Statistics

We know the mean of 10 observations is 5. So, the total sum of the 10 observations is:

$\frac{x_1+x_2+x_3+\ldots+x_8+x_9+x_{10}}{10}=5$

Multiplying both sides by 10 gives:

$x_1+x_2+x_3+\ldots+x_8+x_9+x_{10}=50$

The mean of the first 8 observations is 4, so their total sum is:

$\frac{x_1+x_2+\ldots+x_8}{8}=4$

Therefore,

$x_1+x_2+\ldots+x_8=32$

Subtracting, we get:

$x_9+x_{10}=50-32=18$

Next, we use the formula for variance of the 10 observations:

$\text{Variance} = \frac{\sum\limits_{i=1}^{10} x_i^2}{10} - (\text{mean})^2$

Substitute the given values to find the total of squares:

$\frac{\sum\limits_{i=1}^{10} x_i^2}{10} - 25 = 7$

$\sum\limits_{i=1}^{10} x_i^2 = 320$

Similarly, for the first 8 observations:

$\frac{\sum\limits_{i=1}^{8} x_i^2}{8} - 16 = 3.5$

$\sum\limits_{i=1}^{8} x_i^2 = 156$

Now, subtract the two results:

$x_9^2 + x_{10}^2 = 320 - 156 = 164$

We already know:

$x_9 + x_{10} = 18$

Expanding $ (x_9 + x_{10})^2 $, we get:

$x_9^2 + x_{10}^2 + 2x_9x_{10} = 324$

Substitute $x_9^2 + x_{10}^2 = 164$:

$164 + 2x_9x_{10} = 324$

$x_9x_{10} = 80$

Now, we have two equations:

$x_9 + x_{10} = 18$

$x_9x_{10} = 80$

Solving the quadratic $t^2 - 18t + 80 = 0$, we get $t = 8$ and $10$. Given $x_9 < x_{10}$, so $x_9 = 8$ and $x_{10} = 10$.

Finally,

$3x_9 + 2x_{10} = 3(8) + 2(10) = 44$

$\begin{aligned} &\begin{aligned} & \operatorname{Var}(8,21,34,47, \ldots \ldots, 320) \\ & \operatorname{Var}(0,13,26,39, \ldots \ldots, 312) \\ & 13^2 \cdot \operatorname{Var}(0,1,2, \ldots \ldots, 24) \\ & 13^2 \cdot \operatorname{Var}(1,2,3, \ldots \ldots, 25) \end{aligned}\\ &\text { So, } \sigma^2=13^2 \times\left(\frac{25^2-1}{12}\right)=8788\\ &\text { Alternate solution }\\ &\begin{aligned} & 8+(n-1) 13=320 \\ & 13 n=325 \\ & n=25 \end{aligned} \end{aligned}$

$\begin{aligned} &\text { no. of terms }=25\\ &\begin{aligned} & \text { mean }=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}=\frac{8+21+\ldots+320}{25}=\frac{\frac{25}{2}(8+320)}{25} \\ & \text { variance } \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{\mathrm{n}}-(\text { mean })^2 \\ & =\frac{8^2+21^2+\ldots .+320^2}{13}-(164)^2 \\ & =8788 \end{aligned} \end{aligned}$

$\begin{aligned} & a, b, c \in N \\ & a< b < c \\ & \text { Mean }=18 \\ & \frac{9+25+a+b+c}{5}=18 \\ & 34+a+b+c=90 \\ & a+b+c=56 \end{aligned}$

$\begin{aligned} & \frac{|9-18|+|25-18|+|a-18|+|b-18|+|c-18|}{5}=4 \\ & 9+7+|a-18|+|b-18|+|c-18|=20 \\ & |a-18|+|b-18|+|c-18|=4 \\ & \frac{136}{5}=\frac{706+a^2+b^2+c^2}{5}-(18)^2 \\ & \Rightarrow 136=706+a^2+b^2+c^2-1620 \\ & \Rightarrow a^2+b^2+c^2=1050 \\ & \text { Consider } a<19 < b< c \\ & \text { Solving } a=17, b=19, c=20 \\ & 2 a+b-c \\ & 34+19-20 \\ & =33 \end{aligned}$

Mean $(\mu)=\Sigma x_i P\left(x_i\right)$

Standard deviation $(\sigma)=\sqrt{\left(\Sigma x_i^2 P\left(x_i\right)\right)-\mu^2}$

$\begin{aligned} & \Rightarrow \quad \mu=\frac{1}{3} \alpha+K-\frac{3}{4} \\ & \sigma=\sqrt{\left(\frac{1}{3} \alpha^2+K+0+\frac{9}{4}\right)-\left(\frac{1}{3} \alpha+K-\frac{3}{4}\right)^2} \\ & \because \Sigma P_i=1 \Rightarrow \frac{1}{3}+K+\frac{1}{6}+\frac{1}{4}=1 \\ & \Rightarrow \quad K=\frac{1}{4} \Rightarrow \mu=\frac{1}{3} \alpha-\frac{1}{2} \\ & \because \sigma-\mu=2 \\ & \sigma^2=(\mu+2)^2 \end{aligned}$

