The variance of the data: x: $$1$$ $$a$$ $${ a }^{ 2 }$$ ...... $${ a }^{ n }$$ f: $${ _{ }^{ n }{ C } }_{ 0 }$$ $${ _{ }^{ n }{ C } }_{ 1 }$$ $${ _{ }^{ n }{ C } }_{ 2 }$$ ...... $${ _{ }^{ n }{ C } }_{ n }$$ is

$${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{2n }$$

$${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{ n }$$

$${ \left( \cfrac { 1+a }{ 2 } \right) }^{ 2n }-{ \left( \cfrac { 1+{ { a }^{ 2 } } }{ 2 } \right) }^{ n }$$

MCQ Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 5

If a variable

$x$ takes values

$0,1,2,....n$ with frequencies proportional to the binomial coefficients

${ _{ }^{ n }{ C } }_{ 0 },{ _{ }^{ n }{ C } }_{ 1 },{ _{ }^{ n }{ C } }_{ 2 },......{ _{ }^{ n }{ C } }_{ n }$ , then mean of distribution is

0%
$\dfrac {n(n+1)}{2}$
0%
$\dfrac { n }{ 2 }$
0%
$\dfrac { 2 }{ n }$
0%
$\dfrac {n(n-1)}{2}$

Explanation

Here,

$\displaystyle \mu1 '=\frac{\sum r\frac{n}{r}^{n-1}C_{r-1}}{\sum ^{n}C_{r}}$

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

$\displaystyle \mu 2'=\frac{1}{2^{n}}\sum_{0}^{n}\left\{ \left ( r-1 \right )+r \right \}^{n}C_{r}$

$\displaystyle=\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )^{n}C_{r}+\frac{n}{2}$

$\displaystyle =\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )\frac{n\left ( n-1 \right )}{r\left ( r-1 \right )}^{n-2}C_{r-2}+\frac{n}{2}$

$\displaystyle =\frac{n\left ( n-1 \right)}{2^{n}}.2^{n-2}+\frac{n}{2}$

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

$\therefore$ Variance

$\displaystyle \sigma^2=\mu 2'-\left (\mu 1' \right )^{2}$

$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}-\left ( \frac{n}{2} \right )^{2}=\frac{n}{4}$

Consider the frequency distribution

Class interval:	0-6	6-12	12-18
Frequency:	2	4	6

The variance of the above frequency distribution, is

0%
$24$
0%
$12$
0%
$20$
0%
$25$

Explanation

Class interval	$f_i$	$x_i$	$d_i={x_i-A}$	$d_i^2$	$f_id_i^2$	$f_id_i$
0-6	2	3	-6	36	72	-12
6-12	4	9	0	0	0	0
12-18	6	15	6	36	216	36

Here,

$N=\sum {f_i }=12$

$\sum {f_id_i^2}=288$

$\sum {f_id_i}=24$

Now, variance

$\displaystyle { \sigma }^{ 2 }=\frac { \sum { f_{ i }d_{ i }^{ 2 } } }{ N } -{ \left( \frac { \sum { f_{ i }d_{ i } } }{ N } \right) }^{ 2 }$

$\Rightarrow \displaystyle { \sigma }^{ 2 }=\frac { 288 }{ 12 } -{ \left( \frac { 24 }{ 12 } \right) }^{ 2 }$

$\Rightarrow { \sigma }^{ 2 }=20$

If the mean deviation about the median of the numbers

$a,2a,3a,.....50a$ is

$50$ , then

$\left| a \right|$ equals

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

Given data is

$a,2a, 3a,....49a,50a$
Here,

$n=50 (even)$
So, median

$M= \dfrac{\text{value of }25^{th}\text{observation}+\text{value of }26^{th}\text{observation}}{2}$

$M=\dfrac{25a+26a}{2}=25.5a$

Mean deviation about median

$M.D=\dfrac{\sum |x_i-25.5a|}{50}$

$\Rightarrow 50=\dfrac{|a-25.5a|+|2a-25.5a|+.......+|24a-25.5a|+|25a-25.5a|+|26a-25.5a|+.....|50a-25.5a|}{50}$

