CBSE Class 11 Maths Important Questions - Chapter 13 Statistics

1 Mark Questions

1.In a test with a maximum marks  $ 25 $ , eleven students scored $ 3,\,\,9,\,\,5,\,\,3,\,\,12,\,\,10,\,\,17,\,\,4,\,\,7,\,\,19,\,\,21 $  marks respectively. Calculate the range.

Ans. The marks can be arranged in ascending order as  $ 3,3,4,5,7,9,10,12,17,19,21. $ 

Range  $  =  $ Maximum value – Minimum value

 $  = 21 - 3 $ 

$ = 18$ 

Therefore, the range is  $ 18. $ 

2. Coefficient of variation of two distributions is $ 70{\text{ and }}75 $ , and their standard deviations are  $ 28{\text{ and }}27 $ respectively what are their arithmetic mean?

Ans. Given that, 

Coefficient of variation of first distribution $ {\text{(C}}{\text{.V)}} = {\text{ }}70 $ 

Coefficient of variation of second distribution $ {\text{(C}}{\text{.V)}} = {\text{ }}75 $ 

Standard deviation $ = {\sigma _1} = 28$

For first distribution,

 ${\text{C}}{\text{.V = }}\dfrac{{\sigma 1}}{{\bar x1}} \times 100$

Substitute the values,

$70 = \dfrac{{28}}{{\overline {x1} }} \times 100$

$\bar x = \dfrac{{28}}{{70}} \times 100$

$\bar x = 40$

Similarly for second distribution,

${\text{C}}{\text{.V = }}\dfrac{{{\sigma _2}}}{{{x_2}}} \times 100$

Substitute the values,

$75 = \dfrac{{27}}{{{x_2}}} \times 100$

${\bar x_2} = \dfrac{{27}}{{75}} \times 100$

${\bar x_2} = 36$

Therefore, the arithmetic mean of the first and second distributions are  $ 40{\text{ and }}36 $ respectively.

3. Write the formula for mean deviation.

Ans. The formula for mean deviation is $(\bar x) = \dfrac{{\sum {{f_i}} \left| {{x_i} - \bar x} \right|}}{{\sum {{f_i}} }} = \dfrac{1}{x}\Sigma {f_i}\left| {{x_i} - \bar x} \right|$.

4. Write the formula for variance.

Ans. The formula for variance is ${\sigma ^2} = \dfrac{1}{n}\Sigma {f_i}{\left( {{x_i} - \bar x} \right)^2}$

5. Find the median for the following data.

${x_i}:5,7,9,10,12,15$

${f_i}:8,6,2,2,2,6$

Ans.

$n = 26$. Median is the average of ${13^{{\text{th }}}}$ and ${14^{{\text{th }}}}$ item, both of which lie in the  $ c.f{\text{ }}14 $ .

$\therefore {x_i} = 7$

Thus, Median $ = \dfrac{{13{\text{ observation }} + 14{\text{ th observation }}}}{2}$

$ = \dfrac{{7 + 7}}{2}$

$ = 7$

Therefore, the median is $7$.

6. Write the formula of mean deviation about the median

Ans. The formula of mean deviation about the median is,$MD \cdot (M) = \dfrac{{\sum {{f_i}} \left| {{x_i}M} \right|}}{{\sum {{f_i}} }} = \dfrac{1}{n}\sum {{f_i}} \left| {{x_i} - M} \right|$

7. Find the range of the following series  $ 6,7,10,12,13,4,8,12 $ 

Ans. Range $  =  $  Maximum value – Minimum value

$ = 113 - 4$

$ = 9$

Therefore, the range of the given series is $9$.

8. Find the mean of the following data  $ 3,6,11,12,18 $ 

Ans. Mean $ = \dfrac{{{\text{ Sum of observation }}}}{{{\text{ Total no of observation }}}}$

$ = \dfrac{{50}}{5}$

$ = 10$

Therefore, the mean of given data $10$.

9. Express in the form of $a + ib,\,\,(3i - 7) + (7 - 4i) - (6 + 3i) + {i^{23}}$

Ans. Let ${\text{Z}} = 3i - 7 + 7 - 4i - 6 - 3i + {\left( {{i^4}} \right)^5}.\,{i^3}$

$=-4i-6-i$  $\because [{{i}^{4}}=1  \,\,\,\,\, {{i}^{3}}=-i] $

$ =  - 5i - 6$

$ =  - 6 + ( - 5i)$

10. Find the conjugate of $\sqrt { - 3}  + 4{i^2}$

Ans. Let $z = \sqrt { - 3}  + 4{i^2}$

$ = \sqrt 3 i - 4$

$\bar z =  - \sqrt 3 i - 4$

11. Solve for ${\text{x}}$ and $y,\,\,3x + (2x - y)i = 6 - 3i$

Ans. $3x = 6$

$x = 2$

$2x - y =  - 3$

Substitute the values,

$2 \times 2 - y =  - 3$

$ - y =  - 3 - 4$

$y = 7$

12. Find the value of $1 + {i^2} + {i^4} + {i^6} + {i^8} +  \ldots  + {i^{20}}$

Ans. $1 + {i^2} + {\left( {{i^2}} \right)^2} + {\left( {{i^2}} \right)^3} + {\left( {{i^2}} \right)^4} +  -  -  -  -  + {\left( {{i^2}} \right)^{10}} = 1\left[ {\because {i^2} =  - 1} \right.$

13. Multiply $3 - 2i$ by its conjugate.

Ans. Let $z = 3 - 2i$

The conjugate is,

$\bar z = 3 + 2i$

$z\bar z = (3 - 2i)(3 + 2i)$

$ = 9 + 6i - 6i - 4{{\text{i}}^2}$

$ = 9 - 4( - 1)$

$ = 13$

14. Find the multiplicative inverse $4 - 3i$.

Ans. Let $z = 4 - 3i$

Then, $\bar z = 4 + 3i$

$|z| = \sqrt {16 + 9}  = 5$

${z^{ - 1}} = \dfrac{{\bar z}}{{|z{|^2}}}$

$ = \dfrac{{4 + 3i}}{{25}}$

$ = \dfrac{4}{{25}} + \dfrac{3}{{25}}i$

Therefore, the multiplicative inverse $4 - 3i$ is $\dfrac{4}{{25}} + \dfrac{3}{{25}}i$.

