Maths M.C.Q (1 Marks)

$ \Rightarrow \bar x = \frac{{\sum {{x_i}} }}{n} = 9$

$ \Rightarrow \sum {{x_i}}  = 9 \times 5 = 45$

Now, standard deviation $=0$

$\therefore $ all the five terms are same i.e.;$9$.

Now for changed observation 

${{\bar x}_{new}} = \frac{{36 + {x_5}}}{5} = 10$

$ \Rightarrow {x_5} = 14$

$\therefore {\sigma _{new}} = \sqrt {\frac{{\sum {{{\left( {{x_i} - {{\bar x}_{new}}} \right)}^2}} }}{n}} $

$ = \sqrt {\frac{{4{{\left( {9 - 10} \right)}^2} + {{\left( {14 - 10} \right)}^2}}}{5}} $

$ = 2$

$ \Rightarrow {x_1} + {x_2} + ... + {x_{25}} = 1000$ 

Let $A$ be the age of new teacher.

$\therefore {x_1} + {x_2} + ... + {x_{25}} - 60 + A = 39 \times 25$

$ \Rightarrow 1000 - 60 + A = 975$

$ \Rightarrow A = 975 - 940 = 35$

Variance

${\sigma ^2} = \frac{{\sum {x_i^2} }}{N} - {\left( {\frac{{\sum {{x_i}} }}{N}} \right)^2}$

$ = \frac{{2475}}{{97}} - {\left( {\frac{{388}}{{97}}} \right)^2}$

$ = \frac{{2425 - 1552}}{{97}} = \frac{{873}}{{97}} = 9$

$\frac{49}{4}=\frac{4+9+a^{2}+121}{4}-\left(\frac{16+a}{4}\right)^{2}$

$3 a^{2}-32 a+84=0$

$ = \frac{1}{{101}} \times \frac{{101}}{2}\left[ {1 + \left( {1 + 100d} \right)} \right] = 1 + 50d$

mean deviation from mean

$ = \frac{1}{{101}}[\left| {1 - \left( {1 + 50d} \right)} \right.\left|  +  \right|\left( {1 + d} \right) - $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 + 50d} \right)\left| {......} \right|\,\left. {\left[ {1 + 100d} \right] - \left( {1 + 50d} \right)} \right|]$

$ = \frac{{2\left| d \right|}}{{101}}\left( {1 + 2 + 3... + 50} \right)$

$ = \frac{{2\left| d \right|}}{{101}} \times \frac{{50 \times 51}}{2} = \frac{{2550}}{{101}}\left| d \right|$

$ = \frac{{2550}}{{101}}\left| d \right| = 225 \Rightarrow \left| d \right| = 10.1$

$\bar x = 5$

variance $=124$

${x_1} = 1,{x_2} = 2,{x_3} = 6$

$\bar x = 5$

$\frac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}}}{5} = 5$

$ \Rightarrow {x_4} + {x_5} + 9 = 25$

$ \Rightarrow {x_4} + {x_5} = 16$

$ \Rightarrow {x_4} + {x_5} + 10 - 10 = 16$

$ \Rightarrow \left( {{x_4} - 5} \right) + \left( {{x_5} - 5} \right) = 16 - 10$

$ \Rightarrow \left( {{x_4} - 5} \right) + \left( {{x_5} - 5} \right) = 6$

Mean deviation $ = \frac{{\sum {\left| {{x_i} - \bar x} \right|} }}{N}$

$ = \left| {{x_1} - 5} \right| + \left| {{x_2} - 5} \right| + \left| {{x_3} - 5} \right| + \left| {{x_4} - 5} \right| + \left| {{x_5} - 5} \right|$