$\Rightarrow 2500=|a|+2(1.5+2.5+.....24.5)|a|$

$\Rightarrow 2500=|a|+2\times 12 (3+23)|a|$

$\Rightarrow |a|=\dfrac{2500}{625}=4$

The variance of first

$50$ even natural numbers is

0%
$\displaystyle \frac{833}{4}$
0%
$833$
0%
$437$
0%
$\displaystyle \frac{437}{4}$

Explanation

${\textbf{Step-1: Use mean formula and find the mean.}}$

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

$\Rightarrow n = 50, \sum x_i = 2 +4+6+8+...+100$

${\text{We know that,}}$

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

$\Rightarrow \bar {x} = \dfrac{2 +4+6+8+...+100}{50}$

$\Rightarrow \bar {x} = \dfrac{50 \times 51}{50}$

$[\because \sum 2n = n(n+1)]$

$\Rightarrow \bar {x} = 51$

${\textbf{Step-2: Put the values in variance formula.}}$

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

$= \dfrac{1}{50} (2^2 +4^2+6^2+8^2+...+100^2) - {(51)}^2$

$= \dfrac{1}{50} [({2.1})^2 +({2.2})^2+({2.3})^2+({2.4})^2+...+({2.50})^2] - {(51)}^2$

$= \dfrac{1}{50} 2^2[({1})^2 +({2})^2+({3})^2+({4})^2+...+({50})^2] - {(51)}^2 ...(1)$

${\text{But,}}$

$\Rightarrow ({1})^2 +({2})^2+({3})^2+({4})^2+...+({n})^2 = \dfrac{n(n+1)(2n+1)}{6}$

$\Rightarrow ({1})^2 +({2})^2+({3})^2+({4})^2+...+({50})^2 = \dfrac{50(50+1)(2 \times 50+1)}{6}$

$\Rightarrow ({1})^2 +({2})^2+({3})^2+({4})^2+...+({50})^2 = \dfrac{50(51)(101)}{6}$

${\text{Equation (1) become,}}$

$= \dfrac{1}{50} 2^2[\dfrac{50(51)(101)}{6}] - {(51)}^2$

$= 34 \times 101 -2601$

$= 3434-2601$

$=833$

${\textbf{Thus, option B is correct.}}$

The coefficient of variation for the given distribution is

0%
$28.6$
0%
$0.306$
0%
$30.6$
0%
$306$

Explanation

Lett

$A=10$

$Size$ $x_i$	$Frequency$ $f_i$	$f_ix_i$	$(x_i-A)^2$	$f_i(x_i-A)^2$
$4$	$1$	$4$	$36$	$36$
$6$	$2$	$12$	$16$	$32$
$8$	$3$	$24$	$4$	$12$
$10$	$5$	$50$	$0$	$0$
$12$	$3$	$36$	$4$	$12$
$14$	$2$	$28$	$16$	$32$
$16$	$1$	$16$	$36$	$16$
	$\sum f_i=17$	$\sum f_ix_i=170$		$\sum f_i(x_i-A)^2=140$

$\Rightarrow$

$\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{170}{17}=10$

Now,

$S.D (\sigma)=\sqrt{\dfrac{\sum f_i(x_i-A)^2}{\sum f_i}}$

$=\sqrt{\dfrac{140}{17}}$

$=2.86$

$\Rightarrow$ Coefficient of variation

$=\dfrac{\sigma}{\overline{x}}\times 100$

$=\dfrac{2.86}{10}\times 100$

$=28.6$

The coefficient of dispersion for the given distribution is

0%
3.06
0%
.306
0%
30.6
0%
None of these

Explanation

First of all find the mean of the given frequency distribution after getting mean, then find deviation for each size.

$\begin{matrix} size\left ( x \right ) & frequency\left ( f \right ) & f(x) & Deviation x=X-\bar{x}\left ( say\:\bar{x}=10 \right ) & x^{2} & fx^{2}\\ 4 & 1 & 4 & -6 & 36 & 36\\ 6 & 2 & 12 & -4 & 16 & 32\\ 8 & 3 & 24 & -2 & 4 & 12\\ 10 & 5 & 50 & 0 & 00 & 0\\ 12 & 3 & 36 & +2 & 4 & 12\\ 14 & 2 & 28 & +4 & 16 & 32\\ 16 & 1 & 16 & +6 & 36 & 36\\ & \sum f=N=17 & \sum f\left ( x \right )=170 & & & \sum f(x^{2})=160 \end{matrix}$