15. Express in term of $a + ib,\,\,\dfrac{{(3 + i\sqrt 5 )(3 - i\sqrt 5 )}}{{(\sqrt 3  + \sqrt 2 i) - (\sqrt 3  - i\sqrt 2 )}}$

Ans. $\dfrac{{(3 + i\sqrt 5 )(3 - i\sqrt 5 )}}{{(\sqrt 3  + \sqrt 2 i) - (\sqrt 3  - i\sqrt 2 )}} = \dfrac{{{{(3)}^2} - {{(i\sqrt 5 )}^2}}}{{\sqrt 8  + \sqrt 2 i - \sqrt 8  + i\sqrt 2 }}$ 

$ = \dfrac{{9 + 5}}{{2\sqrt 2 i}}$

$ = \dfrac{{14}}{{2\sqrt 2 i}}$

$ = \dfrac{7}{{\sqrt 2 i}}$

$ = \dfrac{7}{{\sqrt 2 i}} \times \dfrac{{\sqrt 2 i}}{{\sqrt 2 i}}$

$ = \dfrac{{7\sqrt 2 i}}{{ - 2}}$

16. Evaluate ${i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}}$

Ans. ${i^n} + {i^n} \cdot {i^1} + {i^n} \cdot {i^2} + {i^n} \cdot {i^3}$

$={{i}^{n}}+{{i}^{n}}\cdot i-{{i}^{n}}+{{i}^{n}}\cdot (-i) $   $ \because [{i^{3} = -i} \,\,\,\ {i^{2}=-1} ]$

$ = 0$

17. If $1,w,{w^2}$ are three cube root of unity, show that $\left( {1 - w + {w^2}} \right)\left( {1 + w - {w^2}} \right) = 4$

Ans. $\left( {1 - w + {w^2}} \right)\left( {1 + w - {w^2}} \right)$

$\left( {1 + {w^2} - w} \right)\left( {1 + w - {w^2}} \right)$

$=(-w-w)\left( -{{w}^{2}}-{{w}^{2}} \right)$  $\because [{1+w=-{w^{2}}} \,\,\,\,\  {1+{w^{2}}= -w}]$

$( - 2w)\left( { - 2{w^2}} \right)$

$4{w^3}\left[ {{w^3} = 1} \right]$

$4 \times 1$

$ = 4$

Hence, If $1,w,{w^2}$ are three cube root of unity, then $\left( {1 - w + {w^2}} \right)\left( {1 + w - {w^2}} \right) = 4$

18. Find that sum product of the complex number $ - \sqrt 3  + \sqrt { - 2} $ and $2\sqrt 3  - i$

Ans. ${z_1} + {z_2} =  - \sqrt 3  + \sqrt 2 i + 2\sqrt 3  - i$

$ = \sqrt 3  + (\sqrt 2  - 1)i$

${z_1}{z_2} = ( - \sqrt 3  + \sqrt 2 i)(2\sqrt 3  - i)$

$ =  - 6 + \sqrt 3 i + 2\sqrt 6 i - \sqrt 2 {i^2}$

$ =  - 6 + \sqrt 3 i + 2\sqrt 6 i + \sqrt 2 $

$ = ( - 6 + \sqrt 2 ) + (\sqrt 3  + 2\sqrt 6 )i$

19. Write the real and imaginary part $1 - 2{{\mathbf{i}}^2}$

Ans. Let ${\text{z}} = 1 - 2{{\text{i}}^2}$

$ = 1 - 2( - 1)$

$ = 1 + 2$

$ = 3$

$ = 3 + 0.i$

The real part is $\operatorname{Re} (z) = 3$,

The imaginary part is $\operatorname{Im} (z) = 0$.

20. If two complex number ${z_1},\,\,{z_2}$ are such that $\left| {{z_1}} \right| = \left| {{z_2}} \right|$, is it then necessary that ${z_{\text{1}}}{\text{ = }}\,{z_{\text{2}}}$

Ans. Let ${z_1} = a + ib$

$\left| {{z_1}} \right| = \sqrt {{a^2} + {b^2}} $

${z_2} = b + ia$

$\left| {{z_2}} \right| = \sqrt {{b^2} + {a^2}} $

Therefore, $\left| {{z_1}} \right| = \left| {{z_2}} \right|$ but ${z_1} \ne {z_2}$

21. Find the conjugate and modulus of $\overline {9 - i}  + \overline {6 + {i^3}}  - \overline {9 + {i^2}} $

Ans. Let $z = \overline {9 - i}  + \overline {6 - i}  - \overline {9 - 1} $

$ = 9 + i + 6 + i - 0$

$ = 5 + 2i$

$\bar z = 5 - 2i$

$|z| = \sqrt {{{(5)}^2} + {{( - 2)}^2}} $

$ = \sqrt {25 + 4} $

$ = \sqrt {29} $

The conjugate of $\overline {9 - i}  + \overline {6 + {i^3}}  - \overline {9 + {i^2}} $ is $5 - 2i$.

The modulus of $\overline {9 - i}  + \overline {6 + {i^3}}  - \overline {9 + {i^2}} $ is $\sqrt {29} $.

22. Find the number of non zero integral solution of the equation $|1 - i{|^x} = {2^x}$

Ans. $|1 - i{|^x} = {2^x}$

${\left( {\sqrt {{{(1)}^2} + {{( - 1)}^2}} } \right)^x} = {2^x}$

${(\sqrt 2 )^x} = {2^x}$

${(2)^{\dfrac{1}{2}x}} = {2^x}$

$\dfrac{1}{2}x = x$

$\dfrac{1}{2} = 1$

$1 = 2$, Which is false no value of $x$ satisfies.

