Measures of Dispersion Test

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Measures of Dispersion Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

Consider the following statements:
(1) Mean is independent of change in scale and change in origin
(2) Variance is independent of change in scale but not in origin
Which of the above statements is/are correct?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

Mean changes with changes in origin as if $$x$$ is added to all the value, $$nx$$ is added to total value and then, divided by $$n$$ to get an addition of $$x$$.
Variance is independent to the choice of origin as variance is $$(a-\overline{a})^2$$ which negates the addition in observation value and the mean as shown above.
Hence, answer is neither 1 nor 2.
Question 2
1 / -0

The classes in which the lower limit or the upper limit is not specified are known as: _________.

A
Open end classes

B
Close end classes

C
Inclusive classes

D
Exclusive classes
Question 3
1 / -0

Laspeyres Price Index =?

A
$$157.33$$

B
$$153.14$$

C
$$153.33$$

D
$$157.14$$

Solution

Laspeyres Price Index Number
$$P^{L}_{01} =\dfrac{\Sigma P_{1}Q_{0}}{\Sigma P_{0}Q_{0}} \times {100} = \dfrac{23}{15}\times{100} = 153.33$$.
Question 4
1 / -0

Laspeyres Price Index Number = ?

A
$$220.00$$

B
$$216.30$$

C
$$219.12$$

D
$$221.98$$

Solution

Laspeyres Price Index Number
$$P^{l}_{01} =\dfrac{\Sigma P_{1}Q_{0}}{\Sigma P_{0}Q_{0}} \times {100} = \dfrac{404}{182}\times{100} = 221.98$$
Question 5
1 / -0

Given the following set of data, what is the range 12 23 34 54 21 8 9 67:

A
55

B
59

C
8

D
56

Solution

Range is defined as the difference between the highest(or largest ) and lowest(or smallest) observed value in a series. It is the most simple and commonly understandable measures of dispersion. Therefore, it is the most affected measures of dispersion by the extreme values of the series.

Range = H - L

In the given series, H= 67 and L= 12

Range = 67 -12 = 55 .
Question 6
1 / -0

Co-efficient of variation is?

A
Absolute variation

B
Static

C
Relative variation

D
None of the above

Solution

Coefficient of variation is the coefficient of dispersion based on the standard deviation of the statistical series.

Therefore, Coefficient of variation is a unit less or relative measure of dispersion as variation is the absolute measure of dispersion.
Question 7
1 / -0

The sum of squares of deviations for $$10$$ observations taken from mean $$50$$ is $$250 $$. Then Co-efficient of variation is

A
$$10\%$$

B
$$40\%$$

C
$$50\%$$

D
None

Solution

$$\sum(x-\overline{x})^2=250$$, $$\overline{x}=50$$
$$\Rightarrow$$ Standard deviation $$(\sigma)=\sqrt{\dfrac{250}{10}}=\sqrt{25}=5$$
$$\Rightarrow$$ Coefficient of variation $$=\sqrt{\dfrac{\sum(x-\overline{x})^2}{n}}$$
$$=\dfrac{\sigma}{Mean}\times 100$$

$$=\dfrac{5}{50}\times 100$$

$$=10\%$$
Question 8
1 / -0

The mean deviation about the mean of the set of first $$n$$ natural numbers when $$n$$ is an odd number.

