Important Questions for CBSE Class 11 Maths Chapter 15 - 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 $

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