23. If $(a + ib)(c + id)(e + if)(g + ih) = A + iB$ then show that

$\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)\left( {{e^2} + {f^2}} \right)\left( {{g^2} + {h^2}} \right) = {A^2} + {B^2}$

Ans. Given that, $(a + ib)(c + id)(e + if)(g + ih) = A + iB$

$ \Rightarrow |(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|$

$\left( {\sqrt {{a^2} + {b^2}} } \right)\left( {\sqrt {{c^2} + {d^2}} } \right)\left( {\sqrt {{e^2} + {f^2}} } \right)\left( {\sqrt {{g^2} + {h^2}} } \right) = \sqrt {{A^2} + {B^2}} $

Square both side,

$\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)\left( {{e^2} + {f^2}} \right)\left( {{g^2} + {h^2}} \right) = {A^2} + {B^2}$

Hence, If $(a + ib)(c + id)(e + if)(g + ih) = A + iB$ then,

$\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)\left( {{e^2} + {f^2}} \right)\left( {{g^2} + {h^2}} \right) = {A^2} + {B^2}$

4 Mark Questions

1. The mean of  $ 2,7,4,6,8 $  and $p$ is 7 . Find the mean deviation about the median of the

observations.

Ans. Observations are  $ 2,7,4,6,8 $  and $p$ which are  $ 6 $ in numbers 

$\therefore n = 6$

The near of these observations is  $ 7 $ 

That is, $\dfrac{{2 + 7 + 4 + 6 + 8 + p}}{6} = 7$

$ \Rightarrow 27 + p = 42$

$ \Rightarrow p = 15$

Arrange the observations in ascending order  $ 2,4,6,7,8,15 $ 

Median $({\text{M}}) = \dfrac{{\dfrac{n}{2}{\text{ th observation }} + \left( {\dfrac{n}{2} + 1} \right){\text{ th observation }}}}{2}$

$ = \dfrac{{{\text{ 3rd observation }} + 4{\text{ th observation }}}}{2}$

$ = \dfrac{{6 + 7}}{2}$

$ = \dfrac{{13}}{2}$

$ = 6.5$

Calculation of mean deviation about Median:

Mean deviation about median $ = \dfrac{{18}}{6} = 3$.

2. Find the mean deviation about the mean for the following data!

${x_i}:10,30,50,70,90$

${f_i}:4,24,28,16,8$

Ans. To calculate mean, we require ${f_i}xi$ values then for mean deviation, it require 

$\mid xi - \bar x\mid $ values and ${f_i}|xi - \bar x|$ values.

$n = \sum {{f_i}}  = 80\quad $

$\sigma d\sum {{f_i}} xi = 4000$

$\bar x = \dfrac{{\sum {{f_i}} xi}}{n} = \dfrac{{4000}}{{80}} = 50$

Mean deviation about the mean is,

$\operatorname{MD} (\bar x) = \dfrac{{\sum {{f_i}} |xi - \bar x|}}{n} = \dfrac{{1280}}{{80}} = 16$

3. Find the mean, standard deviation and variance of the first $n$ natural numbers.

Ans. The numbers are $1,2,3, \ldots  \ldots ,n$

Mean is,

$\bar x = \dfrac{{\sum n }}{n} = \dfrac{{n(n + 1)}}{{\dfrac{2}{n}}} = \dfrac{{n + 1}}{2}$

Variance is,

$\sigma 2 = \dfrac{{\sum x {i^2}}}{n} - \bar x$

Substitute the values,

$ = \dfrac{{\sum {{n^2}} }}{n} - {\left( {\dfrac{{n + 1}}{2}} \right)^2}$

$ = \dfrac{{n(n + 1)(2n + 1)}}{{6n}} - \dfrac{{{{(n + 1)}^2}}}{4}$

$ = (n + 1)\left[ {\dfrac{{2n + 1}}{6} - \dfrac{{n + 1}}{4}} \right]$

$ = (n + 1)\left( {\dfrac{{n - 1}}{{12}}} \right) = \dfrac{{{n^2} - 1}}{{12}}$

Thus, Standard deviation $\sigma  = \dfrac{{\sqrt {{n^2} - 1} }}{{12}}$

4. Find the mean variance and standard deviation for following data

Ans.

Here, $n = \sum {{f_i}}  = 30$

$\sum {{f_i}} {x_i} = 420$

Mean $\bar x = \dfrac{{\sum {{f_i}} {x_i}}}{n}$

$ = \dfrac{{420}}{{30}}$

$ = 14$

Variance ${\sigma ^2} = \dfrac{1}{n}\sum {{f_i}} {\left( {{x_i} - \bar x} \right)^2}$

$ = \dfrac{1}{{30}} \times 1374$

$ = 45.8$

Standard deviation $\sigma  = \sqrt {45.8} $

$ = 6.77$

5. The mean and standard deviation of  $ 6 $  observations are  $ 8 $  and  $ 4 $  respectively. If  each observation is multiplied by $ 3 $ , find the new mean and new standard deviation of 

the resulting observations.

Ans. Let ${x_i},{x_2} \ldots  \ldots {x_6}$ be the six given observations

Then, $\bar x = 8$ and $\sigma  = 4$

$\bar x = \dfrac{{\sum {{x_i}} }}{n} = 8 = \dfrac{{{x_1} + {x_2} +  \ldots  \ldots  + {x_6}}}{6}$

${x_1} + {x_2} +  \ldots  \ldots {x_6} = 48$

Also ${\sigma ^2} = \dfrac{{\sum {x_1^2} }}{n} - {(\bar x)^2}$

Substitute the values,

$ = {4^2} = \dfrac{{x_1^2 + x_2^2 \ldots  \ldots  + x_6^2}}{6} - {(8)^2}$

$ = x_1^2 + x_2^2 +  \ldots  \ldots x_6^2$

$ = 6 \times (16 + 64) = 480$

As each observation is multiplied by  $ 3 $ , new observations are,

$3{x_1},3{x_2}, \ldots  \ldots 3{x_6}$

Then, $\bar X = \dfrac{{3{x_1} + 3{x_2} +  \ldots .3{x_6}}}{6}$

$ = \dfrac{{3\left( {{x_1} + {x_2} +  \ldots .{x_6}} \right)}}{6}$

$ = \dfrac{{3 \times 48}}{6}$

$ = 24$

Let ${\sigma _1}$ be the new standard deviation, then

$\sigma _1^2 = \dfrac{{{{\left( {3{x_1}} \right)}^2} + {{\left( {3{x_2}} \right)}^2} +  \ldots .. + {{\left( {3{x_6}} \right)}^2}}}{6} - {(\bar X)^2}$

