{\sum x_{\mathrm{i}}}{20}\)
$=\frac{204}{20}=10.2$
since standard deviation \(=2\)
therefore variance \(=(\text { S.D. })^2=2^2=4\)
$\Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{20}-{\left(\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{20}\right)}^2=4$
$\Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{20}-(10{)}^2=4$
$\Rightarrow \frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{20}=104$
$\Rightarrow \Sigma {\mathrm{x}}_{\mathrm{i}}^2=2080$
Now, replaced '8' by '12'
Then, correct $\Sigma {\mathrm{x}}_{\mathrm{i}}^2=2080-8^2+{12}^2=2160$
$\therefore$ correct Variance

$=\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{20}-{\left(\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{20}\right)}^2$
$=\frac{2160}{20}-(10.2{)}^2$
$=108-104.04$
$=3.96$
Correct standard deviation

$=\sqrt{3.96}$

Let the data be:
$6, 4, a, 8, b, 12, 10, 13$

There are $n = 8$ observations, with mean $\bar{x} = 9$ and variance $\sigma^2 = 9.25$.

$\frac{6+4+a+8+b+12+10+13}{8}=9$

Sum of known numbers: $6+4+8+12+10+13=53$.

$\frac{53+a+b}{8}=9\Rightarrow 53+a+b=72\Rightarrow a+b=19.$

${\sigma }^2=\frac{\sum x_i^2}{n}-{\bar{x}}^2$

$9.25=\frac{\sum x_i^2}{8}-9^2=\frac{\sum x_i^2}{8}-81$

$\frac{\sum x_i^2}{8}=90.25\Rightarrow \sum x_i^2=722.$

$529+a^2+b^2=722\Rightarrow a^2+b^2=193.$

$(a+b{)}^2=a^2+b^2+2ab\Rightarrow {19}^2=193+2ab$

$361=193+2ab\Rightarrow 2ab=168\Rightarrow ab=84.$

$a+b+ab=19+84=103.$

$$\begin{gathered} {\sigma }^2=\frac{\sum x^2}{n}-{\left(\frac{\sum x}{n}\right)}^2 \\ \Rightarrow {\sigma }^2=\frac{2^2+3^2+5^2+8^2+{12}^2}{5}-{\left(\frac{2+3+5+8+12}{5}\right)}^2 \\ \Rightarrow {\sigma }^2=13.2 \end{gathered}$$

Data 2, 3, 5, 8, 12
Median = 5
Mean deviation from the median

$$\begin{gathered} M=\frac{\sum \left|x- \mathrm{Median}\right|}{n}=\frac{3+2+0+3+7}{5} \\ M=\frac{15}{5}=3 \\ {\sigma }^2-\mathrm{M}=10.2 \end{gathered}$$