MCQ Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 3

If the expenditure of Company

$Y$ in

$1997$ was

$Rs. 220$ crores, what was its income in

$1997$ ?

0%
$Rs. 312$ crores
0%
$Rs. 297$ crores
0%
$Rs. 283$ crores
0%
$Rs. 275$ crores

Explanation

Profit percent of Company

$Y$ in

$1997 = 35.$
Let the income of Company

$Y$ in

$1997$ be $$Rs. x$$ crores.
Then,

$\displaystyle 35=\frac { x-220 }{ 220 } \times 100\Rightarrow x=297$

$\displaystyle \therefore$ Income of Company

$Y$ in

$1997 = Rs. 297$ crores.

In which of the following years, the difference between the productions of Companies X and Y was the maximum among the given years ?

0%
1997
0%
1998
0%
1999
0%
2000

The difference between the percentage of candidates qualified to appeared was maximum in which of the following pairs of years?

0%
1994 and 1995
0%
1997 and 1998
0%
1998 and 1999
0%
1999 and 2000

What is the difference between the number of vehicles manufactured by Company Y in 2000 and 2001 ?

0%
50000
0%
42000
0%
33000
0%
21000

Explanation

No. of vehicles manufactured by company Y in

$2000 = 128000$

No. of vehicles manufactured by company Y in

$2001 = 107000$

Required difference = (128000 - 107000) = 21000.

In which periodical exams there is a fall in percentage of marks as compared to the previous periodical exams ?

0%
None
0%
June, 01
0%
Oct, 01
0%
Feb, 02

If the profit in 1999 was Rs. 4 crores, what was the profit in 2000?

0%
Rs. 4.2 crores
0%
Rs. 6.6 crores
0%
Rs. 6.8 crores
0%
Cannot be determined

In which year is the expenditure minimum?

0%
$2000$
0%
$1997$
0%
$1996$
0%
Cannot be determined

In how many of the given years were the exports more than the imports ?

0%
1
0%
2
0%
3
0%
4

The percentage increases in sales from 2001 to 2002 was?

0%
115%
0%
128%
0%
122%
0%
118%

The average production of 1996 and 1997 was exactly equal to the average production of which of the following pairs of years?

0%
2000 and 2001
0%
1999 and 2000
0%
1998 and 2000
0%
1995 and 2001

Explanation

Average production (in 10000 tonnes) of 1996 and 1997

$\dfrac{40 + 60}{2} = 50$
We shall find the average production (in 10000 tonnes) for each of the given alternative pairs:
2000 and 2001

$= \dfrac{50 + 75}{2} = 62.5$
1999 and 2000

$= \dfrac{65 + 50}{2} = 57.5$
1998 and 2000

$= \dfrac{45 + 50}{2} = 47.5$
1995 and 1999

$= \dfrac{25 + 65}{2} = 45$
1995 and 2001

$= \dfrac{25 + 75}{2} = 50$

$\therefore$ The average production of 1996 and 1997 is equal to the average production of 1995 and 2001.

The sum of sales of cellular phones in the years 1999 and 2001 is equal to that in?

0%
1997
0%
1998
0%
2000
0%
2002

Total sales of branches B1, B3 and B5 together for both the years (in thousand numbers) is?

0%
250
0%
310
0%
435
0%
560

What percent of the average sales of branches B1, B2 and B3 in 2001 is the average sales of branches B1, B3 and B6 in 2000?

0%
75%
0%
77.5%
0%
82.5%
0%
87.5%

Explanation

Average sales (in thousand number) of branches B1, B3 and B6 in 2000

$= \dfrac{1}{3} \times (80+95+70) = \left( \dfrac{245}{3} \right)$
Average sales (in thousand number) of branches B1, B2 and B3 in 2001

$= \dfrac{1}{3} \times (105 + 65 + 110) = \left( \dfrac{280}{3} \right)$

$\therefore$ Required percentage

$= \left [ \dfrac{245/3}{280/3} \times 100 \right]$ %

$= \left( \dfrac{245}{280} \times 100 \right )$ % = 87.5%

What is the difference between the production of Company Z in 1998 and Company Y in 1996?