$ = \dfrac{{9\left( {x_1^2 + x_2^2 +  \ldots .x_6^2} \right)}}{6} - {(24)^2}$

$ = \dfrac{{9 \times 480}}{6} - 576$

$ = 720 - 576$

$ = 144$

${\sigma _1} = 12$

6. Prove that the standard deviation is independent of any change of origin, but is dependent on the change of scale.

Ans. Use the transformation $u = ax + b$ to change the scale and origin

Now $u = ax + b$

$ = \sum u  = \Sigma (ax + b) = a\sum x  + b.n$

Also $\sigma {u^2} = \dfrac{{\sum {{{(u - \bar u)}^2}} }}{n}$

$ = \dfrac{{\sum {{{(ax + b - a\bar x - b)}^2}} }}{n}$

$ = \dfrac{{\sum {{a^2}} {{(x - \bar x)}^2}}}{n}$

$ = \dfrac{{{a^2}\Sigma {{(x - \bar x)}^2}}}{n}$

$ = {a^2}\sigma {x^2}$

${\sigma ^2}u = a2{\sigma ^2}u$

$ = \sigma u = \mid a\mid \sigma x$

Both $\sigma u,\,\,\sigma x$ are positive which shows that standard deviation is independent of choice of origin, but depends on the scale.

7. Calculate the mean deviation about the mean for the following data:

Expenditure  $ 0 - 100,100 - 200,200 - 300,300 - 400,400 - 500,500 - 600,600 - 700,700 - 800 $  Persons  $ 4,8,9,10,7,5,4,3 $ 

Ans.

$n = \Sigma {f_i} = 50$

$\sum {{f_i}} {x_i} = 17900$

$\therefore $Mean $ = \dfrac{1}{n}\sum {{f_i}} {x_i}$

$ = \dfrac{{17900}}{{50}}$

$ = 358$

$MD(\bar x) = \dfrac{1}{n}\sum f \left| {{x_i} - \bar x} \right|$

$ = \dfrac{{7896}}{{50}}$

$ = 157.92$

8.Find the mean deviation about the median for the following data:

Marks  $ 0 - 10,10 - 20,20 - 30,30 - 40,40 - 50,50 - 60 $ No.of boys  $ 8,10,10,16,4,2 $ 

Ans.

${\left( {\dfrac{n}{2}} \right)^{th}}$or ${25^{{\text{th }}}}$ item $ = 20 - 30$, which is the median class.

Median $ = l + \dfrac{{\dfrac{n}{2} - c}}{f} \times c$

$ = 20 + \dfrac{{25 - 18}}{{10}} \times 10$

$ = 27$

$MD(M) = \dfrac{1}{n}\Sigma {f_i}\left| {{x_i} - M} \right|$

$ = \dfrac{{572}}{{50}}$

$ = 11.44$

9. An analysis of monthly wages points to workers in two firms  $ {\text{A and B}} $ , belonging to the  same industry, given the following result. Find mean deviation about median.

Firm  $ {\text{A}} $ , Firm  $ {\text{B}} $ 

No of wages earns 586, 648

Average monthly wages  $ {\text{Rs 5253, Rs 5253}} $ 

Ans. For firm  $ {\text{A}} $ , number of workers $ = 586$

Average monthly wage is  $ {\text{Rs 5253}} $ 

Total wages $ = \operatorname{Rs} 5253 \times 586$

$ = \operatorname{Rs} 3078258$

For firm $ {\text{B}} $ , total wages $ = {\text{Rs}}\,\,253 \times 648$

$ = \operatorname{Rs} 3403944$

Therefore, the firm  $ {\text{B}} $  pays out an amount of monthly wages.

10. Find the mean deviation about the median of the following frequency distribution -

Class  $ 0 - 6,6 - 12,12 - 18,18 - 24,24 - 30 $ 

Frequency  $ 8,10,12,9,5 $ 

Ans.

$N = 44 = \dfrac{N}{2}$

 $ 12 - 18 $  is the median class

Median $ = l + \dfrac{{\dfrac{N}{2} - F}}{f} \times h$

Here $h = 6,\,\,l = 12,\,\,f = 12,\,\,F = 18$

Median is,

$ = 12 + \dfrac{{22 - 18}}{{12}} \times 6$

$ = 12 + \dfrac{{4 \times 6}}{{12}}$

$ = 14$

Mean deviation about median $ = \dfrac{1}{N}\sum {{f_i}} \left| {{x_i} - 14} \right|$

$ = \dfrac{{278}}{{74}}$

$ = 6.318$

11. Calculate the mean deviation from the median from the following data:

Salary per week(in Rs)  $ 10 - 20,20 - 30,30 - 40,40 - 50,50 - 60,60 - 70,70 - 80 $ 

No. of workers  $ 4,6,10,20,10,6,4 $ 

Ans.

$N = 60$

$\Rightarrow \dfrac{N}{2} = 30$

 $ 40 - 50 $ is the median class

Here $l = 40,\,\,f = 20,\,\,h = 10,\,\,F = 20$

Median $ = \dfrac{{l - \dfrac{N}{2} - F}}{f} \times h$

$= \dfrac{{40 + 30 - 20}}{{20}} \times 10 $

$= 45$

Mean deviation $ = \dfrac{{\sum {{f_i}} \left| {{d_i}} \right|}}{N}$

$ = \dfrac{{680}}{{60}}$

$ = 11.33$

12. Let ${x_1},{x_2} \ldots  \ldots {x_n}$ values of a variable ${\text{Y}}$ and let 'a' be a non zero real number. Then prove that the variance of the observations $a{y_1},a{y_2} \ldots  \ldots a{y_n}$ is ${a^2}\operatorname{var} (Y)$. Also, find their standard deviation.