$ = \frac{{4 + 3 + 1 + 6}}{5} = \frac{{14}}{5} = 2.8$

$ \Rightarrow \sum\limits_{i = 1}^{16} {{x_i}}  = 16 \times 16$

Sum of new observations

$ = \sum\limits_{i = 1}^{18} {{y_i}}  = (16 \times 16 - 16) + (3 + 4 + 5) = 252$

Number of observations $=18$

$\therefore $ New mean $ = \frac{{\sum\limits_{i = 1}^{18} {{y_i}} }}{{18}} = \frac{{252}}{{18}} = 14$

$70 \times  54 + 30 \times  x = 60 \times100$

$x =74$

${\sigma ^2} = \frac{{\sum x_1^2}}{n} - {\left( {\overline {.x} } \right)^2}$

$\frac{{{2^2} + {4^2} + {6^2} + .... + {{100}^2}}}{{50}}$$ - {\left( {\frac{{2 + 4 + 6 + .... + 100}}{{50}}} \right)^2}$

${i_1} = \frac{{{2^2} + {4^2} + {6^2} + .... + {{100}^2}}}{{50}}$

$ = {2^2}\frac{{{1^2} + {2^2} + {3^2} + ... + {{50}^2}}}{{50}}$

$ = \frac{{{2^2}}}{{50}} \times 50\left( {50 + 1} \right)\left( {100 + 1} \right)$

$ = 3434$

${i_2} = {\left( {\frac{{2 + 4 + 6 + ..... + 100}}{{50}}} \right)^2}$

$ = {\left( {\frac{{50 \times \frac{{2 + 100}}{2}}}{{50}}} \right)^2}$

$ = {\left( {51} \right)^2}$

${\sigma ^2} = 3434 - 2661 = 833$

So, maen $ = \sum\limits_{i = 1}^{2n} {\frac{{{x_i}}}{{2n}}} $

Let these observations be divided into two parts ${{x_1},{x_2},......,{x_n}}$ and ${x_{n + 1}},......{x_{2n}}$ 

Each in ${1^{st}}$ part $5$ is added, so total of 

first part is $\sum\limits_{i = 1}^n {{x_i} + 5n} $.

In second part $3$ is subtracted from each 

So, total os second part is $\sum\limits_{i = n + 1}^{2n} {{x_i} - 3n} $

Total of $2n$ terms are

$\sum\limits_{i = 1}^n {{x_i} + 5n}  + \sum\limits_{i = n + 1}^{2n} {{x_i} - 3n}  = \sum\limits_{i = 1}^{2n} {{x_i} + 2n} $

 Mean $ = \sum\limits_{i = 1}^{2n} {\frac{{{x_i} + 2n}}{{2n}}}  = \sum\limits_{i = 1}^{2n} {\frac{{{x_i}}}{{2n}} + 1} $

Let ${\bar X}$ be the mean and $M.D.$be the mean deviation about ${\bar X}$.

If each observation is incerased by $5$

then new mean will be $\bar X + 5$ and new $M.D.$ about new mean wil be $M.D.$

($\because $ Mean $ = \sum\limits_{i = 1}^n {\frac{{{x_i}}}{n}} $)

${\sigma ^2} = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \bar x} \right)}^2}} $

Mean of ${d_1},{d_2},{d_3},......,{d_n}$

$ = \frac{{{d_1} + {d_2} + {d_3} + ...... + {d_n}}}{n}$

$ = \frac{{\left( { - {x_1} - a} \right) + \left( { - {x_2} - a} \right) + \left( { - {x_3} - a} \right) + ..... + \left( { - {x_n} - a} \right)}}{n}$

$ =  - \left[ {\frac{{{x_1} + {x_2} + {x_3} + ... + {x_n}}}{n}} \right] - \frac{{na}}{n}$

$ =  - \bar x - a$

Since, ${d_i} =  - {x_i} - a$ and we multiply or subtract each observation by any number the mode remains the same. Hence mode of $ - {x_i} - a$ i.e. ${d_i}$ and ${x_i}$ are same.