0%
2,00,000 tons
0%
20,00,000 tons
0%
20,000 tons
0%
2,00,00,000 tons

Explanation

Required difference

$= [(45 - 25) \times 1,00,000 ] tons$

$= 20,00,000 tons.$

What is the approximate ratio of the sales of nail enamels in 2000 to the sales of Talcum powders in 1995?

0%
7 : 2
0%
5 : 2
0%
4 : 3
0%
2 : 1

Explanation

Required ratio

$= \dfrac{37.76}{14.97} \approx 2.5 = \dfrac{5}{2}$

What was the percentage increase in production of fertilizers in 2002 compared to that in 1995?

0%
320%
0%
300%
0%
220%
0%
200%

Explanation

Required percentage

$= \left [ \dfrac{(80 - 25)}{25} \times 100 \right ]$ % = 220%

What is the average sales of all the branches (in thousand numbers) for the year 2000?

0%
73
0%
80
0%
83
0%
88

Explanation

Average sales of all the six branches (in thousand numbers) for the year 2000

$=\dfrac{1}{6} \times [80+ 75 + 95+ 85+ 75+ 70]$

$= 80$

The GDP of UAE is what fraction of GDP of the UK for the decade (approximately)?

0%
(1/4)th
0%
(1/5)th
0%
(1/6)th
0%
Data inadequate

Which of the countries listed below accounts for the maximum GDP during the half decade 2006 to 2010?

0%
UAE
0%
US
0%
India
0%
China

What should have been the sales turnover of GM in 2002 - 2003 to have shown an excess of the same quantum over 2001 - 2002 as shown by the sales turnover of Maruti?

0%
953.76
0%
963.76
0%
952.76
0%
962.76

Which of the companies shows the maximum percentage difference in sales turnover between the two years?

0%
Honda
0%
GM
0%
Hyundai
0%
Maruti

Out of every Rs. 10,000 spent during the decade 2001 - 2010 approximately how much was the GDP of Russia during the half decade 2001 - 2005?

0%
Rs. 700
0%
Rs. 1,400
0%
Rs. 2,800
0%
None of these

Which of the countries listed below accounts for the highest GDP during the half decade 2001 to 2005?

0%
Russia
0%
China
0%
India
0%
UAE

What was absolute difference in the FDI to India in between 1996 and 1997?

0%
7.29
0%
7.13
0%
8.13
0%
None of these

What is the approximate difference between the average sales turnover of all the companies put together between the years 2001 - 2002 and 2002 - 2003?

0%
133.45
0%
142.48
0%
117.6
0%
None of these

The ratio of Hindustan Motors production in 2003 - 2004 to Honda's production in 2002 - 2003 is?

0%
0.66
0%
1.5
0%
2
0%
None of these

What was the ratio of investment in 1997 over the investment in 1992?

0%
5.50
0%
5.36
0%
5.64
0%
5.75

Find the mean deviation about the mean of the following data:

$15,17,10,13,7,18,9,6,14,11$ .

0%
$3.1$
0%
$3.2$
0%
$3.3$
0%
$3.4$

Explanation

Let the mean of data be

$u=\dfrac{\sum x_i}{n}=\dfrac {120}{10}=12$
Calculating the deviation :

$|x_i-u|$ we get,

$3,5,-2,1,5,6,3,6,2,1$ .

$\therefore M.D (u)= \dfrac{\sum{|x_i-u|}}{n}= \dfrac{34}{10}=3.4$
Thus,

$M.D(u) =3.4$ .
Ans- Option

$D$

Find odd one out:
Mean, Median, Mode, Variance

0%
Mean
0%
Median
0%
Mode
0%
Variance

Find the variance of the following data:

$5,9,8,12,6,10,6,8$

0%
$4.765$
0%
$4.750$
0%
$4.450$
0%
$4.420$

Explanation

Here

$n=8$ .
Mean,

$u=\dfrac{\sum x_i}{n}=\dfrac{64}{8}=8$
Value of Deviations

$(x_i-u)$ are

$-3,1,0,4,-2,2,-2,0$
Variance,

$\sigma ^2=\dfrac{\sum(x_i-u)^2}{n}=\dfrac{38}{8}=4.75$ .
Ans- Option

$B$ .