Ans. Let ${v_1},{v_2} \ldots  \ldots {v_n}$ be the value of variables $v$ such that ${v_1} = a{y_i},1,2 \ldots  \ldots n$, then

$\bar V = \dfrac{1}{n}\sum\limits_{i = 1}^n {{v_i}}  = \dfrac{1}{n}\sum\limits_{i = 1}^n {(a{y_i})}  = a\left( {\dfrac{1}{n}\sum\limits_{i = 1}^n {{y_i}} } \right) = a\bar y$

${v_i} - \bar V = a{y_i} - a\bar y$

${v_i} - \bar V = a\left( {{y_i} - \bar Y} \right)$

${\left( {{v_i} - \bar V} \right)^2} = {a^2}{\left( {{y_i} - \bar Y} \right)^2}$

$\sum\limits_{i = 1}^n {{{\left( {{v_i} - \bar V} \right)}^2}}  = {a^2}\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{y_i} - \bar Y} \right)}^2}} $

${\text{Var}}(V) = {a^2}\operatorname{Var} (Y)$

${\sigma _u} = \sqrt {\operatorname{var} (v)}  = \sqrt {{a^2}\operatorname{var} (Y)}  = a\mid \sqrt {\operatorname{var} (Y)} $

$ = |a|{\sigma _y}$

13. If $a + ib = \dfrac{{{{(x + i)}^2}}}{{2{x^2} + 1}}$ Prove that ${a^2} + {b^2} = \dfrac{{{{\left( {{x^2} + 1} \right)}^2}}}{{{{\left( {2{x^2} + 1} \right)}^2}}}$

Ans.Given that, $a + ib = \dfrac{{{{(x + i)}^2}}}{{2{x^2} + 1}}$ 

Take conjugate both side,

$a - ib = \dfrac{{{{(x - i)}^2}}}{{2{x^2} + 1}}$ 

$(a + ib)(a - ib) = \left( {\dfrac{{{{(x + i)}^2}}}{{2{x^2} + 1}}} \right) \times \left( {\dfrac{{{{(x - i)}^2}}}{{2{x^2} + 1}}} \right)$

${(a)^2} - {(ib)^2} = \dfrac{{{{\left( {{x^2} - {i^2}} \right)}^2}}}{{{{\left( {2{x^2} + 1} \right)}^2}}}$

${a^2} + {b^2} = \dfrac{{{{\left( {{x^2} + 1} \right)}^2}}}{{{{\left( {2{x^2} + 1} \right)}^2}}}$

14. If ${(x + iy)^3} = u + iv$ then show that $\dfrac{u}{x} + \dfrac{v}{y} = 4\left( {{x^2} - {y^2}} \right)$

Ans. ${(x + iy)^3} = u + iv$

${x^3} + {(iy)^3} + 3{x^2}(iy) + 3 \cdot x{(iy)^2} = u + iv$

${x^3} - i{y^3} + 3{x^2}yi - 3x{y^2} = u + iv$

${x^3} - 3x{y^2} + \left( {3{x^2}y - {y^3}} \right)i = u + iv$

$x\left( {{x^2} - 3{y^2}} \right) + y\left( {3{x^2} - {y^2}} \right)i = u + iv$

$x\left( {{x^2} - 3{y^2}} \right) = u,y\left( {3{x^2} - {y^2}} \right) = v$

${x^2} - 3{y^2} = \dfrac{u}{x}$     $\mid 3{x^2} - {y^2} = \dfrac{v}{y}$

$4{x^2} - 4{y^2} = \dfrac{u}{x} + \dfrac{v}{y}$

$4\left( {{x^2} - {y^2}} \right) = \dfrac{u}{x} + \dfrac{v}{y}$

Hence proved.

15. Solve $\sqrt 3 {x^2} - \sqrt 2 x + 3\sqrt 3  = 0$

Ans.Given that, $\sqrt 3 {x^2} - \sqrt 2 x + 3\sqrt 3  = 0$

Here, $a = \sqrt 3 ,\,\,b =  - \sqrt 2 ,\,\,c = 3\sqrt 3 $

$D = {b^2} - 4ac$

Substitute the values,

$ = {( - \sqrt 2 )^2} - 4 \times \sqrt 3 (3\sqrt 3 )$

$ = 2 - 36$

$ =  - 34$

$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$

Substitute the values,

$ = \dfrac{{ - ( - \sqrt 2 ) \pm \sqrt { - 34} }}{{2 \times \sqrt 3 }}$

$ = \dfrac{{\sqrt 2  \pm \sqrt {34} {\text{i}}}}{{2\sqrt 3 }}$

16. Find the modulus ${i^{25}} + {(1 + 3i)^3}$

Ans.Given, ${i^{25}} + {(1 + 3{\text{i}})^3}$

$ = {\left( {{i^4}} \right)^6} \cdot i + 1 + 27{i^3} + 3(1)(3i)(1 + 3i)$

$ = i + \left( {1 - 27i + 9i + 27{i^2}} \right)$

$ = i + 1 - 18i - 27$

$ =  - 26 - 17i$

$\left| {{i^{25}} + {{(1 + 3i)}^3}} \right| = | - 26 - 17i|$

$ = \sqrt {{{( - 26)}^2} + {{( - 17)}^2}} $

$ = \sqrt {{{( - 26)}^2} + {{( - 17)}^2}} $

$ = \sqrt {676 + 289} $

$ = \sqrt {965} $

17. If $a + ib = \dfrac{{{{(x + i)}^2}}}{{2x - i}}$ prove that ${a^2} + {b^2} = \dfrac{{{{\left( {{x^2} + 1} \right)}^2}}}{{4{x^2} + 1}}$

Ans. Given that, $a + ib = \dfrac{{{{(x + i)}^2}}}{{2x - i}}$

The conjugate is, $a - ib = \dfrac{{{{(x - i)}^2}}}{{2x + i}}$

Take conjugate both side,

$(a + ib)(a - ib) = \dfrac{{{{(x + i)}^2}}}{{(2x - i)}} \times \dfrac{{{{(x - i)}^2}}}{{(2x + i)}}$