$\therefore$

$\displaystyle \bar{x}=\frac{170}{17}=10$
Hence coefficient of dispersion (variation)

$\displaystyle =\frac{\sigma }{\bar{X}}\times 100=0.306$

Standard deviation for first 10 natural numbers is

0%
$5.5$
0%
$3.87$
0%
$2.97$
0%
$2.87$

Explanation

Standard deviation of first

$n$ natural number is

$\displaystyle =\sqrt{\frac{n^2-1}{12}}=\sqrt{\frac{10^2-1}{12}}=2.87$

Standard deviation for first 10 even natural numbers is

0%
$11$
0%
$7.74$
0%
$5.74$
0%
$11.48$

Explanation

We know standard deviation of first

$n$ even natural number is

$, \sigma =\sqrt{\cfrac{n^2-1}{3}}$
Here

$n =10$

$\therefore \sigma = \sqrt{\cfrac{100-1}{3}}=\sqrt{33}=5.74$

The coefficients of variation of two series are

$58$ % and

$69$ %. If their standard deviations are

$21.2$ and

$15.6$ , then their A.M's are

0%
$36.6,22.6$
0%
$34.8,22.6$
0%
$36.6,24.4$
0%
None of these

Explanation

We know that

$\displaystyle C.V.=\frac { \sigma \times 100 }{ \overline { x } } \Rightarrow \overline { x } =\frac { \sigma }{ C.V. } \times 100$

$\therefore$ Mean of first series

$\displaystyle =\frac { 21.2\times 100 }{ 58 } =36.6$
Mean of second series

$\displaystyle =\frac { 15.6\times 100 }{ 69 } =22.6$

The Coefficient of Variation is given by:

0%
$\dfrac{Mean}{\ Standard \ \ deviation } \times 100$
0%
$\dfrac{\ Standard \ \ deviation }{Mean}$
0%
$\dfrac{Standard \ \ deviation }{Mean }\times 100$
0%
$\dfrac{Mean}{Standard \ Deviation}$

Explanation

The coefficient of variation (CV) is a standardized measure of dispersion

. It is defined as the ratio of the standard deviation to the mean.

$CV\quad =\quad \cfrac { \sigma }{ Mean }\times100$

If the coefficient of variation of some observation is 60 and their standard deviation is 20, then their mean is

0%
35
0%
34
0%
33
0%
33.33(nearly)

Explanation

We know coefficient of variation C. V.

$\displaystyle =\frac{\sigma _{x}}{\bar{x}}\times 100$

$\therefore$

$\displaystyle 60=\frac{20}{\bar{x}}\times 100$

$\therefore$

$\bar{x}=33.33$ (nearly)

In a final examination in Statistics the mean marks of a group of

$150$ students were

$78$ and the S.D was

$8.0$ . In Economics, however, the mean marks were

$73$ and the S.S was

$7.6$ . The variability in the two subjects respectively is

0%
$10.3\%,10.4\%$
0%
$95\%,7.9\%$
0%
$11.2\%,10.1\%$
0%
none of these

Explanation

Coefficient of variance

$=\dfrac{\sigma}{\bar{x}} \times 100$

For Statistics,

$\sigma =8, \bar x=78$
So, coefficient of variation

$=\dfrac{8}{78} \times 100=10.3$ %

For economics,

$\sigma =7.6, \bar x=73$
So, coefficient of variation

$=\dfrac{7.6}{73} \times 100=10.4$ %

If mean of a series is 40 and variance 1486, then coefficient of variation is

0%
$0.9021$
0%
$0.9637$
0%
$0.8864$
0%
$0.9853$

Explanation

If mean of the given dist. be

$\bar{x}$ and S.D be

$\sigma$
then given

$\bar{x} = 40, \sigma^2 = 1486$

$\therefore$ Coefficient of variation

$=\cfrac{\sigma}{\bar{x}}=\cfrac{\sqrt{1486}}{40}=.9637$

If the sample size is small, the curve may be__________.