${\sigma}^2$ is known as

0%
mean
0%
variance
0%
Standard deviation
0%
quartile deviation

Explanation

Variance gives the degree to which the numerical data tend to spread about the value.

Variance is denoted by

$\sigma^2$

Variance is given by

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

But mean

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

Therefore, variance is

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

Find odd man out:
Standard Deviation, Variance, Mean, Quartile deviation.

0%
Standard Deviation
0%
Variance
0%
Mean
0%
Quartile deviation

The formula to find SD is

0%
$\sqrt{\dfrac{\sum( x-\overline x)^2}{n}}$
0%
$\sqrt{\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)}$
0%
$\sqrt {\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)^2}$
0%
${\dfrac{\sum x}{n}-\left(\dfrac{\sum x}{2}\right)^2}$

Explanation

$\Rightarrow$ The standard deviation is a statistic that measures the dispersion of a data set relative to its mean and is calculated as the square root of the variance.

$\sigma=\sqrt{\dfrac{\sum (x-\overline{x})^2}{n}}$

Where,

$\sigma=$ Standard deviation

$\overline{x}=$ Mean of data set

$n=$ the number of data point in data set

Standard deviation formula can be written as,

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

Find the variance of first

$10$ multiples of

$3$ .

0%
$72.65$
0%
$74.05$
0%
$74.25$
0%
$73.85$

Explanation

First

$10$ multiples of

$3$ are

$3,6,9...30$ .
This is an A.P.

$sum=\dfrac n2 (a+l)= \dfrac {10}{2} \times (3+30)$

$\therefore sum=165$ .
Mean,

$u=\dfrac {sum}{n}=\dfrac{165}{10}$ .
Variance,

$\sigma ^2=\dfrac{\sum(x_i ^2)}{n}-u^2$ .

$\therefore \sigma ^2=\dfrac{3^2+6^2+...30^2}{10}-{16.5}^2$ .

$\therefore \sigma ^2=\dfrac{3\times(1^2+2^2+...10^2)}{10}-{16.5}^2$ .

$\therefore \sigma ^2=\dfrac{9\times 10\times (10+1)\times (2\times 10+1)}{6\times 10}-{16.5}^2$ .

$\therefore \sigma ^2=346.5-272.25$

$\therefore \sigma ^2=74.25$
Ans-Option

$C$ .

Find the Coefficient of Variation for Factory

$B$ :

0%
$0.15555 \%$
0%
$0.25714 \%$
0%
$0.36934 \%$
0%
$0.42548 \%$

Explanation

Coefficient of Variation,

$C.V=\dfrac{\text{S.D}}{\text{Mean}} \times 100 \%$
So,

$C.V_B=\dfrac{{S.D_B}}{{Mean_B}} \times 100 \%$
Therefore,

$C.V_B=\dfrac{\sqrt {81}}{3500} \times 100 \%$

$\Rightarrow C.V_A=0.25714 \%$ .

Standard Deviation of first n natural numbers.

0%
$\sqrt{\dfrac{n^2+1}{6}}$
0%
$\sqrt{\dfrac{n^2-1}{12}}$
0%
$\sqrt{\dfrac{n^2+1}{12}}$
0%
$\sqrt{\dfrac{n^2-1}{6}}$

Explanation

${\textbf{Step 1: Finding mean}}$

${\text{The first }n\text{ natural numbers are }1,2,3,.....,n.}$

${\text{Their mean, }}\mathop x\limits^\_ {\text{ = }}\dfrac{{\sum x }}{n}{\text{ }}$

${\text{ = }}\dfrac{{1 + 2 + 3 + ....... + n}}{n}$

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

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

${\text{Sum of the square of the first n natural numbers is }}\sum {{x^2}} {\text{ = }}\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}{\text{ }}$

${\textbf{Step 2: Finding standard deviation}}$

${\text{Thus, the standard deviation }}\sigma {\text{ = }}\sqrt {\dfrac{{\sum {{x^2}} }}{n} - {{\left( {\dfrac{{\sum x }}{n}} \right)}^2}}$

