Measures of Dispersion Test

Result

Measures of Dispersion Test - 17

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Question 1
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If the mean deviation about the median of the numbers $$a,2a,3a,.....50a$$ is $$50$$, then $$\left| a \right| $$ equals

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

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

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

$$\Rightarrow 2500=|a|+2\times 12 (3+23)|a|$$
$$\Rightarrow |a|=\dfrac{2500}{625}=4$$
Question 2
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Directions For Questions

The standard deviation of variate $$x$$ is the square root of the A.M. of the squares of all deviations of $$x$$ from the A.M. of observations and we denote it by sigma $$ \displaystyle \left ( \sigma \right ) $$ if $$ \displaystyle x/f_{i}\left ( i = 1,2,3,...,n \right ) $$ is a frequency distribution, then
$$ \displaystyle \sigma =\sqrt{\frac{1}{N}\sum f_{i}\left ( x_{i}-\bar{x} \right )^{2}} $$           ........(i)
where $$ \displaystyle \bar{x} $$ is the A.M.of the distribution & $$ \displaystyle N=\sum_{i=1}^{n}f_{i} $$
The square of the standard deviation is called the variance & given by
$$ \displaystyle \sigma ^{2}=\frac{1}{N}\sum_{i=1}^{n}f_{i}\left ( x_{i}-\bar{x} \right )^{2} $$     ........(ii)
The calculation of coefficient of dispersion & coefficient of variation are count down by $$ \displaystyle \frac{\sigma }{\bar{x}} $$ & $$ \displaystyle \frac{\sigma }{\bar{x}}\times 100 $$ respectively.
If deviation of $$x$$ are measured from an assumed mean $$A$$ then root mean square deviation of $$x$$ is denoted by $$S$$ and given by
$$ \displaystyle S= \sqrt{\frac{1}{N}\sum_{i=1}^{n}f_{i}\left ( x_{i} -A\right )^{2}} $$
which set up the relationship between S.D. and Root mean square deviation given by $$ \displaystyle S^{2}=\sigma ^{2}+d^{2}\left ( Where, d=\sum_{i=1}^{n} x_i-A \right ) $$
Consider the frequency distribution
Size 4 6 8 10 12 14 16
Frequency 1 2 3 5 3 2 1
On the basis of above information answer the following questions.

...view full instructions

The coefficient of dispersion for the given distribution is

A
3.06

B
.306

C
30.6

D
None of these

Solution

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) $$\displaystyle =\frac{\sigma }{\bar{X}}\times 100=0.306$$
Question 3
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Standard deviation for first 10 natural numbers is

A
$$5.5$$

B
$$3.87$$

C
$$2.97$$

D
$$2.87$$

Solution

Standard deviation of first $$n$$ natural number is $$\displaystyle =\sqrt{\frac{n^2-1}{12}}=\sqrt{\frac{10^2-1}{12}}=2.87$$

Question 4

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Directions For Questions

The standard deviation of variate $$x$$ is the square root of the A.M. of the squares of all deviations of $$x$$ from the A.M. of observations and we denote it by sigma $$ \displaystyle \left ( \sigma \right ) $$ if $$ \displaystyle x/f_{i}\left ( i = 1,2,3,...,n \right ) $$ is a frequency distribution, then

$$ \displaystyle \sigma =\sqrt{\frac{1}{N}\sum f_{i}\left ( x_{i}-\bar{x} \right )^{2}} $$ ........(i)

where $$ \displaystyle \bar{x} $$ is the A.M.of the distribution & $$ \displaystyle N=\sum_{i=1}^{n}f_{i} $$

The square of the standard deviation is called the variance & given by

$$ \displaystyle \sigma ^{2}=\frac{1}{N}\sum_{i=1}^{n}f_{i}\left ( x_{i}-\bar{x} \right )^{2} $$ ........(ii)

The calculation of coefficient of dispersion & coefficient of variation are count down by $$ \displaystyle \frac{\sigma }{\bar{x}} $$ & $$ \displaystyle \frac{\sigma }{\bar{x}}\times 100 $$ respectively.