$\begin{aligned} & \frac{1}{3} \alpha^2+\frac{5}{2}-\mu^2=(\mu+2)^2 \\ & \frac{1}{3} \alpha^2+\frac{5}{2}=\left(\frac{1}{3} \alpha-\frac{1}{2}\right)^2+\left(\frac{1}{3} \alpha+\frac{3}{2}\right)^2 \\ & \Rightarrow \alpha=0,6 \end{aligned}$

$\begin{array}{ll} \text { If } \alpha=0, K=\frac{1}{4} & \text { If } \alpha=6, K=\frac{1}{4} \\ \mu=-\frac{1}{2}, \sigma=\frac{3}{2} & \mu=\frac{3}{2}, \sigma=\frac{7}{2} \\ \sigma+\mu=1 & \sigma+\mu=5 \end{array}$

Both (1) and (5) are correct but according to NTA (5) is correct

$\mathrm{x_i}$	$\mathrm{f_i}$	$\mathrm{f_i x_i}$	$\mathrm{f_i x^2_i}$
0	3	0	0
1	2	2	2
5	3	15	75
6	2	12	72
10	6	60	600
12	3	36	432
17	3	51	867
	$\Sigma \mathrm{f}_{\mathrm{i}}=22$		$\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=2048$

$\begin{aligned} & \therefore \quad \Sigma \mathrm{f}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}=176 \\ & \text { So } \overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{176}{22}=8 \\ & \text { for } \sigma^2=\frac{1}{\mathrm{~N}} \sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}{ }^2-(\overline{\mathrm{x}})^2 \\ & =\frac{1}{22} \times 2048-(8)^2 \\ & =93.090964 \\ & =29.0909 \\ & \end{aligned}$

$\begin{aligned} & \overline{\mathrm{x}}=56 \\ & \sigma^2=66.2 \\ & \Rightarrow \frac{\alpha^2+\beta^2+25678}{10}-(56)^2=66.2 \\ & \therefore \alpha^2+\beta^2=6344 \end{aligned}$

Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$

We have

$\mu^{\prime}=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{15}=12 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}=180$

As per given information correct $\Sigma \mathrm{x}_{\mathrm{i}}=180-10+12$

$\Rightarrow \mu(\text { correct mean})=\frac{182}{15}$

Also

$\sigma^{\prime}=\sqrt{\frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{15}-144}=3 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^2=2295$

Correct $\Sigma \mathrm{x}_{\mathrm{i}}{ }^2=2295-100+144=2339$

$\sigma^2(\text { correct variance })=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15} $

Required value

$\begin{aligned} & =15\left(\mu+\mu^2+\sigma^2\right) \\ & =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) \\ & =15\left(\frac{182}{15}+\frac{2339}{15}\right) \\ & =2521 \end{aligned}$

1. The initial mean is given by:

$\bar{x}=50$

So, the total sum of the marks initially was:

$\sum x_i = \bar{x} \times n = 50 \times 10 = 500$

1. We later realize that two marks were incorrectly read as 45 and 50, when they should have been 20 and 25. Therefore, the corrected sum of the marks is:

$\sum x_{i{\text{correct}}} = \sum x_i - 45 - 50 + 20 + 25 = 500 - 45 - 50 + 20 + 25 = 450$

1. The initial variance is given as:

$\sigma^2 = 144$

We know that variance is calculated as the mean of the squares minus the square of the mean. Therefore, rearranging gives:

$\frac{\sum x_i^2}{n} = \sigma^2 + \bar{x}^2 = 144 + 50^2 = 2644$

Then, the sum of the squares of the initial marks is:

$\sum x_i^2 = n \times \frac{\sum x_i^2}{n} = 10 \times 2594 = 26440$

1. The corrected sum of the squares of the marks is calculated by subtracting the squares of the incorrect marks and adding the squares of the correct marks:

$\sum x_{i{\text{correct}}}^2 = \sum x_i^2 - 45^2 - 50^2 + 20^2 + 25^2 = 26400 - 45^2 - 50^2 + 20^2 + 25^2 = 22940$

1. Now we can calculate the corrected variance. The variance is the mean of the squares minus the square of the mean. Using the corrected values gives:

$\sigma_{\text{correct}}^2 = \frac{\sum x_{i{\text{correct}}}^2}{n} - \left(\frac{\sum x{i_{\text{correct}}}}{n}\right)^2 = \frac{22940}{10} - \left(\frac{450}{10}\right)^2 = 2294 - 45^2 = 2294 - 2025 = 269$

Therefore, the correct variance is 269.

$ \begin{aligned} & 5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha} \\\\ & \Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16 \end{aligned} $

$ \begin{aligned} \sum f_i & =80 \\\\ \text { M.D } & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\ & =\frac{4+4+24 \times 2+0+16 \times 2+8 \times 4}{80} \\\\ & =\frac{8}{5} \end{aligned} $

$ \begin{aligned} \sigma^2 & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\ & =\frac{4+16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\frac{22}{5} \end{aligned} $

So, $ \frac{3 \alpha}{m+\sigma^2}=\frac{3 \times 16}{\frac{8}{5}+\frac{22}{5}}=8 $

Since $a_1, a_2, a_3, a_4, a_5$ are in geometric progression, we can write :

$a_2 = r a$,

$a_3 = r^2 a$,

$a_4 = r^3 a$,

$a_5 = r^4 a$.

where $r$ is the common ratio and $a_1$ = $a$ is the first term.

