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CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 5 - MCQExams.com
CBSE
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
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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
Report Question
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
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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
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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
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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
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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
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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
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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
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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:
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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
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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
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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
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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__________.
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a continuous one
0%
more symmetrical
0%
severely skewed to one side
0%
none of the above
Explanation
If the sample size of the population is small, then the curve would be skewed on one side and also as the sample size increases the sampling distribution of the mean gets narrower.
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
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$${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{ n }$$
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
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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
Report Question
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
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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
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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 $$10, 14, 11, 9, 8, 12, 6$$
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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
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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
Report Question
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
Report Question
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
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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
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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
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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
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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
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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____
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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 ...........
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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 $$\dfrac{SD}{AM}\times100=\dfrac{10}{20}\times100=50\%$$
0:0:1
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