Measures of Dispersion Test

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Measures of Dispersion Test - 19

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Question 1
1 / -0

Mean deviations of the series $$a,a+d,a+2d,...,a+2nd$$ from its mean is

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

B
$$\displaystyle \frac{nd}{2n+1}$$

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

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

Solution

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 $$\displaystyle =\frac{1}{2n+1}\sum \left | x-\bar{x} \right |$$ $$\displaystyle =\frac{2}{2n+1}.d\left ( 1+2+3\cdots +n \right )$$ $$\displaystyle =\frac{n\left ( n+1 \right )d}{(2n+1)}$$
Question 2
1 / -0

The mean of a distribution is 4. If its coefficient of variation is 58%. Then the S.D. of the distribution is

A
2.23

B
3.23

C
2.32

D
none of these

Solution

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 $$=\cfrac{\sigma}{\bar{x}}\times 100$$ %
$$\Rightarrow 58 = \cfrac{\sigma}{4}\times 100\Rightarrow \sigma = \cfrac{58\times 4}{100}=2.32$$
Question 3
1 / -0

Standard deviation of the distribution $$1, 3, 5,....., 13$$ will be

A
$$2$$

B
$$4$$

C
$$7$$

D
None

Solution

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 $$4.32$$
Question 4
1 / -0

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

A
2.16

B
1.2

C
5

D
none of these

Solution

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

$$=\dfrac{25+11+3+3+11+25}{65}=1.2$$
Question 5
1 / -0

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

A
47

B
24

C
44

D
50

Solution

Coeffecient of variation $$ = \frac {SD}{AM} \times 100 = \frac {10}{20} \times 100 = 50 $$
Question 6
1 / -0

For the given data, SD $$= 10$$, AM $$= 20$$ the coefficient of variation is ...........

A
$$47$$

B
$$24$$

C
$$44$$

D
$$50$$

Solution

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 $$\dfrac{SD}{AM}\times100=\dfrac{10}{20}\times100=50\%$$
Question 7
1 / -0

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

A
10%

B
40%

C
50%

D
none of these

Solution

Given, $$\sum (x-\bar{x})^2 = 250, n = 10, \bar{x} =50$$
Thus standard deviation $$ = \sqrt{\cfrac{\sum (x-\bar{x})^2}{n}}=\sqrt{25}=5$$
$$\therefore$$ Coefficient of variation $$=\cfrac{\sigma}{\bar{x}}\times 100 =\cfrac{5}{50}\times 100$$ % $$= 10$$%
Question 8
1 / -0

Consider the following groups A and B
A : 3, 4, 5,...... upto n terms
B : 15, 19, 23,..... upto n terms
If the mean deviations of groups A and B about their means are $$\displaystyle \alpha \: \: \: and\: \: \: \beta $$ respectively then

A
$$\displaystyle \beta=5\alpha $$

B
$$\displaystyle \beta=4\alpha+3 $$

C
$$\displaystyle \beta=4\alpha$$

D
None

Solution

$$A$$ $$B$$
$$3$$ $$15$$
$$4$$ $$19$$
$$5$$ $$23$$
$$.$$ $$.$$
$$.$$ $$.$$
$$.$$ $$.$$
$$n$$ $$n$$

$$\dfrac{\sum A}{n}=\alpha$$

$$\dfrac{\sum B}{n}=\beta$$
now for $$A$$
$$a=3$$
$$d=1$$
$$\sum A=\dfrac{n}{2}[2a+(n-1)d]$$

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

=$$\dfrac{n}{2}[5+n]$$

$$\alpha=\dfrac{\sum A}{n}=\dfrac{5+n}{2}$$

=>n=$$2\alpha-5$$
for $$B$$
$$a=15$$
$$d=4$$
$$\sum B=\dfrac{n}{2}[2a+(n-1)d]$$
=$$\dfrac{n}{2}[30+(n-1)4]$$
=$$\dfrac{n}{2}[26+4n]$$
$$\beta=\dfrac{\sum B}{n}=\dfrac{26+4n}{2}$$
$$\beta=\dfrac{26+4(2\alpha-5)}{2}$$
$$2\beta=6+8\alpha$$
$$\beta=4\alpha+3$$
Question 9
1 / -0

Find the mean for the following data using step deviation method.

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

Answer:- By shortcut Method
Class interval width (w) = 2

$$X_i$$ $$F_i$$ $$d=\cfrac{X-A}{i}$$ $$F_i d_i$$
2 10 -1 -10
4=A 20 0 0
6 30 1 30
8 40 2 80
$$\Sigma f = 100$$ $$\Sigma fd = 100$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times w=4+\cfrac{100}{100} \times 2 ={4+2}=6$$
C)6
Question 10
1 / -0

Find the mean for the following data using step deviation method.

A
$$9$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

Answer:- By shortcut Method
Class interval width (w) = 3

X F $$d=\cfrac{X-A}{i}$$ Fd
3 5 -1 -5
6=A 10 -0 -0
9 15 1 15
12 20 2 40
$$\Sigma f = 50$$ $$\Sigma fd = 50$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times i=6+\left(\cfrac{50}{50} \times 3\right) = 6+3 = 9$$
A) 9

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$$X_i$$	$$F_i$$	$$d=\cfrac{X-A}{i}$$	$$F_i d_i$$
2	10	-1	-10
4=A	20	0	0
6	30	1	30
8	40	2	80
	$$\Sigma f = 100$$		$$\Sigma fd = 100$$

X	F	$$d=\cfrac{X-A}{i}$$	Fd
3	5	-1	-5
6=A	10	-0	-0
9	15	1	15
12	20	2	40
	$$\Sigma f = 50$$		$$\Sigma fd = 50$$

Measures of Dispersion Test - 19

Result

Measures of Dispersion Test - 19

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SHARING IS CARING

Question 1

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

$$\displaystyle \frac{nd}{2n+1}$$

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

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

Solution

Question 2

2.23

3.23

2.32

none of these

Solution

Question 3

$$2$$

$$4$$

$$7$$

None

Solution

Question 4

2.16

1.2

5

none of these

Solution

Question 5

47

24

44

50

Solution

Question 6

$$47$$

$$24$$

$$44$$

$$50$$

Solution

Question 7

10%

40%

50%

none of these

Solution

Question 8

$$\displaystyle \beta=5\alpha $$

$$\displaystyle \beta=4\alpha+3 $$

$$\displaystyle \beta=4\alpha$$

None

Solution

Question 9

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 10

$$9$$

$$10$$

$$11$$

$$12$$

Solution

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