Given that the mean of the series is $\frac{31}{10}$, we get

$\frac{1}{5}(a_1 + a_2 + a_3 + a_4 + a_5) = \frac{31}{10}$

Substituting the terms with the values of $a_1$ and $r$ gives

$\frac{1}{5}(a + r a + r^2 a + r^3 a + r^4 a) = \frac{31}{10}$

Simplifying this gives

$a (1 + r + r^2 + r^3 + r^4) = \frac{31}{2}$

$ \begin{aligned} & \frac{a\left(r^5-1\right)}{r-1}=\frac{31}{2} ......(1) \\\\ & \frac{1}{a}\left(1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\frac{1}{r^4}\right)=\frac{31}{40} \cdot 5=\frac{31}{8} \\\\ & \frac{1}{a}\left(\frac{1-\left(\frac{1}{r}\right)^5}{1-\frac{1}{r}}\right)=\frac{31}{8} \\\\ & \text { or } \frac{1}{a}\left(\frac{r^5-1}{r-1}\right) \frac{1}{r^4}=\frac{31}{8} .........(2) \end{aligned} $

From (1) and (2)

$ \frac{1}{a} \cdot \frac{31}{2 a} \cdot \frac{1}{r^4}=\frac{31}{8} $

$ a r^2=2 $

From (1)

$ \begin{aligned} & \frac{2}{r^2}\left(\frac{r^5-1}{r-1}\right)=\frac{31}{2} \\\\ & \frac{1+r+r^2+r^3+r^4}{r^2}=\frac{31}{4} \\\\ & \left(r^2+\frac{1}{r^2}\right)+\left(r+\frac{1}{r}\right)=\frac{27}{4} \\\\ & t^2-2+t=\frac{27}{4} \end{aligned} $

$ \begin{aligned} &4 t^2+4 t-35=0\\\\ &4 t^2+14 t-10 t-35=0\\\\ &(2 t-5)(2 t+7)=0\\\\ &t=\frac{5}{2}, \frac{-7}{2} \Rightarrow r=2\\\\ & \therefore r=2, a=\frac{1}{2} \end{aligned} $

$ \text { Variance of data set }\left\{\frac{1}{2}, 1,2,4,8\right\} $

$ \therefore \sigma^2=\frac{\sum{\mathrm{X}^2}}{\mathrm{~N}}-\left(\frac{\sum{\mathrm{X}}}{\mathrm{N}}\right)^2 $

$ \begin{aligned} & =\frac{\left(\frac{341}{4}\right)}{5}-\left(\frac{31}{10}\right)^2 \\\\ & =\frac{341}{20}-\frac{961}{100}=\frac{1705-961}{100} \\\\ & =\frac{744}{100} \end{aligned} $

$ =\frac{186}{25}=\frac{\mathrm{m}}{\mathrm{n}} \Rightarrow 211=\mathrm{m}+\mathrm{n} $

Given mean is 28

$ \begin{array}{ll} \text { So, } \frac{2 \times 5+3 \times 15+x \times 25+5 \times 35+4 \times 45}{14+x}=28 \\\\ \Rightarrow \frac{10+45+25 x+175+180}{14+x}=28 \\\\ \Rightarrow 310+25 x=392+28 x \\\\ \Rightarrow 3 x=18 \Rightarrow x=6 \end{array} $

$ \begin{aligned} & \therefore \text { Variance }=\left(\frac{\sum x_i^2 f_i}{\sum f_i}\right)-(\text { mean })^2 \\\\ & =\left(\frac{2 \times 5^2+3 \times 15^2+6 \times 25^2+5 \times 35^2+4 \times 45^2}{20}\right)-(28)^2 \\\\ & =\left(\frac{50+675+3750+6125+8100}{20}\right)-(28)^2 \\\\ & =\left(\frac{18700}{20}\right)-(28)^2 \\\\ & =935-784=151 \end{aligned} $

$ \begin{array}{|c|c|c|} \hline x_i & (x_i-\bar{x}) & (x_i-\bar{x})^2 \\ \hline x & x-9 & (x-9)^2 \\ \hline y & y-9 & (y-9)^2 \\ \hline 10 & 1 & 1 \\ \hline 12 & 3 & 9 \\ \hline 6 & -3 & 9 \\ \hline 12 & 3 & 9 \\ \hline 4 & -5 & 25 \\ \hline 8 & -1 & 1 \\ \hline x+y+92 & & (x-9)^2+(y-9)^2+54 \\ \hline \end{array} $

Now, mean $(\bar{x})=9$

$ \begin{aligned} & \Rightarrow \frac{x+y+52}{8}=9 \\\\ & \Rightarrow x+y=20 \end{aligned} $