If deviation of $$x$$ are measured from an assumed mean $$A$$ then root mean square deviation of $$x$$ is denoted by $$S$$ and given by

$$ \displaystyle S= \sqrt{\frac{1}{N}\sum_{i=1}^{n}f_{i}\left ( x_{i} -A\right )^{2}} $$

which set up the relationship between S.D. and Root mean square deviation given by $$ \displaystyle S^{2}=\sigma ^{2}+d^{2}\left ( Where, d=\sum_{i=1}^{n} x_i-A \right ) $$

Consider the frequency distribution

Size	4	6	8	10	12	14	16
Frequency	1	2	3	5	3	2	1

On the basis of above information answer the following questions.

...view full instructions

The coefficient of variation for the given distribution is

$$28.6$$

$$0.306$$

$$30.6$$

$$306$$

Solution

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

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

$$=28.6$$

Question 5
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Standard deviation for first 10 even natural numbers is

A
$$11$$

B
$$7.74$$

C
$$5.74$$

D
$$11.48$$

Solution

We know standard deviation of first $$n$$ even natural number is $$, \sigma =\sqrt{\cfrac{n^2-1}{3}}$$
Here $$n =10$$
$$\therefore \sigma = \sqrt{\cfrac{100-1}{3}}=\sqrt{33}=5.74$$
Question 6
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If the coefficient of variation of some observation is 60 and their standard deviation is 20, then their mean is

A
35

B
34

C
33

D
33.33(nearly)

Solution

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)
Question 7
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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

A
$$36.6,22.6$$

B
$$34.8,22.6$$

C
$$36.6,24.4$$

D
None of these

Solution

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 $$\displaystyle =\frac { 15.6\times 100 }{ 69 } =22.6$$
Question 8
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The Coefficient of Variation is given by:

A
$$\dfrac{Mean}{\ Standard \ \ deviation } \times 100$$

B
$$\dfrac{\ Standard \ \ deviation }{Mean}$$

C
$$\dfrac{Standard \ \ deviation }{Mean }\times 100$$

D
$$\dfrac{Mean}{Standard \ Deviation}$$

Solution

The coefficient of variation (CV) is a standardized measure of dispersion
. It is defined as the ratio of the standard deviation to the mean.
$$CV\quad =\quad \cfrac { \sigma }{ Mean }\times100 $$
Question 9
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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

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

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

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

D
none of these

Solution

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

$$Variance = (S.D)^2$$
Question 10
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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

A
$$10.3\%,10.4\%$$

B
$$95\%,7.9\%$$

C
$$11.2\%,10.1\%$$

D
none of these

Solution

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

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

Measures of Dispersion Test - 17

Result

Measures of Dispersion Test - 17

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Rank

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SHARING IS CARING

Question 1

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 2

3.06

.306

30.6

None of these

Solution

Question 3

$$5.5$$

$$3.87$$

$$2.97$$

$$2.87$$

Solution

Question 4

$$28.6$$

$$0.306$$

$$30.6$$

$$306$$

Solution

Question 5

$$11$$

$$7.74$$

$$5.74$$

$$11.48$$

Solution

Question 6

35

34

33

33.33(nearly)

Solution

Question 7

$$36.6,22.6$$

$$34.8,22.6$$

$$36.6,24.4$$

None of these

Solution

Question 8

$$\dfrac{Mean}{\ Standard \ \ deviation } \times 100$$

$$\dfrac{\ Standard \ \ deviation }{Mean}$$

$$\dfrac{Standard \ \ deviation }{Mean }\times 100$$

$$\dfrac{Mean}{Standard \ Deviation}$$

Solution

Question 9

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

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

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

none of these

Solution

Question 10

$$10.3\%,10.4\%$$

$$95\%,7.9\%$$

$$11.2\%,10.1\%$$

none of these

Solution

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