Now variance of ${d_1},{d_2},......,{d_n}$

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left[ {{d_i} - \left( { - \bar x - a} \right)} \right]}^2}} $

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left[ { - {x_i} - a + \bar x - a} \right]}^2}} $

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( { - {x_i} + \bar x} \right)}^2}} $

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\bar x - {x_i}} \right)}^2}}  = {\sigma ^2}$

Given mean $=7$

$\therefore 7 = \frac{{6 + 7 + 8 + 10 + x}}{5}$

$ \Rightarrow x = 4$ Now, Variance 

$ = \sqrt {\frac{{{{\left( {6 - 7} \right)}^2} + {{\left( {7 - 7} \right)}^2} + {{\left( {8 - 7} \right)}^2} + {{\left( {10 - 7} \right)}^2} + {{\left( {4 - 7} \right)}^2}}}{5}} $

$ = \sqrt {\frac{{{1^2} + {0^2} + {1^2} + {3^2} + {3^2}}}{5}}  = \sqrt {\frac{{20}}{5}}  = \sqrt 4  = 2$

$\sigma  = \sqrt {\frac{{\sum \left( {x - \overline {.x} } \right)}}{n}} $

$ = \sqrt {\frac{{\sum \left( {y - 10 - \overline y  + 10} \right)}}{n}} $

$ = \sqrt {\frac{{\sum \left( {y - \overline y } \right)}}{n}} $

variance does not change

Nearest option : $(a) 41$

$\therefore y > \frac{x}{2} \Rightarrow x - y < \frac{x}{2}$

$\therefore x - y < y < x < 2x + y$

Hence median $ = \frac{{y + x}}{2} = 10$

$ \Rightarrow x + y = 20\,\,\,\,\,\,\,\,....\left( i \right)$

And range $ = \left( {2x + y} \right) - \left( {x - y} \right) = x + 2y$

But range $=28$

$\therefore x + 2y = 28\,\,\,\,\,\,\,.....\left( {ii} \right)$

From equation $(i)$ and $(ii)$,

$x = 12,y = 8$

$\therefore $ Mean

$ = \frac{{\left( {x - y} \right) + y + x + \left( {2x + y} \right)}}{4} = \frac{{4x + y}}{4}$

         $ = x + \frac{y}{4} = 12 + \frac{8}{4} = 14$

Now, standard deviation $\sigma  = \sqrt {\frac{{\sum {{{\left( {x - A} \right)}^2}} }}{{2n}}} $

$2 = \sqrt {\frac{{{{\left( {a - 0} \right)}^2} + {{\left( {a - 0} \right)}^2} + ... + {{\left( {0 - a} \right)}^2} + ...}}{{2n}}} $

          $ = \sqrt {\frac{{{a^2}.2n}}{{2n}}}  = \left| a \right|$

Hence, $\left| a \right| = 2$

$\sum_{ x =0}^4( mx + c )=1 \Rightarrow 10 m +5 c =1 \Rightarrow 2 m + c =\frac{1}{5}$   $. . . (1)$

$\text { mean }=\sum x _{ i } P _{ i }=\sum_{ i =0}^4\left( mx _{ i }+ c \right) x _{ i }=30 m +10 c =\frac{5}{2}$

$\therefore 3 m + c =\frac{1}{4} \ldots(2)$

$\text { from (1) and (2) m= } \frac{1}{20}, c =\frac{1}{10}$

$\sum P _{ i } x _{ i }^2=\sum_{ i =0}^4\left( mx _{ i }+ c \right) x _1^2$

$=\sum_{ i =0}^4\left( mx _{ i }^3+ cx _{ i }^2\right) \Rightarrow 100 m +30 c (\text { Now putting } m \text { and } c )$

$\Rightarrow \Sigma P _{ i }^2=5+3=8$

$\text { Variance }=\Sigma P _{ i } x _{ i }^2-\left(\Sigma P _{ i } x _{ i }\right)^2=8-\left(\frac{5}{2}\right)^2=\frac{7}{4}$