Also, variance $=9.25$

$ \begin{aligned} & \Rightarrow \frac{(x-9)^2+(y-9)^2+54}{8}=9.25 \\\\ & \Rightarrow x^2+y^2+81+81-2 \times 9(x+y)=20 \\\\ & \Rightarrow x^2+y^2-18 \times 20=-142 \\\\ & \Rightarrow x^2+y^2=218 \\\\ & \Rightarrow x^2+(20-x)^2=218 \\\\ & \Rightarrow x^2+400+x^2-40 x=218 \\\\ & \Rightarrow 2 x^2-40 x+182=0 \\\\ & \Rightarrow x=\frac{40 \pm 12}{4} \\\\ & \Rightarrow x=13 \text { or } x=7 \Rightarrow y=7 \text { or } y=13 \\\\ & \text { But } x>y \\\\ & \therefore x=13 \text { and } y=7 \\\\ & \text { So, } 3 x-2 y=39-14=25 \end{aligned} $

Concept :

(a) Mean $=\frac{\Sigma x_i}{n}$

(b) Variance $=\frac{\Sigma\left(x_i-\bar{x}\right)^2}{n}$

$ \begin{array}{lllll} \hline x_i & f_i & x_i^2 & f_i x_i & f_i x_i^2 \\ \hline 2 & 4 & 4 & 8 & 16 \\ \hline 4 & 4 & 16 & 16 & 64 \\ \hline 6 & \alpha & 36 & 6 \alpha & 36 \alpha \\ \hline 8 & 15 & 64 & 120 & 960 \\ \hline 10 & 8 & 100 & 80 & 800 \\ \hline 12 & \beta & 144 & 12 \beta & 144 \beta \\ \hline 14 & 4 & 196 & 56 & 784 \\ \hline 16 & 5 & 256 & 80 & 1280 \\ \hline \end{array} $

$ \therefore $ $\begin{aligned} & \Sigma f_i=40 +\alpha+\beta\end{aligned}$

$\begin{aligned} & \Sigma f_i x_i=360+ 6 \alpha+12 \beta\end{aligned}$

$\begin{aligned} & \Sigma f_i x_i^2=3904 +36 \alpha+144 \beta\end{aligned}$

Given, mean $=9$

$ \begin{aligned} & \Rightarrow \frac{\Sigma f_i x_i}{\Sigma f_i}=9 \\\\ & \Rightarrow \frac{360+6 \alpha+12 \beta}{40+\alpha+\beta}=9 \\\\ & \Rightarrow 360+6 \alpha+12 \beta=9(40+\alpha+\beta) \\\\ & \Rightarrow 3 \beta=3 \alpha \\\\ & \Rightarrow \alpha=\beta .........(i) \end{aligned} $

$\begin{aligned} & \text { Variance }=15.08 \\\\ & \Rightarrow \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2=15.08 \\\\ & \Rightarrow \frac{3904+36 \alpha+144 \beta}{40+\alpha+\beta}-(9)^2=15.08 \\\\ & \Rightarrow \frac{3904+180 \alpha}{40+2 \alpha}=81+15.08 \quad[\because \alpha=\beta] \\\\ & \Rightarrow 3904+180 \alpha=96.08(40+2 \alpha) \\\\ & \Rightarrow 3904+180 \alpha=3843.2+192.16 \alpha \\\\ & \Rightarrow 60.8=12.16 \alpha \\\\ & \Rightarrow \alpha=5=\beta \\\\ & \therefore \alpha^2+\beta^2-\alpha \beta=25+25-25=25\end{aligned}$

$\sigma_{\mathrm{x}}^{2}=\sigma_{\mathrm{d}}^{2}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}^{2}}{\sum \mathrm{f}_{\mathrm{i}}}-\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\right)^{2}$

$=\frac{150}{45+\alpha}-0=3$

$\Rightarrow 150=135+3 \alpha$

$\Rightarrow 3 \alpha=15 \Rightarrow \alpha=5$

$\sum {{x_i} = 7 \times 8 = 56} $

${{\sum {x_i^2} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2} = 16$

${{\sum {x_i^2} } \over 7} - 64 = 16$

$\sum {x_i^2 = 560} $

when 14 is omitted

$\sum {{x_i} = 56 - 14 = 42} $

New mean $ = a = {{\sum {{x_i}} } \over 6} = 7$

$\sum {x_i^2 = 560 - 196 = 364} $

new variance, $b = {{\sum {x_i^2} } \over 6} - {\left( {{{\sum {{x_i}} } \over 6}} \right)^2}$

$ = {{364} \over 6} - 49 = {{35} \over 3}$

$3b = 35$

$a + 3b - 5 = 7 + 35 - 5 = 37$

$x = \{ 11,12,13\,....,40,41\} $

$y = \{ 61,62,63\,....,90,91\} $

$\overline x = {{{{31} \over 2}(11 + 41)} \over {31}} = {1 \over 2} \times 52 = 26$

$\overline y = {{{{31} \over 2}(61 + 91)} \over {31}} = {1 \over 2} \times 152 = 76$

${\sigma ^2} = {{\sum {x_i^2 + \sum {y_i^2} } } \over {62}} - {\left( {{{\sum {x + \sum y } } \over {62}}} \right)^2}$