0%
a continuous one
0%
more symmetrical
0%
severely skewed to one side
0%
none of the above

The variance of the data:

x:	$1$	$a$	${ a }^{ 2 }$	......	${ a }^{ n }$
f:	${ _{ }^{ n }{ C } }_{ 0 }$	${ _{ }^{ n }{ C } }_{ 1 }$	${ _{ }^{ n }{ C } }_{ 2 }$	......	${ _{ }^{ n }{ C } }_{ n }$

0%
${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{ n }$
0%
${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{2n }$
0%
${ \left( \cfrac { 1+a }{ 2 } \right) }^{ 2n }-{ \left( \cfrac { 1+{ { a }^{ 2 } } }{ 2 } \right) }^{ n }$
0%
none of these

Explanation

$\displaystyle \bar { X } =\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } \\ \displaystyle =\frac { ^{ n }C_{ 0 }+a.^{ n }C_{ 1 }+{ a }^{ 2 }.^{ n }C_{ 2 }+....{ a }^{ n }.^{ n }C_{ n } }{ ^{ n }C_{ 0 }+^{ n }C_{ 1 }+^{ n }C_{ 2 }+....^{ n }C_{ n } } \\ \displaystyle =\frac { { \left( 1+a \right) }^{ n } }{ { 2 }^{ n } }$
.

$\displaystyle S.D.=\sqrt { \frac { \sum { { f }_{ i }{ { x }_{ i } }^{ 2 } } }{ \sum { { f }_{ i } } } -{ \left( \frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } \right) }^{ 2 } } \\ \displaystyle =\sqrt { \frac { ^{ n }C_{ 0 }+{ a }^{ 2 }.^{ n }C_{ 1 }+{ a }^{ 4 }.^{ n }C_{ 2 }+....{ a }^{ 2n }.^{ n }C_{ n } }{ ^{ n }C_{ 0 }+^{ n }C_{ 1 }+^{ n }C_{ 2 }+....^{ n }C_{ n } } -{ \left[ \frac { { \left( 1+a \right) }^{ n } }{ { 2 }^{ n } } \right] }^{ 2 } } \\ \displaystyle \sqrt { \frac { { \left( 1+{ a }^{ 2 } \right) }^{ n } }{ { 2 }^{ n } } -{ \left( \frac { 1+a }{ 2 } \right) }^{ 2n } }$

$Variance = (S.D)^2$

Mean deviation of the observations 70, 42, 63,34, 44, 54, 55, 46, 38, 48 from median is

0%
$7.8$
0%
$8.6$
0%
$7.6$
0%
$8.8$

Explanation

Arranging the given data in ascending order-

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

$\because n = 10$

Therefore,

Median

$\left( M \right) = \cfrac{{\left( \cfrac{n}{2} \right)}^{th} \text{ term } + {\left( \cfrac{n}{2} + 1 \right)}^{th} \text{ term}}{2}$

$\Rightarrow M = \cfrac{46 + 48}{2} = 47$

Therefore,

${x}_{i}$	$\left\| {x}_{i} - M \right\|$
$34$	13
$38$	9
$42$	5
$44$	3
$46$	1
$48$	1
$54$	7
$55$	8
$63$	16
$70$	23
	$\sum{\left\| {x}_{i} - M \right\|} = 86$

Therefore,

Mean deviation

$= \cfrac{\sum{\left| {x}_{i} - M \right|}}{n} \\ = \cfrac{86}{10} = 8.6$

Hence the mean deviation of the observations is

$8.6$ .

The S.D. of the following frequency distribution is

Class	0-10	10-20	20-30	30-40
$\displaystyle f_{i}$	1	3	4	2

0%
$7.8$
0%
$9$
0%
$8.1$
0%
$0.9$

Explanation

Class	$x_i$	$f_i$	$u_i=\dfrac{x_i-a}{h}$	$f_iu_i$	$f_iu_i^2$
0-10	5	1	-2	-2	4
10-20	15	3	-1	-3	3
20-30	25	4	0	0	0
30-40	35	2	1	2	2
		$N=10$		$\sum f_iu_i=-3$	$\sum f_iu_i^2=9$

$\displaystyle \sigma =h\sqrt{\frac{ \Sigma f_{1}u_{1}^{2}}{N}-\left ( \frac{\Sigma f_{1}u_{1}}{N} \right )^{2}}$

$\displaystyle =10\sqrt{\frac{9}{10}-\left ( \frac{-3}{10} \right )^{2}}=9$

The sum of the squares of deviation of 10 observations from their mean 50 is 250, then coefficient of varition is

