MCQ Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 10

The mean of

$100$ observations is

$18.4$ and sum of squares of deviations from mean is

$1444$ , the Co-efficient of variation is _______.

0%
$30.6$
0%
$35.6$
0%
$20.6$
0%
$10.6$

The standard deviation of

$5$ items is found to be $$15$. What will be the standard deviation if the values of all the items are increased?

0%
$15$
0%
$20$
0%
$10$
0%
None of the above

The relation between variance and standard deviation is ________.

0%
variance is the square root of standard Deviation
0%
square of the standard deviation is equal to Variance
0%
variance is equal to standard deviation
0%
standard deviation is the square of the variance

The standard deviation of a set of

$50$ items is

$8$ , what is the standard deviation, if each item is multiplied by

$2$ ?

0%
$32$
0%
$8$
0%
$4$
0%
$16$

If each value of a series is multiplied by a constant, the coefficient of variation as compared to original value is _______.

0%
increased
0%
unaltered
0%
decreased
0%
zero

The coefficient of variation is?

0%
The same as the variance
0%
A measure of central tendency
0%
A measure of absolute variability
0%
A measure of relative variability

Each outcome of a random experiment is called ______.

0%
primary event
0%
probable event
0%
complementary event
0%
none of them

Which of the following statements is true of a measure of dispersion?

0%
Mean deviation does not follow algebraic value
0%
Range is crudest measure
0%
Coefficient of variation is a relative measure
0%
All the above statements

Probability can take values from ______.

0%
-3 to 3
0%
-3 to 1
0%
0 to 1
0%
-1 to 1

Co-efficient of variation is?

0%
Absolute variation
0%
Static
0%
Relative variation
0%
None of the above

Probability is expressed as _______.

0%
percentage
0%
ratio
0%
proportion
0%
all (a), (b), (c)

If A and B are events, the probability of occurrence of A and B simultaneously is given as _______.

0%
P(A) + P(B)
0%
$P(A\cup{B})$
0%
P(AB)
0%
P(A) P(B)

Classical probability is measured in terms of _________.

0%
absolute value and ratio both
0%
a ratio
0%
an absolute value
0%
none of the above

If A and B are two events, the probability of occurrence of either A or B is given as _______.

0%
P(A) + P(B)
0%
$P(A\cup{B})$
0%
P(A)P(B)
0%
P(AB)

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?

0%
1 only
0%
2 only
0%
Both 1 and 2
0%
Neither 1 nor 2

Explanation

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.

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

0%
Open end classes
0%
Close end classes
0%
Inclusive classes
0%
Exclusive classes

Laspeyres Price Index =?

0%
$157.33$
0%
$153.14$
0%
$153.33$
0%
$157.14$

Explanation

Laspeyres Price Index Number

$P^{L}_{01} =\dfrac{\Sigma P_{1}Q_{0}}{\Sigma P_{0}Q_{0}} \times {100} = \dfrac{23}{15}\times{100} = 153.33$ .

Laspeyres Price Index Number = ?

0%
$220.00$
0%
$216.30$
0%
$219.12$
0%
$221.98$

Explanation

Laspeyres Price Index Number

$P^{l}_{01} =\dfrac{\Sigma P_{1}Q_{0}}{\Sigma P_{0}Q_{0}} \times {100} = \dfrac{404}{182}\times{100} = 221.98$

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

0%
55
0%
59
0%
8
0%
56

The sum of squares of deviations for

$10$ observations taken from mean

$50$ is

$250$ . Then Co-efficient of variation is

0%
$10\%$
0%
$40\%$
0%
$50\%$
0%
None

Explanation

$\sum(x-\overline{x})^2=250$ ,

$\overline{x}=50$

$\Rightarrow$ Standard deviation

$(\sigma)=\sqrt{\dfrac{250}{10}}=\sqrt{25}=5$

$\Rightarrow$ Coefficient of variation

$=\sqrt{\dfrac{\sum(x-\overline{x})^2}{n}}$

$=\dfrac{\sigma}{Mean}\times 100$

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

$=10\%$

The sum of the deviation of the individual data elements from their mean is always_________.

0%
Equal to zero
0%
Equal to one
0%
Negative
0%
Positive

Which of the following relations among the location parameters does not hold?

0%
$Q_2 = Median$
0%
$P_50 = Median$
0%
$D_5 = Median$
0%
$D_4 = Median$

The mean deviation about the mean of the set of first

$n$ natural numbers when

$n$ is an odd number.

0%
$\dfrac{n^{2}-1}{4n}$
0%
$\dfrac{n}{4}$
0%
$\dfrac{n^{2}+1}{4n}$
0%
$\dfrac{n^{2}-1}{12}$

Explanation

$\overline { X } =\cfrac { \sum _{ i=1 }^{ n }{ i } }{ n } =\cfrac { n(n+1) }{ 2n } =\cfrac { n+1 }{ 2 }$

$\left| { d }_{ i } \right| =\left| r-\left( \cfrac { n+1 }{ 2 } \right) \right|$ $

$\sum { \left| { d }_{ i } \right| } =\sum _{ r=1 }^{ n }{ \left| r-\left( \cfrac { n+1 }{ 2 } \right) \right| } =\sum _{ r=1 }^{ \cfrac { n+1 }{ 2 } }{ \cfrac { n+1 }{ 2 } } =r+\sum _{ r=\cfrac { n+1 }{ 2 } }^{ n }{ r-\left( \cfrac { n+1 }{ 2 } \right) }$

$\Rightarrow \sum { = } \left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+1 }{ 2 } \right) -\cfrac { 1 }{ 2 } \left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+3 }{ 2 } \right) -\left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+1 }{ 2 } \right) +\cfrac { n+1 }{ 4 } \left( \cfrac { n+1 }{ 2 } +n \right) \quad$

$\Rightarrow \cfrac { n+1 }{ 4 } \left( \cfrac { 3n+1 }{ 2 } \right) -\cfrac { \left( n+1 \right) \left( n+3 \right) }{ 8 } \Rightarrow \cfrac { n+1 }{ 4 } \left( \cfrac { 3n+1 }{ 2 } -\cfrac { n+3 }{ 2 } \right)$

$\cfrac { n+1 }{ 4 } \left( \cfrac { 2n-2 }{ 2 } \right) =\cfrac { { n }^{ 2 }-1 }{ 4 }$

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

0%
$\cfrac { 3 }{ 2 }$
0%
5
0%
$\cfrac { 5 }{ \sqrt { 2 } }$
0%
$\sqrt { \cfrac { 81 }{ 34 } }$

Find variance of the following data.