$ = 705$

Now,

$\left| {\overline x + \overline y - {\sigma ^2}} \right|$

$ = |26 + 76 - 705| = 603$

Given $\sum\limits_{{{i = 1} \over {20}}}^{20} {{x_i} = 15 \Rightarrow \sum\limits_{i = 1}^{20} {{x_i} = 300} } $

and $\sum\limits_{{{i = 1} \over {20}}}^{20} {x_i^2 - {{\left( {\overline x } \right)}^2} = 9 \Rightarrow \sum\limits_{i = 1}^{20} {x_i^2 = 4680} } $

Mean $ = {{{{({x_i} + \alpha )}^2} + {{({x_2} + \alpha )}^2}\, + \,.....\, + \,{{({x_{20}} + \alpha )}^2}} \over {20}} = 178$

$ \Rightarrow {{\sum\limits_{i = 1}^{20} {x_i^2 + 2\alpha \sum\limits_{i = 1}^{20} {{x_i} + 20{\alpha ^2}} } } \over {20}} = 178$

$ \Rightarrow 4680 + 600\alpha + 20{\alpha ^2} = 3560$

$ \Rightarrow {\alpha ^2} + 30\alpha + 56 = 0$

$ \Rightarrow {\alpha ^2} + 28\alpha + 2\alpha + 56 = 0$

$ \Rightarrow (\alpha + 28)(\alpha + 2) = 0$

${\alpha _{\max }} = - 2 \Rightarrow \alpha _{\max }^2 = 4.$

${x_1},{x_2},{x_3},\,.....,\,{x_{20}}$ are in G.P.

${x_1} = 3,\,r = {1 \over 2}$

$\overline x = {{\sum {x_i^2 - 2{x_i}i + {i^2}} } \over {20}}$

$ = {1 \over {20}}\left[ {12\left( {1 - {1 \over {{2^{40}}}}} \right) - 6\left( {4 - {{11} \over {{2^{18}}}}} \right) + 70 \times 41} \right]$

$\left\{ {\matrix{ {S = 1 + 2\,.\,{1 \over 2} + 3\,.\,{1 \over {{2^2}}}\, + \,....} \cr {{S \over 2} = {1 \over 2} + {2 \over {{2^2}}}\, + \,....} \cr } } \right.$

$\left. {{S \over 2} = 2\left( {1 - {1 \over {{2^{20}}}}} \right) - {{20} \over {{2^{20}}}} = 4 - {{11} \over {{2^{18}}}}} \right\}$

$\therefore$ $[\overline x ] = \left[ {{{2858} \over {20}} - \left( {{{12} \over {240}} - {{66} \over {{2^{18}}}}} \right)\,.\,{1 \over {20}}} \right]$

$ = 142$

Given ${{\sum\limits_{i = 1}^{10} {{x_i}} } \over {10}} = 15$ ..... (1)

$ \Rightarrow \sum\limits_{i = 1}^{10} {{x_i} = 150} $

and ${{\sum\limits_{i = 1}^{10} {x_i^2} } \over {10}} - {15^2} = 15$

$ \Rightarrow \sum\limits_{i = 1}^{10} {x_i^2 = 2400} $

Replacing 25 by 15 we get

$\sum\limits_{i = 1}^9 {{x_i} + 25 = 150} $

$ \Rightarrow \sum\limits_{i = 1}^9 {{x_i} = 125} $

$\therefore$ Correct mean $ = {{\sum\limits_{i = 1}^9 {{x_i} + 15} } \over {10}} = {{125 + 15} \over {10}} = 14$

Similarly, $\sum\limits_{i = 1}^2 {x_i^2 = 2400 - {{25}^2} = 1775} $

$\therefore$ Correct variance $ = {{\sum\limits_{i = 1}^9 {x_i^2 + {{15}^2}} } \over {10}} - {14^2}$

$ = {{1775 + 225} \over {10}} - {14^2} = 4$

$\therefore$ Correct $S.D = \sqrt 4 = 2$.

$\mu = {{\sum {{x_i}} } \over {40}} = 30 \Rightarrow \sum {{x_i} = 1200} $

${\sigma ^2} = {{\sum {x_i^2} } \over {40}} - {(30)^2} = 25 \Rightarrow \sum {x_i^2 = 37000} $

After omitting two wrong observations

$\sum {{y_i} = 1200 - 12 - 10 = 1178} $

$\sum {y_i^2 = 37000 - 144 - 100 = 36756} $

Now ${\sigma ^2} = {{\sum {y_i^2} } \over {38}} - {\left( {{{\sum {{y_i}} } \over {38}}} \right)^2}$

$ = {{36756} \over {38}} - {\left( {{{1178} \over {38}}} \right)^2} = - {31^2}$

$ = 38{\sigma ^2} = 36756 - 36518 = 238$

According to given data

${{\sum\limits_{i = 1}^7 {{{({x_i} - 62)}^2}} } \over 7} = 20$

$ \Rightarrow \sum\limits_{i = 1}^7 {{{({x_i} - 62)}^2} = 140} $

So for any x_i, ${({x_i} - 62)^2} \le 140$

$ \Rightarrow {x_i} > 50\,\forall i = 1,2,3,\,\,.....\,\,7$

So no student is going to score less than 50.