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

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

$= \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left[ {\dfrac{{2n + 1}}{3} - \dfrac{{n + 1}}{2}} \right]}$

$= \sqrt {\left( {\dfrac{{n + 1}}{2}} \right)\left[ {\dfrac{{2(2n + 1) - 3(n + 1)}}{6}} \right]}$

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

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

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

${\text{Hence, the S}}{\text{.D}}{\text{. of the first n natural numbers is }}\sqrt {\dfrac{{{n^2} - 1}}{{12}}}$

${\textbf{ Hence, the correct answer is option B}}{\text{.}}$

Which factory has more variation in wages?

0%
$A$
0%
$B$
0%
Equal Variation
0%
Cannot be determined.

Explanation

$(\sigma _A) ^2=64$ and

$(\sigma _B) ^2=81$

$\because S.D=\sqrt {\sigma ^2}$

$\therefore S.D_A=8$ and

$S.D_B=9$
Since monthly mean wages in two factories are same, and

$S.D_A < S.D_B$ .
Thus, the factory

$B$ has more variation in wages.
Ans-Option

$B$ .

Variance of first

$n$ natural numbers.

0%
$\dfrac{n^2-1}{6}$
0%
$\dfrac{n^2+1}{6}$
0%
$\dfrac{n^2-1}{12}$
0%
$\dfrac{n^2+1}{12}$

Explanation

${\textbf{Step - 1: Finding mean of first n natural numbers}}$

${\text{We know that, mean of n observations = }}\dfrac{{{\text{Sum of n observations}}}}{{\text{n}}}$

$\therefore {\text{ Mean = }}\dfrac{{{\text{1 + 2 + 3 + }}.....{\text{ + n}}}}{{\text{n}}}$

${\text{We know that, sum of first n natural numbers = }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)}}{{\text{2}}}$

$\Rightarrow {\text{ Mean = }}\dfrac{{\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)}}{{\text{2}}}}}{{\text{n}}}$

$\Rightarrow {\text{ Mean = }}\dfrac{{{\text{n + 1}}}}{{\text{2}}}$

${\textbf{Step - 2: Calculating variance}}$

${\text{We know that, variance = }}\dfrac{{\sum {{{\text{x}}_{\text{i}}}^{\text{2}}} }}{{\text{n}}}{\text{ - }}{\left( {{\text{Mean}}} \right)^{\text{2}}}$

$\therefore {\text{ Variance = }}\dfrac{{{{\text{1}}^{\text{2}}}{\text{ + }}{{\text{2}}^{\text{2}}}{\text{ + }}{{\text{3}}^{\text{2}}}{\text{ + }}....{\text{ + }}{{\text{n}}^{\text{2}}}}}{{\text{n}}}{\text{ - }}{\left( {\dfrac{{{\text{n + 1}}}}{{\text{2}}}} \right)^{\text{2}}}$

${\text{Also, sum of squares of first n natural numbers = }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)\left( {{\text{2n + 1}}} \right)}}{{\text{6}}}$

$\Rightarrow {\text{ Variance = }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)\left( {{\text{2n + 1}}} \right)}}{{{\text{6n}}}}{\text{ - }}\dfrac{{{{{\text{(n + 1)}}}^{\text{2}}}}}{{\text{4}}}$

$\Rightarrow {\text{ Variance = }}\dfrac{{\left( {{\text{n + 1}}} \right)\left( {{\text{2n + 1}}} \right)}}{{\text{6}}}{\text{ - }}\dfrac{{{{{\text{(n + 1)}}}^{\text{2}}}}}{{\text{4}}}$

$\Rightarrow {\text{ Variance = }}\dfrac{{{\text{2}}{{\text{n}}^{\text{2}}}{\text{ + 3n + 1}}}}{{\text{6}}}{\text{ - }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{ + 2n + 1}}}}{{\text{4}}}$

$\Rightarrow {\text{ Variance = }}\dfrac{{{\text{4}}{{\text{n}}^{\text{2}}}{\text{ + 6n + 2 - 3}}{{\text{n}}^{\text{2}}}{\text{ - 6n - 3}}}}{{{\text{12}}}}$

$\Rightarrow {\text{ Variance = }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{ - 1}}}}{{{\text{12}}}}$

$\mathbf{{\text{Thus, the variance of first n natural numbers is }}\dfrac{{{{\text{n}}^{\text{2}}}{\text{ - 1}}}}{{{\text{12}}}}.}$

Find the Standard Deviation of of first

$10$ multiples of

$3$ .