$ \Rightarrow {a^2} + {b^2} = \dfrac{{{{\left( {{x^2} + 1} \right)}^2}}}{{4{x^2} + 1}}$ 

18. Evaluate ${\left[ {{i^{18}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right]^3}$

Ans.Given that, ${\left[ {{i^{18}} + {{\left( {\dfrac{1}{i}} \right)}^{25}}} \right]^3}$

$ \Rightarrow {\left[ {{{\left( {{i^4}} \right)}^4} \cdot {i^2} + \dfrac{1}{{{i^{25}}}}} \right]^3}$

$ \Rightarrow {\left[ {{i^2} + \dfrac{1}{{{{\left( {{i^4}} \right)}^6}.i}}} \right]^{ - 3}}$

$ \Rightarrow {\left[ { - 1 + \dfrac{1}{i}} \right]^3}$

$ \Rightarrow {\left[ { - 1 + \dfrac{{{i^3}}}{{{i^4}}}} \right]^3}$

$ \Rightarrow {[ - 1 - i]^3} =  - {(1 + i)^3}$

$ =  - \left[ {{1^3} + {i^3} + 3.1i(1 + i)} \right]$

$ =  - \left[ {1 - i + 3i + 3{i^2}} \right]$

$ =  - [1 - i + 3i - 3]$

$ =  - [ - 2 + 2i] = 2 - 2i$

19. Find that modulus and argument $\dfrac{{1 + i}}{{1 - i}}$

Ans. $\dfrac{{1 + i}}{{1 - i}} = \dfrac{{1 + i}}{{1 - i}} \times \dfrac{{1 + i}}{{1 + i}}$

$ = \dfrac{{{{(1 + i)}^2}}}{{{1^2} - {i^2}}}$

$ = \dfrac{{1 + {i^2} + 2i}}{{1 + 1}}$

$ = \dfrac{{2i}}{2}$

$ = i$

$z = 0 + i$

$r = |z| = \sqrt {{{(0)}^2} + {{(1)}^2}}  = 1$

Let $\alpha $ be the acute $\angle {\text{s}}$

$\tan \alpha  = \left| {\dfrac{1}{0}} \right|$

$\alpha  = \pi /2$

$\arg (z) = \pi /2$

$r = 1$

20. For what real value of $x$ and $y$ are numbers equal $(1 + i){y^2} + (6 + i)$ and $(2 + i)x$

Ans. $(1 + i){y^2} + (6 + i) = (2 + i)x$

${y^2} + i{y^2} + 6 + i = 2x + xi$

$\left( {{y^2} + 6} \right) + \left( {{y^2} + 1} \right)i = 2x + xi$

${y^2} + 6 = 2x$

${y^2} + 1 = x$

${y^2} = x - 1$

$x - 1 + 6 = 2x$

$5 = x$

Substitute the value of  $ x $ in ${y^2} = x - 1$

$y =  \pm 2$

21. If $x + iy = \sqrt {\dfrac{{1 + i}}{{1 - i}}} $, prove that ${x^2} + {y^2} = 1$

Ans. Given that, $x + iy = \sqrt {\dfrac{{1 + i}}{{1 - i}}} \quad $ 

Taking conjugate both side,

$(x + iy)(x - iy) = \sqrt {\dfrac{{1 + i}}{{1 - i}}}  \times \sqrt {\dfrac{{1 - i}}{{1 + i}}} $

${(x)^2} - {(iy)^2} = 1$

${x^2} + {y^2} = 1$

Hence proved.

22. Convert in the polar form $\dfrac{{1 + 7i}}{{{{(2 - i)}^2}}}$

Ans. $\dfrac{{1 + 7i}}{{{{(2 - i)}^2}}} = \dfrac{{1 + 7i}}{{4 + {i^2} - 4i}} = \dfrac{{1 + 7i}}{{3 - 4i}}$

Take conjugate on both side,

$ = \dfrac{{1 + 7i}}{{3 - 4i}} \times \dfrac{{3 + 4i}}{{3 + 4i}}$

$ = \dfrac{{3 + 4i + 21i + 28{i^2}}}{{9 + 16}}$

$ = \dfrac{{25i - 25}}{{25}} = i - 1$

$ =  - 1 + i$

$r = |z| = \sqrt {{{( - 1)}^2} + {1^2}}  = \sqrt 2 $

Let $\alpha $ be the acute $\angle {\text{s}}$

tan $\alpha  = \left| {\dfrac{1}{{ - 1}}} \right|$

$\alpha  = \pi /4$

since $\operatorname{Re} (z) < 0,\operatorname{Im} (z) > 0$

$\theta  = \pi  - \alpha $

$ = \pi  - \dfrac{\pi }{4} = 3\pi /4$

$z = r(\operatorname{Cos} \theta  + i\operatorname{Sin} \theta )$

$ = \sqrt 2 \left( {\operatorname{Cos} \dfrac{{3\pi }}{4} + i\sin \dfrac{{3\pi }}{4}} \right)$

23. Find the real values of $x$ and $y$ if $(x - iy)(3 + 5i)$ is the conjugate of $ - 6 - 24i$

Ans. $(x - iy)(3 + 5i) =  - 6 + 24i$

Expand brackets,

$3x + 5xi - 3yi - 5y{i^2} =  - 6 + 24i$

Group terms,

$(3x + 5y) + (5x - 3y)i =  - 6 + 24i$

$3x + 5y =  - 6$

$5x - 3y = 24$

$x = 3$

$y =  - 3$

24. If $\left| {{z_1}} \right| = \left| {{z_2}} \right| = 1$, prove that $\left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}}} \right| = \left| {{z_1} + {z_2}} \right|$

Ans. Given that, $\left| {{z_1}} \right| = \left| {{z_2}} \right| = 1\quad $ 

$ \Rightarrow {\left| {{z_1}} \right|^2} = {\left| {{z_2}} \right|^2} = 1$

$ \Rightarrow {z_1}\overline {{z_1}}  = 1$

$\overline {{z_1}}  = \dfrac{1}{{{z_1}}} \to \left( 1 \right)$

${z_2}\overline {{z_2}}  = 1$

$\overline {{z_2}}  = \dfrac{1}{{{z_2}}} \to (2)$   $\left[ {\because z\bar z = |z{|^2}} \right.$

$\left| {\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}}} \right| = \left| {\overline {{z_1}}  + \overline {{z_2}} } \right|$

$ = \left| {\overline {{z_1} + {z_2}} } \right|$

$ = \left| {{z_1} + {z_2}} \right|$   $[\because |\bar z| = |z|\quad $ 

6 Mark Questions

1.Calculate the mean, variance and standard deviation of the following data:

Ans.

