Measures of Dispersion Test

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Measures of Dispersion Test - 37

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Weekly Quiz Competition

Question 1
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Standard deviation of four observations $$-1, 0, 1$$ and k is $$\sqrt{5}$$ then k will be?

A
$$2\sqrt{6}$$

B
$$1$$

C
$$2$$

D
$$\sqrt{6}$$

Solution

$$\sigma^2=\dfrac{\displaystyle\sum x^2_i}{n}-\left(\dfrac{\displaystyle\sum x_i}{n}\right)^2$$
$$\Rightarrow 5=\dfrac{1+0+1+k^2}{4}-\left(\dfrac{-1+0+1+k}{4}\right)^2$$
$$\Rightarrow 5=\dfrac{k^2+2}{4}=\dfrac{k^2}{16}$$
$$\Rightarrow 80=4k^2+8-k^2$$
$$\Rightarrow 72=3k^2$$
$$\Rightarrow k=2\sqrt{6}$$.
Question 2
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The standard deviation of the data $$6,5,9, 13, 12, 8, 10$$ is

A
$$\dfrac{\sqrt{52}}{7}$$

B
$$\dfrac{52}{7}$$

C
$$\dfrac{\sqrt{53}}{7}$$

D
$$\dfrac{53}{7}$$

E
$$6$$

Solution

Given data $$6,5,9,13,12,8,10$$ Mean of the given data $$(\bar{x})$$
$$=\dfrac{6+5+9+13+12+8+10}{7}$$
$$=\dfrac{63}{7}=9$$
The deviation of the respective data from the mean i.e. $$(x_i-\bar{x})$$ are
$$6-9,5-9,9-9,13-9,12-9,8-9,10-9$$
$$(x_i-\bar{x})=-3,-4,0,4,3,-1,1$$
$$(x_1-\bar{x})^2=9,16,0,16,9,1,1$$
$$\displaystyle \sum^7_{i=i} (x_1-\bar{x})^2=9+16+0+16+9+1+1=52$$
$$\therefore$$ standard deviation $$(\sigma)$$
$$\displaystyle =\sqrt{\dfrac{1}{n}\sum^7_{i=1} (x_1-\bar{x})^2}=\sqrt{\dfrac{52}{7}}$$
Question 3
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If the mean deviation about the median of the numbers $$a,2a,....,50a$$ is $$50$$, then $$|a|$$ equals:-

A
$$4$$

B
$$5$$

C
$$2$$

D
$$3$$

Solution

Given series:- $$a, 2a, 3a, ....., 50a$$
Median $$= \cfrac{25a + 26a}{2} = 25.5 a$$
Mean deviation about median $$= 50 \quad \left( \text{Given} \right)$$
Mean deviation $$= \cfrac{\sum_{i = 1}^{50}{\left| {x}_{i} - 25.5a \right|}}{50}$$
$$\therefore \cfrac{24.5a + 23.5a + 22.5a + ..... + 23.5a + 24.5a}{50} = 50$$
$$\Rightarrow a + 3a + 5a + ..... + 47a + 49a = 2500$$
$$\Rightarrow \cfrac{25}{2} \left( 2a + \left( 25 - 1 \right) 2a \right) = 2500$$
$$\Rightarrow 2a \times 25 = 200$$
$$\Rightarrow a = \cfrac{200}{50} = 4$$
Question 4
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Standard deviation of first $$n$$ odd natural numbers is

A
$$\sqrt{n}$$

B
$$\sqrt{\dfrac{(n+2)(n+1)}{3}}$$

C
$$\sqrt{\dfrac{n^2-1}{3}}$$

D
$$n$$

Solution

Stndard deviation, $$\sigma = \sqrt{\dfrac{\sum x_i^2}{N}-(\bar{x})^2}$$

$$\therefore \bar{x} = \dfrac{\sum x_i}{N}$$

$$= \dfrac{1 + 3 +5 + ...(2n-1)}{n}$$

$$= \dfrac{\dfrac{n}{2}[1+2n-1]}{n}$$

$$=\dfrac{n^2}{n} = n$$

Again, $$\sum x_i^2 = 1^2 + 3^2 + 5^2 + ...(2n - 1)^2$$

$$=\sum (2n-1)^2$$

$$=\sum(4n^2 - 4n+1)$$

$$= 4\sum n^2 - 4\sum n + \sum 1$$

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

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

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

$$= \dfrac{n}{3} [ 4n^2 + 6n + 2 - 6n - 6 + 3]$$

$$= \dfrac{n}{3} [4n^2 - 1]$$

$$\therefore \delta = \sqrt{\dfrac{n(4n^2 - 1)}{3n} - n^2}$$

$$= \sqrt{\dfrac{4n^2-1}{3}-n^2}$$

$$= \sqrt{\dfrac{4n^2-1-3n^2}{3}}$$

$$= \sqrt{\dfrac{n^2 - 1}{3}}$$
Question 5
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For the observations $${ x }_{ 1, }{ x }_{ 2 },{ x }_{ 3 },........{ x }_{ 18, }$$ it is given that $$\sum _{i =1 }^{ 18 }{ ({ x }_{ i }-8)=9 } $$ and $$\sum _{ j=1}^{ 18 }{ { ({ x }_{ j}-8) }^{ 2 }=45 } $$ then the standard deviation of these eighteen observations is