0%
10%
0%
40%
0%
50%
0%
None of these

Explanation

Given

$\displaystyle \Sigma \left ( x_{i}-\overline{x} \right )^{2}=250$ ,

$n=10,\overline{x}=50$

Now,

$\sigma=\sqrt{\dfrac{1}{n}\Sigma \left ( x_{i}-\overline{x} \right )^{2}}$

$= \sqrt{\dfrac{1}{10}\times 250}=5$
Hence coefficient of variation

$\displaystyle =\dfrac{\sigma }{\overline{x}}\times 100=\dfrac{5}{50}\times 100=10$ %

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

0%
40
0%
30
0%
20
0%
None of these

Explanation

We know if a distribution having mean

$\bar{x}$ and standard deviation

$\sigma$
then coefficient of variation

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

$\therefore \cfrac{20}{\bar{x}}\times 100=50\Rightarrow \bar{x} = 40$
Hence required mean is

$=40$

The coefficient of range of the following distribution

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

0%
$0.4$
0%
$2.5$
0%
$8$
0%
$0.9$

Explanation

Here greatest term in the given observation is,

$x_m =14$
and least term is,

$x_l = 6$
Hence coefficient of range is

$=\cfrac{x_m-x_l}{x_m+x_l}=\cfrac{8}{20}=0.4$

The coefficient of mean deviation from median of observations

$40, 62, 54, 90, 68, 76$ is

0%
$2.16$
0%
$0.2$
0%
$5$
0%
None of these

Explanation

Arrange the given observations in ascending order

$40,54,62,68,76,90$
Here, number of terms

$n=6 (even)$

$\displaystyle \therefore$ Median (M)

$\displaystyle =\frac{\left ( \frac{n}{2} \right )th\:term+\left ( \frac{n}{2}+1 \right )th\:term}{2}=\frac{62+68}{2}=65$

$\Sigma \left | x_{i}-M \right |=25+11+3+3+11+25=78$
Mean deviation from median

$\displaystyle =\frac{\Sigma \left | x_{i}-M \right |}{n}=\frac{78}{6}=13$

$\therefore$ Coefficient of M.D.=

$\displaystyle =\frac{M.D.}{median}=\frac{13}{65}=0.2$

The S.D. of the following freq. dist.

Class	0-10	10-20	20-30	30-40
$f_i$	1	3	4	2

0%
$7.8$
0%
$9$
0%
$8.1$
0%
$0.9$

Explanation

Class	$x_i$	$f_i$	$u_i=\dfrac{x_i-A}{h}$	$u_i^2$	$f_iu_i$	$f_iu_i^2$
0-10	5	1	-2	4	-2	4
10-20	15	3	-1	1	-3	3
20-30	25	4	0	0	0	0
30-40	35	2	1	1	2	2
Total		10			-3	9

$S.D.$

$=h\sqrt { \cfrac { \sum { { f }_{ i }{ u }_{ i }^{ 2 } } }{ \sum { { f }_{ i } } } -{ \left( \cfrac { \sum { f } _{ i }{ u }_{ i } }{ \sum { { f }_{ i } } } \right) }^{ 2 } }$

$S.D.=10\sqrt { \cfrac { 9 }{ 10 } -{ \left( \cfrac { -3 }{ 10 } \right) }^{ 2 } } =9$

The mean of a dist. is

$4$ . if its coefficient of variation is

$58\%$ . Then the S.D. of the dist. is

0%
$2.23$
0%
$3.23$
0%
$2.32$
0%
None of these

Explanation

We know if a distribution having mean

$\bar{x}$ and standard deviation

$\sigma$
then coefficient of variation

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

$\therefore \cfrac{\sigma}{4}\times 100=58\Rightarrow \sigma = \cfrac{58}{25}=2.32$
Hence required standard deviation is

$=2.32$

The mean deviation from mean of observations 5, 10, 15, 20, ......85 is

0%
$43.71$
0%
$21.17$
0%
$38.7$
0%
None of these

Explanation

Given series is

$5,10,15,....,85$ , which is an AP.
Last term

$85=a+(n-1)d$

$\Rightarrow n=17$
Sum of

$n$ terms of AP

$=\dfrac{17}{2}(5+85)$

$\sum x_i=765$

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

$\bar x=\dfrac{765}{17}=45$

So, deviations from mean are

$40,35,25,20,15,10,5,0,5,10,15,20,25,30,35,40$
Sum of deviations

$=360$

Mean deviation from mean

$=\dfrac{360}{17}=21.17$

Mean deviations of the series

$a,a+d,a+2d,...,a+2nd$ from its mean is

0%
$\displaystyle \frac{n\left ( n+1 \right )d}{\left ( 2n+1 \right )}$
0%
$\displaystyle \frac{nd}{2n+1}$
0%
$\displaystyle \frac{\left ( n+1 \right )d}{2n+1}$
0%
$\displaystyle \frac{\left ( 2n+1 \right )d}{n\left ( n+1 \right )}$