Here $n = \Sigma {f_i} = 50,\Sigma {f_i}{x_i} = 3100$

Thus, Mean $\bar x = \dfrac{{\sum {{f_i}} {x_i}}}{n} = \dfrac{{3100}}{{50}} = 62$

Variance ${\sigma ^2} = \dfrac{1}{n}\sum {{f_i}} {(xi - \bar x)^2}$

$ = \dfrac{1}{{50}} \times 10050$

$ = 201$

Standard deviation $\sigma  = \sqrt {201}  = 14.18$

2. The mean and the standard deviation of  $ 100 $  observations were calculated as  $ 40 $  and  $ 5.1 $  respectively by a student who mistook one observation as  $ 50 $  instead of  $ 40 $ . What are the correct mean and standard deviation?

Ans. Given that, $n = 100$

Incorrect mean $\bar x = 40$,

Incorrect S.D $(\sigma ) = 5.1$

As $\bar x = \dfrac{{\sum {{x_i}} }}{n}$

$40 = \dfrac{{\sum {{x_i}} }}{{100}} \Rightarrow \sum {{x_i}}  = 4000$

 $  \Rightarrow  $ Incorrect sum of observation $ = 4000$

 $  \Rightarrow  $ Correct sum of observations $ = 4000 - 50 + 40$

$ = 3990$

Now, Correct mean $ = \dfrac{{3990}}{{100}} = 39.9$

Also, $\sigma  = \sqrt {\dfrac{1}{n}\sum {x_i^2}  - {{(\bar x)}^2}} $

Use incorrect values,

$5.1 = \sqrt {\dfrac{1}{{100}}\sum {x_i^2}  - {{(40)}^2}} $

$ \Rightarrow 26.01 = \left[ {\dfrac{1}{{100}}\sum {x_i^2}  - 1600} \right]$

$ = 162601$

Incorrect $\sum {x_i^2}  = 162601$

Correct $\sum {x_i^2}  = 162601 - {(50)^2} + {(40)^2}$

$ = 162601 - 2500 + 1600 = 161701$

Correct $\sigma  = \sqrt {\dfrac{1}{{100}}{\text{ correct }}\sum {x_i^2}  - {{({\text{ correct }}\bar x)}^2}} $

$ = \sqrt {\dfrac{1}{{100}}(161701) - {{(39.9)}^2}} $

$ = \sqrt {1617.01 - 1592.01} $

$ = \sqrt {25}  = 5$

Therefore, the correct mean is  $ 39.9 $  and the correct standard deviation is  $ 5. $ 

3. Given,  $ 200 $  candidates the mean and standard deviation was found to be  $ 10{\text{ and 15}} $  respectively. After that if was found that the scale  $ 43 $  was misread as  $ 34. $  Find the correct mean and correct S.D

Ans. Given that,  $n = 200,\,\,\bar X = 40,\,\,\sigma  = \overline {15} $

$ \bar X = \dfrac{1}{n}\sum {{x_i}}$

$= \sum {{x_i}}$

$= n\bar X$

${\text{Substitute}}\,{\text{the}}\,{\text{values,}}$

$= 200 \times 40$

$= 8000$

Corrected $\sum {{x_i}}  = $ Incorrect $\sum {{x_i}}  - $ (Sum of incorrect  $  +  $  Sum of correct value)

$ = 8000 - 34 + 43 = 8009$

Thus, Corrected mean $ = \dfrac{{{\text{ Corrected }}\sum {{x_i}} }}{n}$

$= \dfrac{{8009}}{{200}}$

$= 40.045$

$\sigma  = 15$

${15^2} = \dfrac{1}{{200}}\left( {\sum {x_i^2} } \right) - {\left( {\dfrac{1}{{200}}\sum {{x_i}} } \right)^2}$

Substitute the values,

$225 = \dfrac{1}{{200}}\left( {\sum {x_i^2} } \right) - {\left( {\dfrac{{8000}}{{200}}} \right)^2}$

$225 = \dfrac{1}{{200}} \times 1825 = 365000$

Incorrect $\sum {x_i^2}  = 365000$

Corrected $\sum {x_i^2}  = $ (incorrect $\left. {\sum {x_i^2} } \right) - $(Sum of squares of incorrect values)  $  +  $ (Sum of square of 

correct values)

$ = 365000 - {(34)^2} + {(43)^2} = 365693$

Corrected $\sigma  = \sqrt {\dfrac{1}{n}\sum {x_i^2}  - {{\left( {\dfrac{1}{n}\sum {{x_i}} } \right)}^2}} $

$ = \sqrt {\dfrac{{365693}}{{200}} - {{\left( {\dfrac{{8009}}{{200}}} \right)}^2}} $

$ = \sqrt {1828.465 - 1603.602} $

$ = 14.995$

4. Find the mean deviation from the mean  $ 6,7,10,12,13,4,8,20 $ 

Ans. Let $\bar X$ be the mean

$\bar X = \dfrac{{6 + 7 + 10 + 12 + 13 + 4 + 8 + 20}}{8} = 10$

$\sum {{d_i}}  = 30$ and $n = 8$

Thus, $MD = \dfrac{1}{n}\sum {\left| {{d_i}} \right|}  = \dfrac{{30}}{8} = 3.75$

$MD = 3.75$

5. Find two numbers such that their sum is  $ 6 $  and the product is  $ 14. $ 

Ans. Let $x$ and $y$ be the no.