0%
$8.61$
0%
$8.33$
0%
$9.09$
0%
$9.69$

Explanation

First

$10$ multiples of

$3$ are

$3,6,9...30$ .
This is an A.P.

$sum=\dfrac n2 (a+l)= \dfrac {10}{2} \times (3+30)$

$\therefore sum=165$ .
Mean,

$u=\dfrac {sum}{n}=\dfrac{165}{10}$ .
Variance,

$\sigma ^2=\dfrac{\sum(x_i ^2)}{n}-u^2$ .

$\therefore \sigma ^2=\dfrac{3^2+6^2+...30^2}{10}-{16.5}^2$ .

$\therefore \sigma ^2=\dfrac{3\times(1^2+2^2+...10^2}{10}-{16.5}^2$ .

$\therefore \sigma ^2=\dfrac{9\times 10\times (10+1)\times (2\times 10+1)}{6\times 10}-{16.5}^2$ .

$\therefore \sigma ^2=346.5-272.25$

$\therefore \sigma ^2=74.25$
Standard Deviation,

$S.D=\sqrt{ \sigma ^2}$

$\therefore S.D=\sqrt{74.25}$
Thus,

$S.D=8.61$
Ans-Option

$A$ .

The variance of six observations is given as

$16$ and mean is

$8$ . If each observation is multiplied by

$3$ .
Find the new variance:

0%
$155$
0%
$124$
0%
$177$
0%
$144$

Explanation

Mean,

$u=8$ .
So,

$\dfrac{x_1+x_2...x_6}{6}=8$ .

$\therefore x_1+x_2...x_6=48$ .
Multiplying the above equation by

$3$ on both the sides we get,

$3x_1+3x_2...3x_6=48\times 3$

$\therefore u'=\dfrac{3x_1+3x_2....3x_6}{6}=\dfrac{48 \times 3}{6}$

$\therefore u'=24$
Thus, new mean,

$u'=24$ .

$\therefore$ New variance,

$\sigma ^2=\dfrac{(3x_1)^2+...(3x_6)^2}{6}-(24)^2$

$\therefore \sigma ^2=9\times \dfrac{(x_1)^2+....(x_6)^2}{6}-576$

$\therefore \sigma ^2=9\times \dfrac{480}{16}-576$

$\therefore \sigma ^2=720-576$

$\therefore \sigma ^2=144$
Thus, New Variance is

$144$ .
Ans-Option

$D$

Find the Coefficient of Variation of Factory

$A$ .

0%
$0.22857 \%$
0%
$0.25758 \%$
0%
$0.39462 \%$
0%
$0.82547 \%$

Explanation

Coefficient of Variation,

$C.V=\dfrac{S.D}{Mean} \times 100 \%$
So,

$C.V_A=\dfrac{S.D_A}{Mean_A} \times 100 \%$

$\therefore C.V_A=\dfrac{\sqrt {64}}{3500} \times 100 \%$ .

$\therefore C.V_A=0.22857 \%$ .
Ans-Option

$A$ .

Which of the following is not a measure of central location?

0%
Mean
0%
Median
0%
Mode
0%
Variance

Find the Standard Deviation of the following data:

$5,9,8,12,6,10,6,8$

0%
$2.14$
0%
$2.16$
0%
$2.15$
0%
$2.17$

Explanation

Here

$n=8$ .
Mean,

$u=\dfrac{\sum x_i}{n}=\dfrac{64}{8}=8$
Value of Deviations

$(x_i-u)$ are

$-3,1,0,4,-2,2,-2,0$
Variance,

$\sigma ^2=\dfrac{\sum(x_i-u)^2}{n}\\=\dfrac{(-3-8)^2-(1-8)^2-(0-8)^2-(4-8)^2-(-2-8)^2-(2-8)^2-(-2-8)^2-(0-8)^2}{8}\\=\dfrac{38}{8}=4.75$ .
Standard Deviation,

$S.D=\sqrt{\sigma ^2}$

$\therefore S.D=\sqrt{4.75}=2.17$
Ans- Option

$D$ .