A
$$\cfrac { 3 }{ 2 } $$

B
5

C
$$\cfrac { 5 }{ \sqrt { 2 } } $$

D
$$\sqrt { \cfrac { 81 }{ 34 } } $$

Solution
Question 6
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Directions For Questions

The daily wages (in $$Rs$$) of $$11$$ persons given as $$140, 145, 130, 165, 160, 125, 150, 170, 175, 120, 180$$ then

...view full instructions

the coefficient of quartile deviation is

A
$$\dfrac{40}{3}$$

B
$$\dfrac{15}{2}$$

C
$$20$$

D
$$150$$

Solution

$$\begin{array}{l} 120,125,130,140,145,150,160,165,170,175,180 \\ n=11 \\ \Rightarrow { Q_{ 1 } }=\left( { \frac { { n+1 } }{ 4 } } \right) th\, \, term=\left( { \frac { { 12 } }{ 4 } } \right) th\, \, term=3th\, \, term \\ \Rightarrow { Q_{ 3 } }=130 \\ and \\ { Q_{ 3 } }=3\left( { \frac { { n+1 } }{ 4 } } \right) th\, \, term\, =3\left( { \frac { { 12 } }{ 4 } } \right) th\, \, term=9th\, \, term=170 \\ Then,\, \, coefficient\, \, of\, \, quartiledeviation\, \, =\frac { { \left( { { Q_{ 3 } }-{ Q_{ 1 } } } \right) } }{ { \left( { { Q_{ 3 } }+{ Q_{ 1 } } } \right) } } \times 100 \\ =100\times \frac { { \left( { 170-130 } \right) } }{ { \left( { 170+130 } \right) } } =\frac { { 40 } }{ { 300 } } \times 100=\frac { { 40 } }{ 3 } \, \, \end{array}$$
Question 7
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If $$ N = 10 \sum x = 120 $$ and $$ \sigma _{x} = 60 $$ , the variation coefficient is -

A
$$5$$

B
$$50$$

C
$$500$$

D
$$0.5$$

Solution

Variation coefficient
$$= \dfrac{\sigma _{x}}{Mean} \times 100$$
$$ = \dfrac{60 \times 10}{120} \times = 500$$
Question 8
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Mean of variable series $$\bar{x} = 773$$ and mean deviation 64.4 , then coefficient of mean deviation is

A
$$0.065$$

B
$$12.003$$

C
$$0.083$$

D
$$0.073$$

Solution

$$\dfrac{Mean deviation}{Mean} = \dfrac{64.4}{773} = 0.083 $$
Thus (C) is correct
Question 9
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Which one of the following is a false description?

A
In a moderately asymmetrical distribution, the empirical relationship between Mean, Mode and Median suggested by Karl Pearson is Mean Mode = 3 (Mean Median)

B
Coeffcient of variation is an absolute measure of dispersion

C
Measure of skewness in the distribution of numerical values in the data set

D
Kurtosis refers to the degree of flatness or peakedness in the region around the mode of a frequency curve
Question 10
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Mean and standard deviation of a data are $$48$$ and $$12$$ respectively. The coefficient of variation is.

A
$$42$$

B
$$25$$

C
$$28$$

D
$$48$$

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Re-Attempt Weekly Quiz Competition

Measures of Dispersion Test - 37

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Measures of Dispersion Test - 37

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SHARING IS CARING

Question 1

$$2\sqrt{6}$$

$$1$$

$$2$$

$$\sqrt{6}$$

Solution

Question 2

$$\dfrac{\sqrt{52}}{7}$$

$$\dfrac{52}{7}$$

$$\dfrac{\sqrt{53}}{7}$$

$$\dfrac{53}{7}$$

$$6$$

Solution

Question 3

$$4$$

$$5$$

$$2$$

$$3$$

Solution

Question 4

$$\sqrt{n}$$

$$\sqrt{\dfrac{(n+2)(n+1)}{3}}$$

$$\sqrt{\dfrac{n^2-1}{3}}$$

$$n$$

Solution

Question 5

$$\cfrac { 3 }{ 2 } $$

5

$$\cfrac { 5 }{ \sqrt { 2 } } $$

$$\sqrt { \cfrac { 81 }{ 34 } } $$

Solution

Question 6

$$\dfrac{40}{3}$$

$$\dfrac{15}{2}$$

$$20$$

$$150$$

Solution

Question 7

$$5$$

$$50$$

$$500$$

$$0.5$$

Solution

Question 8

$$0.065$$

$$12.003$$

$$0.083$$

$$0.073$$

Solution

Question 9

In a moderately asymmetrical distribution, the empirical relationship between Mean, Mode and Median suggested by Karl Pearson is Mean Mode = 3 (Mean Median)

Coeffcient of variation is an absolute measure of dispersion

Measure of skewness in the distribution of numerical values in the data set

Kurtosis refers to the degree of flatness or peakedness in the region around the mode of a frequency curve

Question 10

$$42$$

$$25$$

$$28$$

$$48$$

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