A
$$\dfrac{n^{2}-1}{4n}$$

B
$$\dfrac{n}{4}$$

C
$$\dfrac{n^{2}+1}{4n}$$

D
$$\dfrac{n^{2}-1}{12}$$

Solution

$$\overline { X } =\cfrac { \sum _{ i=1 }^{ n }{ i } }{ n } =\cfrac { n(n+1) }{ 2n } =\cfrac { n+1 }{ 2 } $$
$$\left| { d }_{ i } \right| =\left| r-\left( \cfrac { n+1 }{ 2 } \right) \right| $$$
$$\sum { \left| { d }_{ i } \right| } =\sum _{ r=1 }^{ n }{ \left| r-\left( \cfrac { n+1 }{ 2 } \right) \right| } =\sum _{ r=1 }^{ \cfrac { n+1 }{ 2 } }{ \cfrac { n+1 }{ 2 } } =r+\sum _{ r=\cfrac { n+1 }{ 2 } }^{ n }{ r-\left( \cfrac { n+1 }{ 2 } \right) } $$
$$\Rightarrow \sum { = } \left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+1 }{ 2 } \right) -\cfrac { 1 }{ 2 } \left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+3 }{ 2 } \right) -\left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+1 }{ 2 } \right) +\cfrac { n+1 }{ 4 } \left( \cfrac { n+1 }{ 2 } +n \right) \quad $$
$$\Rightarrow \cfrac { n+1 }{ 4 } \left( \cfrac { 3n+1 }{ 2 } \right) -\cfrac { \left( n+1 \right) \left( n+3 \right) }{ 8 } \Rightarrow \cfrac { n+1 }{ 4 } \left( \cfrac { 3n+1 }{ 2 } -\cfrac { n+3 }{ 2 } \right) $$
$$\cfrac { n+1 }{ 4 } \left( \cfrac { 2n-2 }{ 2 } \right) =\cfrac { { n }^{ 2 }-1 }{ 4 } $$

Question 9

1 / -0

Find variance of the following data.

Class interval	Frequency
$$4-8$$	$$3$$
$$8-12$$	$$6$$
$$12-16$$	$$4$$
$$16-20$$	$$7$$

$$19$$

$$20$$

$$21$$

$$23$$

Solution

$$C.I$$	$$x_i$$	$$f_i$$	$$f_ix_i$$	$$(x_i-\overline{x})^2$$	$$f_i(x_i-\overline{x})^2$$
$$4-8$$	$$6$$	$$3$$	$$18$$	$$49$$	$$147$$
$$8-12$$	$$10$$	$$6$$	$$60$$	$$9$$	$$54$$
$$12-16$$	$$14$$	$$4$$	$$56$$	$$1$$	$$4$$
$$16-20$$	$$18$$	$$7$$	$$126$$	$$25$$	$$175$$
		$$\sum f_i=20$$	$$\sum f_ix_i=260$$	$$\sum(x_i-\overline{x})^2=84$$	$$\sum f_i(x_i-\overline{x})^2=380$$

$$\Rightarrow$$ $$\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{260}{20}=13$$

$$\Rightarrow$$ $$Variance(\sigma^2)\dfrac{\sum f_i(x_i-\overline{x})^2}{\sum f_i}=\dfrac{380}{20}=19$$

Question 10
1 / -0

If the sum and sum of squares of $$10$$ observations are $$12$$ and $$18$$ resp., then, The $$S.D$$ of observations is :-

A
$$\dfrac{1}{5}$$

B
$$\dfrac{2}{5}$$

C
$$\dfrac{3}{5}$$

D
$$\dfrac{4}{5}$$

Solution

$$\sum x=12,\sum x^2=18,N=10$$
$$SD=\sqrt{\cfrac{\sum x^2}{N}-(\cfrac{\sum x}{N})^2}$$
$$\implies SD=\sqrt{\cfrac{18}{10}-(\cfrac{12}{10})^2}=\cfrac{3}{5}$$

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Re-Attempt Weekly Quiz Competition

Measures of Dispersion Test - 36

Result

Measures of Dispersion Test - 36

Score

-

Rank

-

SHARING IS CARING

Question 1

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 2

Open end classes

Close end classes

Inclusive classes

Exclusive classes

Question 3

$$157.33$$

$$153.14$$

$$153.33$$

$$157.14$$

Solution

Question 4

$$220.00$$

$$216.30$$

$$219.12$$

$$221.98$$

Solution

Question 5

55

59

8

56

Solution

Question 6

Absolute variation

Static

Relative variation

None of the above

Solution

Question 7

$$10\%$$

$$40\%$$

$$50\%$$

None

Solution

Question 8

$$\dfrac{n^{2}-1}{4n}$$

$$\dfrac{n}{4}$$

$$\dfrac{n^{2}+1}{4n}$$

$$\dfrac{n^{2}-1}{12}$$

Solution

Question 9

$$19$$

$$20$$

$$21$$

$$23$$

Solution

Question 10

$$\dfrac{1}{5}$$

$$\dfrac{2}{5}$$

$$\dfrac{3}{5}$$

$$\dfrac{4}{5}$$

Solution

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