${{\sum {x_i^2} } \over {15}} - {8^2} = 9 \Rightarrow \sum {x_i^2 = 15 \times 73 = 1095} $

Let ${\overline x _c}$ be corrected mean ${\overline x _c}$ = 9

$\sum {x_c^2 = 1095 - 25 + 400 = 1470} $

Correct variance $ = {{1470} \over {15}} - {(9)^2} = 98 - 81 = 17$

Mean $ = {{n{{(n + 1)} \over 2}} \over n} = {{n + 1} \over 2}$

M.D. $ = {{2\left( {{{n - 1} \over 2} + {{n - 3} \over 2} + {{n - 5} \over 2} + \,\,\,...\,\,\,0} \right)} \over n} = {{5(n + 1)} \over n}$

$ \Rightarrow ((n - 1) + (n - 3) + (n - 5) + \,\,...\,\,0) = 5(n + 1)$

$ \Rightarrow \left( {{{n + 1} \over 4}} \right)\,.\,(n - 1) = 5(n + 1)$

So, $n = 21$

7 $\times$ 8, 10 $\times$ 10, 13 $\times$ 12, 16 $\times$ 14 ........

T_n = (3n + 4) (2n + 6) = 2(3n + 4) (n + 3)

= 2(3n² + 13n + 12) = 6n² + 26n + 24

S₁₀ = $\sum\limits_{n = 1}^{10} {{T_n}} = 6\sum\limits_{n = 1}^{10} {{n^2}} + 26\sum\limits_{n = 1}^{10} n + 24\sum\limits_{n = 1}^{10} 1 $

$ = {{6(10 \times 11 \times 21)} \over 6} + 26 \times {{10 \times 11} \over 2} + 24 \times 10$

$ = 10 \times 11(21 + 13) + 240$

= 3980

Mean $ = {{{S_{10}}} \over {10}} = {{3980} \over {10}} = 398$

$\sigma _b^2$ = 2 (variance of boys)

n₁ = no. of boys

${\overline x _b}$ = 12

n₂ = no. of girls

$\sigma _g^2$ = 2

${\overline x _g}$ = ${{50 \times 15 - 12 \times {\sigma _b}} \over {30}} = {{750 - 12 \times 20} \over {30}} = 17 = \mu $

variance of combined series

$\sigma _{}^2 = {{{n_1}\sigma _b^2 + {n_2}\sigma _g^2} \over {{n_1} + {n_2}}} + {{{n_1}.\,{n_2}} \over {{{({n_1} + {n_2})}^2}}}{\left( {{{\overline x }_b} - {{\overline x }_g}} \right)^2}$

$\sigma _{}^2 = {{20 \times 2 + 30 \times 2} \over {20 + 30}} + {{20 \times 30} \over {{{(20 + 30)}^2}}}{(12 - 17)^2}$

$\sigma$² = 8

$\Rightarrow$ $\mu$ + $\sigma$² = 17 + 8 = 25

${{{n^2} - 1} \over {12}} = 14 \Rightarrow n = 13$

$5 = {{3 + 7 + x + y} \over 4} \Rightarrow x + y = 10$

Var(x) = $10 = {{{3^2} + {7^2} + {x^2} + {y^2}} \over 4} - 25$

$140 = 49 + 9 + {x^2} + {y^2}$

${x^2} + {y^2} = 82$

x + y = 10

$\Rightarrow$ (x, y) = (9, 1)

Four numbers are 21, 9, 10, 8

Mean = ${{48} \over 4}$ = 12

$\because$ Sum of frequencies = 584

$\Rightarrow$ $\alpha$ + $\beta$ = 390

Now, median is at ${{584} \over 2}$ = 292^th

$\because$ Median = 45 (lies in class 40 - 50)

$\Rightarrow$ $\alpha$ + 110 + 54 + 15 = 292

$\Rightarrow$ $\alpha$ = 113, $\beta$ = 277

$\Rightarrow$ | $\alpha$ $-$ $\beta$ | = 164

Class	Frequency	${x_i}$	${f_i}{x_i}$
0-6	a	3	3a
6-12	b	9	9b
12-18	12	15	180
18-24	9	21	189
24-30	5	27	135
	$N = (26 + a + b)$		$(504 + 3a + 9b)$

Mean = ${{3a + 9b + 180 + 189 + 135} \over {a + b + 26}} = {{309} \over {22}}$

$ \Rightarrow 66a + 198b + 11088 = 309a + 309b + 8034$

$ \Rightarrow 243a + 111b = 3054$

$ \Rightarrow 81a + 37b = 1018$ $\to$ (1)

Now, Median $ = 12 + {{{{a + b + c} \over 2} - (a + b)} \over {12}} \times 6 = 14$

$ \Rightarrow {{13} \over 2} - \left( {{{a + b} \over 4}} \right) = 2$

$ \Rightarrow {{a + b} \over 4} = {9 \over 2}$

$ \Rightarrow a + b = 18$ $\to$ (2)

From equation (1) $ (2)

a = 8, b = 10

$\therefore$ ${(a - b)^2} = {(8 - 10)^2}$ = 4

Mean $\left( {\overline x } \right)$ = ${{{x_1} + {x_2}..... + {x_n}} \over n}$ = ${{\sum x } \over n}$

Here, Mean = 40 of 25 teachers

$\therefore$ 40 = ${{\sum x } \over {25}}$

$ \Rightarrow $ $\sum x $ = 40 $ \times $ 25 = 1000

After retireing of a 60 year old teacher, total age of 24 teachers,

x₁ + x₂ + . . . . . .x₂₄ = 1000 $-$ 60 = 940

Now a new teacher of age A year is appointed.