$\therefore 24 \alpha=42$

$\mathrm{f}_{\mathrm{i}}$ $\ \ 5 \ \ 4 \ \ 4 \ \ 2 \ \ 2 \ \ 3$

($P$) Mean

($Q$) Median

($R$) Mean deviation about mean

($S$) Mean deviation about median

($P$) Mean $=\frac{\Sigma x_i f_i}{\Sigma f_i}=\frac{120}{20}=6$

($Q$) Median $=\left(\frac{20}{2}\right)^{\text {th }}$ observation $=10^{\text {th }}$ observation $=5$

($R$) Mean deviation about mean $=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mid \mathrm{x}_{\mathrm{i}}-\text { Mean } \mid}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{54}{20}=2.70$

($S$) mean deviation about median $=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mid \mathrm{x}_{\mathrm{i}}-\text { Median } \mid}{\Sigma \mathrm{f}_{\mathrm{i}}}=\frac{48}{20}=2.40$

==> $30 = 24 + x$

==> $x = 6$.

New mean = $\frac{{\Sigma \lambda {x_i}}}{n}$

$ = \lambda \frac{{\Sigma {x_i}}}{n}$$ = \lambda \bar x$.

Then, $\bar x = \frac{1}{n}\Sigma {x_i}$ let ${y_i} = \frac{{{x_i}}}{\alpha } + 10$

then, $\frac{1}{n}\sum\limits_{i = 1}^n {{y_i}} = \frac{1}{\alpha }$ $\left( {\frac{1}{n}\Sigma {x_i}} \right) + \frac{1}{n}(10n)$

==> $\bar y = \frac{1}{\alpha }\bar x + 10$

$ = \frac{{\bar x + 10\alpha }}{\alpha }$.

$ = \frac{{a({x_1} + {x_2} + ..... + {x_n}) + nb}}{n} $

$= a\bar x + b$

Then, $M = \frac{{a + 4{{\bar x}_1}}}{{(n - 4) + 4}}$

==> $\overline {{x_1}} = \frac{{nM - a}}{4}$.

If $ n$ is odd then,

Median = value of ${\left( {\frac{{n + 1}}{2}} \right)^{th}}$ term

Median $ = {\left( {\frac{{7 + 1}}{2}} \right)^{th}}$term

$ = {4^{th}}$ term $ = 10$.

$\alpha - \frac{7}{2},\alpha - 3,\alpha - \frac{5}{2},\alpha - 2,\alpha - \frac{1}{2},\alpha + \frac{1}{2},\alpha + 4,\alpha + 5$

Median $ = \frac{1}{2}[{\rm{value\ of\ }}{{\rm{4}}^{{\rm{th}}}}{\rm{\ item}} + {\rm{value \ of\ }}{{\rm{5}}^{{\rm{th}}}}{\rm{\ item]}}$

Median $ = \frac{{\alpha - 2 + \alpha - \frac{1}{2}}}{2}$

$ = \frac{{2\alpha - \frac{5}{2}}}{2}$= $\alpha - \frac{5}{4}$.

Mode = $3$ median -$2$ mean

==> $6\lambda $ = $3$ median -$18$ $\lambda $

==> median = $8\ \lambda $.

i.e., $\mathop {\underline {\begin{array}{*{20}{c}}{nM}\\{nM - {x_n}}\\{nM - {x_n} + x'}\end{array}} }\limits_n \begin{array}{*{20}{c}} = \\ = \\ = \end{array}\mathop {\underline {\begin{array}{*{20}{c}}{{x_1} + {x_2} + {x_3} + ......{x_{n - 1}} + {x_n}}\\{{x_1} + {x_2} + {x_3} + ......{x_{n - 1}}\;\;\;\;\;}\\{{x_1} + {x_2} + {x_3} + ......{x_{n - 1}} + x'}\end{array}} }\limits_n $

New average $ = \frac{{nM - {x_n} + x'}}{n}$.