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

0%
$19$
0%
$20$
0%
$21$
0%
$23$

Explanation

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

$\Rightarrow$

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

$\Rightarrow$

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

the coefficient of quartile deviation is

0%
$\dfrac{40}{3}$
0%
$\dfrac{15}{2}$
0%
$20$
0%
$150$

Explanation

$\begin{array}{l} 120,125,130,140,145,150,160,165,170,175,180 \\ n=11 \\ \Rightarrow { Q_{ 1 } }=\left( { \frac { { n+1 } }{ 4 } } \right) th\, \, term=\left( { \frac { { 12 } }{ 4 } } \right) th\, \, term=3th\, \, term \\ \Rightarrow { Q_{ 3 } }=130 \\ and \\ { Q_{ 3 } }=3\left( { \frac { { n+1 } }{ 4 } } \right) th\, \, term\, =3\left( { \frac { { 12 } }{ 4 } } \right) th\, \, term=9th\, \, term=170 \\ Then,\, \, coefficient\, \, of\, \, quartiledeviation\, \, =\frac { { \left( { { Q_{ 3 } }-{ Q_{ 1 } } } \right) } }{ { \left( { { Q_{ 3 } }+{ Q_{ 1 } } } \right) } } \times 100 \\ =100\times \frac { { \left( { 170-130 } \right) } }{ { \left( { 170+130 } \right) } } =\frac { { 40 } }{ { 300 } } \times 100=\frac { { 40 } }{ 3 } \, \, \end{array}$

If the sum and sum of squares of

$10$ observations are

$12$ and

$18$ resp., then, The

$S.D$ of observations is :-

0%
$\dfrac{1}{5}$
0%
$\dfrac{2}{5}$
0%
$\dfrac{3}{5}$
0%
$\dfrac{4}{5}$

Explanation

$\sum x=12,\sum x^2=18,N=10$

$SD=\sqrt{\cfrac{\sum x^2}{N}-(\cfrac{\sum x}{N})^2}$

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

$5$ students of a class have an average height

$150\ cm$ and variance

$18\ cm^{2}$ . A new student, whose height is

$156\ cm$ , joined them. The variance (in

$cm^{2})$ of the height of these six students is

0%
$22$
0%
$20$
0%
$16$
0%
$18$

Explanation

Given

$\bar {x} = \dfrac {\sum x_{i}}{5} = 150$

$\Rightarrow \displaystyle \sum_{i = 1}^{5} x_{i} = 750 .... (i)$

$\sigma^2=18$

$\dfrac {\sum x_{i}^{2}}{5} - (\bar {x})^{2} = 18$

$\dfrac {\sum x_{i}^{2}}{5} - (150)^{2} = 18$

$\sum x_{i}^{2} = 112590 ..... (ii)$
Given height of new student

$x_{6} = 156$
Now,

$\bar {x}_{new} = \dfrac {\displaystyle \sum_{i = 1}^{5}x_{i}}{6} = \dfrac {750 + 156}{6} = 151$
Also,

$\sigma_{new}^2= \dfrac {\displaystyle \sum_{i = 1}^{5}x_{i}^{2}}{6} - (\overline {x}_{new})^{2}$

$= \dfrac {112590 + (156)^{2}}{6} - (151)^{2}$

$= 22821 - 22801 = 20$ .

Standard deviation of four observations

$-1, 0, 1$ and k is

$\sqrt{5}$ then k will be?

0%
$2\sqrt{6}$
0%
$1$
0%
$2$
0%
$\sqrt{6}$

Explanation

$\sigma^2=\dfrac{\displaystyle\sum x^2_i}{n}-\left(\dfrac{\displaystyle\sum x_i}{n}\right)^2$

$\Rightarrow 5=\dfrac{1+0+1+k^2}{4}-\left(\dfrac{-1+0+1+k}{4}\right)^2$

$\Rightarrow 5=\dfrac{k^2+2}{4}=\dfrac{k^2}{16}$

$\Rightarrow 80=4k^2+8-k^2$

$\Rightarrow 72=3k^2$

$\Rightarrow k=2\sqrt{6}$ .

A student scores the following marks in five test:

$45, 54, 41, 57, 43$ . His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six test is

0%
$\dfrac{10}{\sqrt{3}}$
0%
$\dfrac{100}{\sqrt{3}}$
0%
$\dfrac{130}{3}$
0%
$\dfrac{10}{3}$

Explanation

Let

$x$ be the

$6^{th}$ observation

$\Rightarrow 45 + 54 + 41 + 57 + 43 + x = 48 \times 6 = 288$

$\Rightarrow x = 48$
Variance

$= \left(\dfrac{\sum x_i^2}{6} - (\bar{x})^2\right)$

$\Rightarrow$ variance

$=\dfrac{14024}{6} - (48)^2$

$=\dfrac{100}{3}$

$\Rightarrow$ standard deviation

$=\dfrac{10}{\sqrt{3}}$

CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 10 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 10 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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