$\therefore$ Now total age of this 25 teachers

x₁ + x₂ + x₃ + . . . . . + x₂₅ = 940 + A

$\therefore$ Mean age = ${{940 + A} \over {25}}$

According to question,

${{940 + A} \over {25}}$ = 39

$ \Rightarrow $ A = 35

Let first 2n observations are x₁, x₂ ...................., x_2n

and last n observations are y₁, y₂ ....................., y_n

Now, ${{\sum {{x_i}} } \over {2n}} = 6$, ${{\sum {{y_i}} } \over n} = 3$

$ \Rightarrow \sum {{x_i}} = 12n,\sum {{y_i}} = 3n$

$ \therefore $ ${{\sum {{x_i}} + \sum {{y_i}} } \over {3n}} = {{15n} \over {3n}} = 5$

Now, ${{\sum {x_i^2} + \sum {y_i^2} } \over {3n}} - {5^2} = 4$

$ \Rightarrow \sum {x_i^2} + \sum {y_i^2} = 29 \times 3n = 87n$

Now, mean is ${{\sum {({x_i} + 1) + \sum {({y_i} - 1)} } } \over {3n}} = {{15n + 2n - n} \over {3n}} = {{16} \over 3}$

Now, variance is ${{{{\sum {{{({x_i} + 1)}^2} + \sum {({y_i} - 1)} } }^2}} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$

$ = {{\sum {x_i^2 + \sum {y_i^2} + 2\left( {\sum {{x_i}} - \sum {{y_i}} } \right) + 3n} } \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$

$ = {{87n + 2(9n) + 3n} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$

= $29 + 6 + 1 - {\left( {{{16} \over 3}} \right)^2}$

$ = {{324 - 256} \over 9} = {{68} \over 9} = k$

$ \Rightarrow $ 9k = 68

Therefore, the correct answer is 68.

For group - 1 : ${{\sum {{x_i}} } \over {10}} = 2 \Rightarrow \sum {{x_i}} = 20$

${{\sum {{x_i^2}} } \over {10}} - {(2)^2} = 2 \Rightarrow \sum {x_i^2} = 60$

For group - 2 : ${{\sum {{y_i}} } \over n} = 3 \Rightarrow \sum {{y_i}} = 3n$

${{\sum {y_i^2} } \over n} - {3^2} = 1 \Rightarrow \sum {y_i^2} = 10n$

Now, combined variance

${\sigma ^2} = {{\sum {\left( {x_i^2 + y_i^2} \right)} } \over {10 + n}} - {\left( {{{\sum {\left( {{x_i} + {y_i}} \right)} } \over {10 + n}}} \right)^2}$

$ \Rightarrow {{17} \over 9} = {{60 + 10n} \over {10 + n}} - {{{{(20 + 3n)}^2}} \over {{{(10 + n)}^2}}}$

$ \Rightarrow $ 17 (n² + 20n + 100) = 9(n² + 40n + 200)

$ \Rightarrow $ 8n² $-$ 20n $-$ 100 = 0

$ \Rightarrow $ 2n² $-$ 5n $-$ 25 = 0 $ \Rightarrow $ n = 5

Given, $\sum\limits_{i = 1}^{18} {({x_1} - \alpha ) = 36} $

$ \Rightarrow \sum {{x_i} - 18\alpha = 36} $

$ \Rightarrow \sum {{x_i} = 18(\alpha + 2)} $ .... (1)

Also, $\sum\limits_{i = 1}^{18} {{{({x_1} - \beta )}^2} = 90} $

$ \Rightarrow \sum {x_i^2 + 18{\beta ^2} - 2\beta \sum {{x_i} = 90} } $

$ \Rightarrow \sum {x_i^2 + 18{\beta ^2} + 2\beta \times 18(\alpha + 2) = 90} $ (using equation (1))

$ \Rightarrow \sum {x_i^2 = 90} - 18{\beta ^2} + 36\beta (\alpha + 2)$

Given, ${\sigma ^2} = 1 \Rightarrow {1 \over {18}}{\sum {x_i^2 - \left( {{{\sum {{x_i}} } \over {18}}} \right)} ^2} = 1$

$ = {1 \over {18}}(90 - 18{\beta ^2} + 36\alpha \beta + 72\beta ) - {\left( {{{18(\alpha + 2)} \over {18}}} \right)^2} = 1$

$ \Rightarrow 90 - 18{\beta ^2} + 36\alpha \beta + 72\beta - 18{(\alpha + 2)^2} = 18$