Mean = ($2$ mean) $k$

==> $k = \frac{1}{2}$,

[ Mode = $3$ median -$2$ mean].

Mode = $3$ median -$2$ mean

==> $7 = 3$ median -$2× 4$

==> $15 = 3$ median

 Median = $5$.

Hence $ M.D.$ (about mean) $ = \frac{{| - 1 - 1| + |0 - 1| + |4 - 1|}}{3} = 2$.

In this case median is $\frac{{101 + 1}}{2}^{th}$

$i.e.$, $51^{st}$ item

$i.e.$, ${x_{51}}$.

We have $34, 38, 42, 44, 46, 48, 54, 55, 63, 70$

Here, median = $M$ = $\frac{{46 + 48}}{2} = 47$

$(\because n = 10,$ median is the mean of $5^{th}$ and $6^{th}$ items)

$\therefore $ Mean deviation $ = \frac{{\Sigma |{x_i} - M|}}{n}$$ = \frac{{\Sigma |{x_i} - 47|}}{{10}}$

$ = \frac{{13 + 9 + 5 + 3 + 1 + 1 + 7 + 8 + 16 + 23}}{{10}} = 8.6$.

 ${\rm{S}}{\rm{.D}}{\rm{.}} = \frac{3}{2} \times {\rm{Q}}{\rm{.D}}{\rm{.}}$

$ = \frac{3}{2} \times 10$

==> ${\rm{S}}{\rm{.D}}{\rm{.}} = 15$.

$ = \frac{{19.76}}{{35.16}} \times 100$.

Variane of $a x_1 a x_2, \ldots a x_n=$ ?

varience $=\sigma^2=\frac{1}{n} \sum \limits_{i=1}^r y_i\left(n_i-\bar{x}\right)^2$

If each obs is weltiplied $2 y$ a the $y_i=a x_i \quad i . e \quad x_i=\frac{1}{a} y_i$

$y_i=a x_i n$

$\therefore \bar{y}=\frac{1}{n} \sum\limits_{i=1}^n y_i=\frac{1}{n} \sum\limits_{i=1}^n a x_i=\frac{a}{n} \sum\limits_{i=1}^n x_i=a \bar{x} .$

${\left[\because \bar{x}=\frac{1}{n} \sum\limits_{i=1}^n x_i\right]}$

$(1) \Rightarrow \quad \sigma^2=\frac{1}{A} \sum_{i=1}^n\left(\frac{1}{a} y_i-\frac{1}{a} \bar{y}\right)^2$

$\Rightarrow a a^2 \sigma^2=\frac{1}{n} \sum_{i=1}^n\left(y_i-\bar{y}\right)^2$

These varieme of new obs' is $a^2 \sigma^2$

$(\sigma ) = \sqrt {\frac{{250}}{{10}}} = \sqrt {25} = 5$

Hence, coefficient of variation $ = \frac{\sigma }{{{\rm{mean}}}} \times 100$

$ = \frac{5}{{50}} \times 100 = 10\%$.

All negative terms $\underbrace {\,\,O\,\,}_{{{(n + 1)}^{th}}\ term}$ All positive terms

The median of given observations $ = {(n + 1)^{th}}$ term = $0$

$ S. D. > M .D.$

i.e., $y = Ax + B$, where $A = \frac{a}{c}$,$B = \frac{b}{c}$

 $\bar y = A\bar x + B$

 $y - \bar y = A(x - \bar x)$ ==> ${(y - \bar y)^2} = {A^2}{(x - \bar x)^2}$

==> $\sum {(y - \bar y)^2} = {A^2}\sum {(x - \bar x)^2}$

==> $n.\sigma _y^2 = {A^2}.n\sigma _x^2$ ==> $\sigma _y^2 = {A^2}\sigma _x^2$

==> ${\sigma _y} = \,|A|{\sigma _x}$ ==> ${\sigma _y} = \,\left| {\frac{a}{c}} \right|{\sigma _x}$

Thus, new $S.D$. $ = \left| {\frac{a}{c}} \right|\,\sigma $.