Explanation

Clearly given observation is in A.P, and have

$2n+1$ terms.

$\therefore$ Mean

$\displaystyle =\frac{1}{2}\left \{ a+{a+2nd}\right \}=a+nd$

Use for

$x = a$ ,

$|x-\bar{x}| =|a+nd-a|=nd$

$x = a+d$ ,

$|x-\bar{x}| =|a+nd-a-d|=(n-1)d$

Similarly, write till

$x=a+2nd$ , to get sum of

$\sum \left | x-\bar{x} \right |$

And thus mean deviation about mean

$\displaystyle =\frac{1}{2n+1}\sum \left | x-\bar{x} \right |$

$\displaystyle =\frac{2}{2n+1}.d\left ( 1+2+3\cdots +n \right )$

$\displaystyle =\frac{n\left ( n+1 \right )d}{(2n+1)}$

The mean of a distribution isIf its coefficient of variation is 58%. Then the S.D. of the distribution is

0%
2.23
0%
3.23
0%
2.32
0%
none of these

Explanation

Given, mean

$\bar{x} = 4,$ and coefficient of variation

$=58$ %
If S.D of the given distribution is

$\sigma$ then we know that,
Coefficient of variation

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

$\Rightarrow 58 = \cfrac{\sigma}{4}\times 100\Rightarrow \sigma = \cfrac{58\times 4}{100}=2.32$

The coefficient of mean deviation from median of observations 40, 62, 54, 90, 68, 76 is

0%
2.16
0%
1.2
0%
5
0%
none of these

Explanation

Arranging the given data in ascending order
40,54,62,68,76,90
Here,

$n=6 (even)$

$M= \dfrac{\text{value of }3^{rd}\text{observation}+\text{value of }4^{th}\text{observation}}{2}$
Median

$M=\dfrac{62+68}{2}=65$

Mean deviation about median

$M.D=\dfrac{|40-65|+|54-65|+|62-65|+|68-65|+|76-65|+|90-65|}{65}$

$=\dfrac{25+11+3+3+11+25}{65}=1.2$

Standard deviation of the distribution

$1, 3, 5,....., 13$ will be

0%
$2$
0%
$4$
0%
$7$
0%
None

Explanation

Mean of the given numbers is

$\bar x =\dfrac{1+3+5+7+9+11+13}{7}=\dfrac{49}{7}=7$

total number of values

$n=7$

Standard deviation is given by

$SD=\sqrt{\dfrac{\sum_{i=1}^n (x_i-\bar x)^2}{n-1}}$

$\implies SD=\sqrt{\dfrac{(1-7)^2+(3-7)^2+(5-7)^2+(7-7)^2+(9-7)^2+(11-7)^2+(13-7)^2}{7-1}}$

$\implies SD=\sqrt{\dfrac{36+16+4+0+4+16+36}{6}}$

$\implies SD=\sqrt{\dfrac{112}{6}}$

$\implies SD=\sqrt{18.6667}=4.32$

Therefore the standard deviation for the given values is

$4.32$

For the given data, SD = 10, AM = 20, the coefficient
of variation is____

0%
47
0%
24
0%
44
0%
50

Explanation

Coeffecient of variation

$= \frac {SD}{AM} \times 100 = \frac {10}{20} \times 100 = 50$

For the given data, SD

$= 10$ , AM

$= 20$ the coefficient of variation is ...........

0%
$47$
0%
$24$
0%
$44$
0%
$50$

Explanation

Coefficient of variation is the ratio of standard deviation to the mean.

Given that

$SD=10$ and

$AM=20$

Therefore of coefficient of variation is

$\dfrac{SD}{AM}\times100=\dfrac{10}{20}\times100=50\%$

CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 5 - MCQExams.com

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CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 5 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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