MCQ Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 2

__________ is the positive square root of the mean of squared deviations from mean.

0%
Mean Deviation
0%
Standard Deviation
0%
Quartile Deviation
0%
None of the above

What is standard error of mean defined as?

0%
Standard deviation of the sampling distribution of mean
0%
Standard deviation of data
0%
Inter quartile range of the data
0%
None of these

If the coefficient of correlation between

$x,y$ is

$0.7$ and covariance is

$35$ find the standard deviation of

$y$ if the standard deviation of

$x$ is

$5$ ____.

0%
$10$
0%
$9$
0%
$8$
0%
$11$

The formula Of

$X^2$ distribution is______________.

0%
${X^2} = \dfrac {{S}^{2}_{1}}{{S}^{2}_{2}}$
0%
${X^2} = \dfrac {E}{{\Sigma}(O - E)^2}$
0%
${X^2} = {\Sigma} \dfrac {E}{(E - O)^2}$
0%
${X^2} = {\Sigma} \dfrac {(O - E)^2}{E}$

Range =

0%
Largest value - Smallest value
0%
Largest value divided by 2
0%
Both A and B
0%
None of the above

The coefficient correlation between x and y is 0.8, the covariance beingIf the standard deviation of x is 4 find the standard deviation of y___.

0%
6.25
0%
10
0%
11.25
0%
9.10

If a population has a standard deviation of

$2.25$ , how large the sample should be to allow a maximum error of

$0.33$ with

$95\%$ confidence level?

0%
$252$
0%
$201$
0%
$220$
0%
$178$

If the coefficient of correlation between x,y is 0.8 and covariance is 32 find the standard deviation of X if the standard deviation of y is 4___.

0%
10
0%
9
0%
8
0%
11

If Quartile deviation of a data is

$8$ , find Mean deviation____.

0%
$9.0$
0%
$9.6$
0%
$9.90$
0%
$10.10$

The standard deviation of sample of 30 observations isIf the value of each item of the observation is increased byThe new standard deviation will be ______.

0%
same as original
0%
increased by 5
0%
reduced by 5
0%
reduced by 5 times.

If the size of the sample increases, the curve becomes___________.

0%
continuous
0%
severely skewed to one side
0%
more symmetrical
0%
intermittent

Find the mean and standard deviation of the following observations:

$X = 2, 5, 7, 8, 13$ .

0%
$7$ and $3.63$
0%
$3.63$ and $7$
0%
$7.63$ and $3$
0%
$3$ and $7.63$

The geometric mean of

$3,7,11,15,24,28,30,0$ is

0%
$6$
0%
$0$
0%
$8$
0%
$9$

A frequency distribution having two modes is said to be ________.

0%
unimodal
0%
bimodal
0%
trimodal
0%
without mode

Geometric mean of two observations can be calculated only if ________.

0%
both the observations are positive
0%
one of the two observations is zero
0%
one of them is negative
0%
both of them are zero

If the two observations are

$10$ and

$0$ then their geometric mean is ______.

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

Which of the measure of central tendency is not affected by extreme values?

0%
Mode
0%
Median
0%
Sixth decile
0%
All the above

Shoe size of most of the people in a city is

$No.7$ . Which measure of central value does it represent?

0%
Mean
0%
Median
0%
Eighth decile
0%
Mode

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

$C$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$A$ .

If the coefficient of variation and standard deviation of a distribution are 50% and 20 respectively, then its mean is

0%
40
0%
30
0%
20
0%
none of these

Explanation

Given

$\sigma = 20$ , coefficient of variation

$=50$ %
We know coefficient of variation

$=\cfrac{\sigma }{\bar{x}}\times 100=50$

$\Rightarrow \bar{x} = 2\times \sigma = 40$

The variance of first n natural numbers is

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

Explanation

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

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

$\text{The Sum of } n \text{ natural number is } \dfrac{{n(n+1)}}{{2}}$

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

$\textbf{Step - 2: Finding the Variance}$

$\text{Variance} = \dfrac{{\sum(x_i)^2}}{{n}} - (\text{Mean})^2$

$\sum(x_i)^2 = \dfrac{{1^2 +2^2 + ...+ n^2}}{{n}}$

$\text{Since } 1^2 +2^2 + ...+ n^2 = \dfrac{{n(n+1)(2n+1)}}{{6}}$

$\dfrac{{\sum(x_i)^2}}{{n}} = \dfrac{{n(n+1)(2n+1)}}{{6n}}$

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

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

$= \dfrac{{(n+1)}}{{2}}(\dfrac{{4n+2-3n-3}}{{6}})$

$=\dfrac{{n+1}}{{2}}\times\dfrac{{n-1}}{{6}}$

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

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

The mean deviation about median from the data

$340,150,210,240,300,310,320$ is

0%
$50$
0%
$52.8$
0%
$55$
0%
$45$

Explanation

Arrange in increasing order
150,210,240,300,310,320,340 N=number of data = 7
median (M) =300
The mean deviation about median

$= \frac{150+90+60+0+10+20+40}{7 }= 52.857$

Coefficient of deviation is calculated by the formula:

0%
$\cfrac { \bar { X } }{ \sigma } \times 100$
0%
$\cfrac { \bar { X } }{ \sigma }$
0%
$\cfrac { \sigma } {\bar { X }} \times 100$
0%
$\cfrac{ \sigma } { \bar { X }}$

Explanation

It is a fundamental concept.
coefficient of deviation

$=\cfrac{\sigma}{\bar{x}}\times 100$
where

$\sigma$ and

$\bar{x}$ are standard deviation and mean respectively.

When all the observations are same, the relation between A.M., G.M and H.M is _________.

0%
A.M. = G.M. = H.M.
0%
A.M. = G.M. > H.M.
0%
A.M. > G.M. > H.M.
0%
A.M. < G.M. < H.M.

The mean of distribution is

$22.2$ and its mode is

$23.3$ . The median is _______.

0%
$26.5$
0%
$22.6$
0%
$28.5$
0%
$26.6$

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

$D$ .

Standard Error of Mean is defined as ________.

0%
Standard deviation of the sampling distribution of mean
0%
Average of sampling distribution of mean
0%
Inter-Quartile range of sampling distribution of mean
0%
Correlation co-efficient between the sampling distribution of mean and mean

CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 2 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 2 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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