$ \Rightarrow 5 - {\beta ^2} + 2\alpha \beta + 4\beta - {(\alpha + 2)^2} = 1$

$ \Rightarrow 5 - {\beta ^2} + 2\alpha \beta + 4\beta - {\alpha ^2} - 4 - 4\alpha = 1$

$ \Rightarrow {\alpha ^2} - {\beta ^2} + 2\alpha \beta + 4\beta - 4\alpha = 0$

$ \Rightarrow (\alpha - \beta )(\alpha - \beta + 4) = 0$

$ \Rightarrow \alpha - \beta = - 4$

$ \therefore $ $|\alpha - \beta |\, = 4$ $(\alpha \ne \beta )$

${\sigma ^2} = {{\sum {{x^2}} } \over n} - {\left( {{{\sum x } \over n}} \right)^2}$

${\sigma ^2} = {{(9 + {k^2})} \over {10}} - {\left( {{{9 + k} \over {10}}} \right)^2} < 10$

$(90 + {k^2})10 - (81 + {k^2} + 8k) < 1000$

$90 + 10{k^2} - {k^2} - 18k - 81 < 1000$

$9{k^2} - 18k + 9 < 1000$

${(k - 1)^2} < {{1000} \over 9} \Rightarrow k - 1 < {{10\sqrt {10} } \over 3}$

$k < {{10\sqrt {10} } \over 3} + 1$

k $ \le $ 11

Maximum integral value of k = 11.

Mean = ${{\sum {{x_1}.{f_1}} } \over {\sum {{f_1}} }}$

= ${{0.{}^n{C_0} + 2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}} \over {{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}}$

We know,

(1 + x)ⁿ = ${{}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}}$ ...(1)

Put x = 2, at (1) we get
$ \Rightarrow $ 3ⁿ - 1 = ${2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}}$

And Putting x = 1 in (1), we get

2ⁿ = ${{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}$

$ \therefore $ Mean = ${{{3^n} - 1} \over {{2^n}}}$

According to question,

${{{3^n} - 1} \over {{2^n}}}$ = ${{728} \over {{2^n}}}$

$ \Rightarrow $ 3ⁿ = 729

$ \Rightarrow $ n = 6

x_i = midpoint of class interval


Variance($\sigma^{2})$$ =\frac{\sum f_{i}\left( x_{i}-\bar{x} \right)^{2} }{\sum f_{i}} $

Also, $\bar{x} =\frac{\sum f_{i}x_{i}}{\sum f_{i}} $

= $\frac{30+25x+70}{2+2+x} $ = 25

Given, Variance = 50

$ \therefore $ 50 = $\frac{200+0+200}{2+2+x} $

$ \Rightarrow $ x = 4

Let the common difference = d

and ${b_1} = a$
${b_2} = a + d$
${b_3} = a + 2d$
... ${b_{11}} = a + 10d$

Variance = ${{\sum {a_i^2} } \over {11}} - {\left( {{{\sum {{a_i}} } \over {11}}} \right)^2} = 90$

$ \Rightarrow {{{a^2} + {{\left( {a + d} \right)}^2} + ... + {{\left( {a + 10d} \right)}^2}} \over {11}} - {\left( {{{a + \left( {a + d} \right) + ... + \left( {a + 10d} \right)} \over {11}}} \right)^2} = 90$

$ \Rightarrow 11\left[ {11{a^2} + 385{d^2} + 110ad} \right] - {\left[ {11a + 55d} \right]^2} = 10890$

$ \Rightarrow 1210{d^2} = 10890$

$ \Rightarrow {d^2} = 9$

$ \Rightarrow d = \pm 3$

As A.P is increasing so d should be positive

$ \therefore $ d = 3

Mean = ${{3 + 7 + 9 + 12 + 13 + 20 + x + y} \over 8}$ = 10

16 = x + y ....(1)

Variance (${\sigma ^2}$) = 25

$ \Rightarrow $ ${{{3^2} + {7^2} + {9^2} + {{12}^2} + {{13}^2} + {{20}^2} + {x^2} + {y^2}} \over 8}$ - 100 = 25

$ \Rightarrow $ 125 × 8 = 9 + 49 + 81 + 144 + 169 + 400 + x² + y² - 800

$ \Rightarrow $ x² + y² = 148

We know, (x + y)² = x² + y² + 2xy

$ \Rightarrow $ 256 = 148 + 2xy

$ \Rightarrow $ x.y = 54

Variance ${\sigma ^2} = {{\sum {x_i^2} } \over N} - {\mu ^2}$

variance of (1, 2, ….. n)

10 = ${{{1^2} + {2^2} + .... + {n^2}} \over n} - {\left( {{{1 + 2 + 3 + .... + n} \over n}} \right)^2}$

on solving we get n = 11

variance of 2, 4, 6…….2m = 16

$ \Rightarrow $ ${{{2^2} + {4^2} + .... + {{\left( {2m} \right)}^2}} \over m} - {\left( {m + 1} \right)^2}$ = 16

$ \Rightarrow $ m² = 49

$ \Rightarrow $ m = 7

$ \therefore $ m + n = 18