Given that, $x + y = 6$

$xy = 14$

${x^2} - 6x + 14 = 0$

$D =  - 20$

We have , $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$

Substitute the values,

$x = \dfrac{{ - ( - 6) \pm \sqrt { - 20} }}{{2 \times 1}}$

$ = \dfrac{{6 \pm 2\sqrt 5 i}}{2}$

$ = 3 \pm \sqrt 5 {\text{i}}$

When $x = 3 + \sqrt 5 {\text{i}}$

$y = 6 - (3 + \sqrt 5 {\text{i}})$

$ = 3 - \sqrt 5 {\text{i}}$

When $x = 3 - \sqrt 5 {\text{i}}$

$y = 6 - (3 - \sqrt 5 {\text{i}})$

$ = 3 + \sqrt 5 {\text{i}}$

Therefore, the numbers are $3 + \sqrt 5 {\text{i}}$ and $3 - \sqrt 5 {\text{i}}$.

6. Convert into polar form $z = \dfrac{{i - 1}}{{\cos \dfrac{\pi }{3} + i\operatorname{Sin} \dfrac{\pi }{3}}}$

Ans. Given that, $z = \dfrac{{i - 1}}{{\cos \dfrac{\pi }{3} + i\operatorname{Sin} \dfrac{\pi }{3}}}$

Substitute the known values,

 $z = \dfrac{{i - 1}}{{\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i}}$

$ = \dfrac{{2(i - 1)}}{{1 + \sqrt 3 i}} \times \dfrac{{1 - \sqrt 3 i}}{{1 - \sqrt 3 i}}$

$z = \dfrac{{\sqrt 3  - 1}}{2} + \dfrac{{\sqrt 3  + 1}}{2}i$

$r = |z| = {\left( {\dfrac{{\sqrt 3  - 1}}{2}} \right)^2} + {\left( {\dfrac{{\sqrt 3  + 1}}{2}} \right)^2}$

$r = 2$

Let $\alpha $ be the acute $\angle {\text{s}}$,

$\tan \alpha  = \left| {\dfrac{{\dfrac{{\sqrt 3  + 1}}{2}}}{{\dfrac{{\sqrt 3  - 1}}{2}}}} \right|$

$ = \left| {\dfrac{{\sqrt 3 \left( {1 + \dfrac{1}{{\sqrt 3 }}} \right)}}{{\sqrt 3 \left( {1 - \dfrac{1}{{\sqrt 3 }}} \right)}}} \right|$

$ = \left| {\dfrac{{\tan \dfrac{\pi }{4} + \tan \dfrac{\pi }{6}}}{{1 - \tan \dfrac{\pi }{4}\tan \dfrac{\pi }{6}}}} \right|$

$\tan \alpha  = \left| {\tan \left( {\dfrac{\pi }{4} + \dfrac{\pi }{6}} \right)} \right|$

$\alpha  = \dfrac{\pi }{4} + \dfrac{\pi }{6} = \dfrac{{5\pi }}{{12}}$

$z = 2\left( {\operatorname{Cos} \dfrac{{5\pi }}{{12}} + i\operatorname{Sin} \dfrac{{5\pi }}{{12}}} \right)$

7. If $\alpha $ and $\beta $ are different complex number with $|\beta | = 1$. Then find $\left| {\dfrac{{\beta  - \alpha }}{{1 - \bar \alpha \beta }}} \right|$

Ans. ${\left| {\dfrac{{\beta  - \alpha }}{{1 - \bar \alpha \beta }}} \right|^2} = \left( {\dfrac{{\beta  - \alpha }}{{1 - \bar \alpha \beta }}} \right)\left( {\dfrac{{\overline {\beta  - \alpha } }}{{1 - \bar \alpha \beta }}} \right)\quad \,\,\left[ {\because |z{|^2} = z\bar z} \right.$

$ = \left( {\dfrac{{\beta  - \alpha }}{{1 - \bar \alpha \beta }}} \right)\left( {\dfrac{{\bar \beta  - \bar \alpha }}{{1 - \alpha \bar \beta }}} \right)$

$ = \left( {\dfrac{{\beta \bar \beta  - \beta \bar \alpha  - \alpha \bar \beta  + \alpha \bar \alpha }}{{1 - \alpha \beta  - \alpha \beta  + \alpha \bar \alpha \beta \bar \beta }}} \right)$

$\left. { = \dfrac{{|\beta {|^2} - \beta \bar \alpha  - \alpha \bar \beta  + |\alpha {|^2}}}{{1 - \alpha \bar \beta  - \bar \alpha \beta  + |\alpha {|^2}|\beta {|^2}}}} \right\}$

$ 1-\beta \bar{\alpha }-\alpha \bar{\beta }+|\alpha {{|}^{2}} $

$ 1-\alpha \bar{\bar{\beta }}-\overline{\bar{\alpha }\beta }\beta +|\alpha {{|}^{2}} $

$ \because$ $ |\beta |=1$

$ =1 $

$ \left| \dfrac{\beta -\alpha }{1-\alpha \beta } \right|=\sqrt{1} $

$ \left| \dfrac{\beta -\alpha }{1-\vec{\alpha }\beta } \right|=1 $

5 Important Formulas of Class 11 Maths Chapter 13 Statistics

Conclusion

Practising important questions from Class 11 Maths Chapter 13, Statistics, is crucial for understanding the core concepts like mean, variance, standard deviation, and data interpretation. These topics are not only significant for exams but also help build a strong foundation in analysing and interpreting data in real-life scenarios. By focusing on these questions, you can improve your problem-solving skills and gain confidence to tackle similar problems with ease. Consistent practice and clarity in concepts are key to mastering this chapter.

Important Study Materials for Class 11 Maths Chapter 13 Statistics

CBSE Class 11 Maths Chapter-wise Important Questions

CBSE Class 11 Maths Chapter-wise Important Questions and Answers cover topics from all 14 chapters, helping students prepare thoroughly by focusing on key topics for easier revision.

Additional Study Materials for Class 11 Maths