$ = \sqrt {\frac{{n(n + 1)(2n + 1)}}{{6n}} - {{\left[ {\frac{{n(n + 1)}}{{2n}}} \right]}^2}} $

$ = \sqrt {\frac{{(n + 1)(2n + 1)}}{6} - {{\left( {\frac{{n + 1}}{2}} \right)}^2}} = \sqrt {\frac{{n + 1}}{2}\left( {\frac{{2n + 1}}{3} - \frac{{n + 1}}{2}} \right)} $

$ = \sqrt {\frac{{n + 1}}{2}\left( {\frac{{4n + 2 - 3n - 3}}{6}} \right)} $

$ = \sqrt {\frac{{{n^2} - 1}}{{12}}} $.

Then,${\rm{S}}{\rm{.D}}{\rm{.}} = \sqrt {\frac{1}{N}\sum\limits_{i = 1}^n {{f_i}{{({x_i} - \bar x)}^2}} } $

and ${\rm{M}}{\rm{.D}}{\rm{.}} = \frac{1}{N}\sum\limits_{i = 1}^n {{f_i}|{x_i}} - \bar x|$

Let $|{x_i} - \bar x| = {z_i};i = 1,2,.....n$ .

Then,

$(S.D.)2 -(M.D.)2$ $ = \frac{1}{N}\sum\limits_{i = 1}^n {{f_i}z_i^2 - {{\left( {\frac{1}{N}\sum\limits_{i = 1}^n {{f_i}{z_i}} } \right)}^2}} $

$ = \sigma _z^2 \ge 0$==> S. D. $ \ge $ $M.D.$

Let ${y_1} = {x_1} + 1,\;{y_2} = {x_2} + 2,\;{y_3} = {x_3} + 3,..,{y_n} = {x_n} + n$

Then the mean of the new series is $\frac{1}{n}\Sigma {y_i} = \frac{1}{n}\sum\limits_{i = 1}^n {({x_i} + i)} $

$ = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} + \frac{1}{n}(1 + 2 + 3 + ..... + n)$

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

$ = \bar x + \frac{{n + 1}}{2}$.

$\bar x = \frac{{0.1 + {{1.}^n}{C_1} + {{2.}^n}{C_2} + {{3.}^n}{C_3} + ...... + n{.^n}{C_n}}}{{1{ + ^n}{C_1}{ + ^n}{C_2} + ....{ + ^n}{C_n}}}$

$ = \frac{{\sum\limits_{r = 0}^n {r.\,{\,^{n}}{C_r}} }}{{\sum\limits_{r = 0}^n {^n{C_r}} }} = \frac{{\sum\limits_{r = 1}^n {r.\frac{n}{r}\,{\,^{n - 1}}{C_{r - 1}}} }}{{\sum\limits_{r = 0}^n {^n{C_r}} }}$= $\frac{{n\sum\limits_{r = 1}^n {^{n - 1}{C_{r - 1}}} }}{{\sum\limits_{r = 0}^n {^n{C_r}} }}$

$ = \frac{{n{{.2}^{n - 1}}}}{{{2^n}}}$ $ = \frac{n}{2}$.

Here, $N = \sum\limits_{i = 1}^6 {{f_i}}  = 100,\sum\limits_{i = 1}^6 {{f_i}{x_i}}  = 12530$

$\therefore \bar x = \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}{x_i}}  = \frac{1}{{100}} \times 12530 = 125.3$

$M.D.\left( {\bar x} \right) = \frac{1}{N}\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - \bar x} \right|}  = \frac{1}{{100}} \times 1128.8 = 11.28$