3 weeks?

0%
$7cm$
0%
$8cm$
0%
$9cm$
0%
$10cm$

Explanation

In the given graph, The straight line represents the growth of the Plant

$B$ .

From the graph, we can observe that after

$3$ weeks Plant

$B$ has grown

$10\space cm$

Quartile deviation of the following numbers

$10, 8, 12, 11, 14, 9, 6$ is?

0%
$2$
0%
$10$
0%
$1/2$
0%
$1$

Explanation

Arranging given numbers in ascending order,

$6,8,9,10,11,12,14$

Number of observation

$(N)=7$

$Q_i=\left\{i\left(\dfrac{N+1}{4}\right)\right\}^{th}term$

$Q_1=\left\{1.\left(\dfrac{7+1}{4}\right)\right\}^{th}term=2^{nd}term$

$2^{nd}term=8$

$\therefore$

$Q_1=8$

$Q_3=\left\{3.\left(\dfrac{7+1}{4}\right)\right\}^{th}term=6^{th}term$

$6^{th}term=12$

$\therefore$

$Q_3=12$

$\therefore$ Quartile deviation

$=\dfrac{1}{2}(Q_3-Q_1)$

$=\dfrac{1}{2}(12-8)$

$=\dfrac{4}{2}$

$=2$

The variance of

$10, 10, 10, 10, 10$ , is

0%
$10$
0%
$\sqrt {10}$
0%
$0$
0%
$5$

If t is the standard deviation of x, y, z then the standard deviation of x + 5, y + 5, z + 5 is

0%
$\dfrac{t}{3}$
0%
$t + 5$
0%
$t$
0%
$xyz$

Explanation

$\text{Standard deviation is invariant under change of origin}$

$\text{Hence, standard deviation of x+5,y+5, z+5 is same as x, y, z}$

$\Rightarrow \text{ standard deviation of x+5,y+5, z+5 is t}$ .
Option C is correct.

The given pictograph shows the number of wall clocks sold by a company during a week..Each clock in the pictograph represents

$5$ wall clocks. How many more wall clocks were sold on Thursday than on Tuesday?

0%
$9$
0%
$8$
0%
$5$
0%
$10$

Explanation

Number of wall clocks sold on Tuesday

$5$

$\times$

$5 = 25$
Number of wall clocks sold on Thursday =

$5$

$\times$

$7 = 35$

$\therefore$ Required difference =

$35 - 25 = 10$

What is the variance of the first

$11$ natural numbers?

0%
$10$
0%
$11$
0%
$12$
0%
$13$

Explanation

Sum of first

$11$ natural numbers

$=\dfrac { n(n+1) }{ 2 } =\dfrac { 11\times 12 }{ 2 } =66$

So, mean

$=\dfrac { 66 }{ 11 } =6$

Variance

$= { \dfrac { (1-6)^{ 2 }+(2-6)^{ 2 }+(3-6)^{ 2 }+........+(11-6)^{ 2 } }{ 11 } }$

$= { \dfrac { 25+16+9+4+1+0+1+4+9+16+25 }{ 11 } }=\dfrac { 110 }{ 11 } =10$

Hence, A is correct.

If the standard deviation of

$0,1,2,3......,9$ is

$K$ , then the standard deviation of

$10,11,12,13,........19$ is

0%
$K$
0%
$K+10$
0%
$K+\sqrt { 10 }$
0%
$10K$

Explanation

As the standard deviation only depend upon total no of values and the difference between mean and each value,

Both sequence have same standard deviation.

Note:

$1^{st}$ sequence is

$x_i$ and

$2^{nd}$ sequence is

$y_i$ ,

$y_i=10+x_i\Rightarrow \overline y=10+\overline x$

So,

$\overline y-y_i=\overline x-x_i$

ie, Option A is correct answer.

CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 3 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 3 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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