Statistics Test

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Statistics Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

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 2
1 / -0

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 3
1 / -0

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 4
1 / -0

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 5
1 / -0

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 $$ =\dfrac{7.6}{73} \times 100=10.4$$%
Question 6
1 / -0

If mean of a series is 40 and variance 1486, then coefficient of variation is

A
$$0.9021$$

B
$$0.9637$$

C
$$0.8864$$

D
$$0.9853$$

Solution

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 $$=\cfrac{\sigma}{\bar{x}}=\cfrac{\sqrt{1486}}{40}=.9637$$
Question 7
1 / -0

The mean of first five prime numbers is

A
$$5.6$$

B
$$3.6$$

C
$$6.83$$

D
$$5.2$$

Solution

First five prime numbers are: $$2,3,5,7,11$$
Mean $$= \cfrac{\text{Sum}}{\text{Count of Numbers}}$$
Mean $$= \cfrac{2 + 3 + 5 + 7 + 11}{5}$$
Mean $$= \cfrac{28}{5}$$
Mean $$= 5.6$$
Question 8
1 / -0

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

A
$$7.8$$

B
$$9$$

C
$$8.1$$

D
$$0.9$$

Solution

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
$$N=10$$ $$\sum f_iu_i=-3$$ $$\sum f_iu_i^2=9$$
$$\displaystyle \sigma =h\sqrt{\frac{ \Sigma f_{1}u_{1}^{2}}{N}-\left ( \frac{\Sigma f_{1}u_{1}}{N} \right )^{2}}$$

$$\displaystyle =10\sqrt{\frac{9}{10}-\left ( \frac{-3}{10} \right )^{2}}=9$$
Question 9
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The marks of a class test are given below.
$$28, 26, 17, 12, 14, 19, 27, 26, 21, 16, 15$$. Find the median.

A
$$15$$

B
$$17$$

C
$$19$$

D
$$21$$

Solution

The marks in the test are: $$28, 26, 17, 12, 14, 19, 27, 26, 21, 16, 15$$
Arranging the marks in the test in ascending order: $$12, 14, 15, 16, 17, 19, 21, 26, 26, 27, 28$$
There are $$11$$ values. Since, it is an odd number, median will be the $$6^{th}$$ number after arranging the marks in ascending order.
Hence, median $$= 19$$
Question 10
1 / -0

What is the mean of first 8 whole numbers?

A
$$3$$

B
$$3.5$$

C
$$4$$

D
$$4.5$$

Solution

First $$8$$ whole numbers are $$ 0,1,2,3,4,5,6,7$$

We know that, $$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Total number of observations}}$$

$$\therefore \ \text{Mean}=\dfrac {0+1+2+3+4+5+6+7}{8}$$
$$\Rightarrow \dfrac {28}{8}=3.5$$
Hence, the mean of first $$8$$ whole numbers is $$3.5$$.

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Re-Attempt Weekly Quiz Competition

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

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
		$$N=10$$		$$\sum f_iu_i=-3$$	$$\sum f_iu_i^2=9$$

Statistics Test - 26

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Statistics Test - 26

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Question 1

35

34

33

33.33(nearly)

Solution

Question 2

$$36.6,22.6$$

$$34.8,22.6$$

$$36.6,24.4$$

None of these

Solution

Question 3

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

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

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

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

Solution

Question 4

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

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

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

none of these

Solution

Question 5

$$10.3\%,10.4\%$$

$$95\%,7.9\%$$

$$11.2\%,10.1\%$$

none of these

Solution

Question 6

$$0.9021$$

$$0.9637$$

$$0.8864$$

$$0.9853$$

Solution

Question 7

$$5.6$$

$$3.6$$

$$6.83$$

$$5.2$$

Solution

Question 8

$$7.8$$

$$9$$

$$8.1$$

$$0.9$$

Solution

Question 9

$$15$$

$$17$$

$$19$$

$$21$$

Solution

Question 10

$$3$$

$$3.5$$

$$4$$

$$4.5$$

Solution

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