MCQ Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 4

Mean is a measure of _______.

0%
location (central value)
0%
dispersion
0%
correlation
0%
none of the above

The median of the following observations is

$10, 8, -9, -12, 15, 0, 23, -3, -2, 13, -24, 28, 35, 42$

0%
$4$
0%
$-2$
0%
$7$
0%
$-9.75$

What is the standard deviation of

$5, 5, 9, 9, 9, 10, 5, 10, 10$ ?

0%
$2.29$
0%
$6.48$
0%
$4.50$
0%
$8.00$

If a constant

$25$ is added to each observation of a set, the mean is _____.

0%
increased by $25$
0%
decreased by $25$
0%
$25$ times the original mean
0%
not affected

Which measure of dispersion is the quickest to compute?

0%
Standard deviation
0%
Quartile deviation
0%
Mean deviation
0%
Range

If all the observations are increased by

$10$ , then

0%
SD would be increased by $10$
0%
Mean deviation would be increased by $10$
0%
Both (A) and (B)
0%
None of above

The standard deviation of

$10, 16, 10, 16, 10, 10, 16, 16$ is?

0%
$4$
0%
$6$
0%
$3$
0%
$0$

Explanation

Assume

$A=13;$ hence

$d=A-13$

x	d	$d^2$
$10$	$-3$	$9$
$16$	$3$	$9$
$10$	$-3$	$9$
$16$	$3$	$9$
$10$	$-3$	$9$
$10$	$-3$	$9$
$16$	$-3$	$9$
$16$	$3$	$9$
	$\displaystyle\sum d=0$	$\displaystyle\sum d^2=72$

Use following formula:

$\sqrt{\displaystyle\frac{\displaystyle\sum d^2}{n}-\left[\displaystyle\frac{\displaystyle\sum d}{n}\right]^2}$

$\sqrt{\displaystyle\frac{72}{8}-\left[\displaystyle\frac{0}{8}\right]^2}$

$\sqrt{9}$

$=3$ .

Class Intervals	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$120$	?	$200$	?	$185$
Cumulative Frequency	$120$	?	?	?	$900$

Cumulative frequency for class

$50-60=$ ?

0%
$715$
0%
$465$
0%
$145$
0%
$265$

Explanation

Class Intervals Wages	Frequency(f)	Cumulative Frequency
$30-40$	$120$	$120$
$40-50$	$f_1$	$120+f_1$
$50-60$	$200$	$320+f_1$
$60-70$	$f_2$	$320+f_1+f_2$
$70-80$	$185$	$900$

Since median is given to be

$59.25$ , the median class is

$50-60$ . Thus, we can write, solving this for

$f_1$ , we get,

$f_1=145$ ; Further,

$f_2=900-(120+145+200+185)=250$ .

The geometric mean of

$3, 6, 24 and 48$ is ______.

0%
$32$
0%
$12$
0%
$14$
0%
$24$

Sum of the deviations about mean is ______.

0%
zero
0%
minimum
0%
maximum
0%
one

Harmonic mean gives more weight age to _______.

0%
small values
0%
large values
0%
positive values
0%
negative values

In a frequency distribution with open ends, one cannot find out ______.

0%
mean
0%
median
0%
mode
0%
all of the above

The middle value of an ordered series is called _______.

0%
$2nd$ quartile
0%
$5th$ decile
0%
$50th$ percentile
0%
All the above

Class Intervals	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$120$	?	$200$	?	$185$
Cumulative Frequency	$120$	?	?	?	$900$

Cumulative Frequency for class

$60-70=$ ?

0%
$715$
0%
$465$
0%
$145$
0%
$265$

Explanation

Class Intervals Wages	Frequency(f)	Cumulative Frequency
$30-40$	$120$	$120$
$40-50$	$f_1$	$120+f_1$
$50-60$	$200$	$320+f_1$
$60-70$	$f_2$	$320+f_1+f_2$
$70-80$	$185$	$900$

Since median is given to be

$59.25$ , the median class is

$50-60$ . Thus, we can write, Solving this for

$f_1$ we get,

$f_1=145$ ; Further

$f_2=900-(120+145+200+185)=250$ .

Extreme value have no effect on _______.

0%
average
0%
median
0%
geometric mean
0%
harmonic mean

Paasche Price Index Number = ?

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

Explanation

Paasche Price Index Number

$P^{p}_{01} =\dfrac{\Sigma P_{1}Q_{1}}{\Sigma P_{0}Q_{1}} \times {100} = \dfrac{398}{184}\times {100} = 216.30$ .

Find the present value of Rs.

$10,000$ to be required after

$5$ years if the interest rate be

$9\%$ . Given that

$(1.09)^5=1.5386$ .

0%
$6,994.42$
0%
$6,949.24$
0%
$6,449.24$
0%
$6,499.42$

Explanation

Here,

$i=0.09=9\%$

$n=5$

$A_n=10,000$
Required present value

$=\displaystyle\frac{A_n}{(1+i)^n}$

$=\displaystyle\frac{10,000}{(1+0.09)^5}$

$=Rs. 6499.42$ .

Calculate standard deviation of the following data.

X	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
f	$10$	$8$	$15$	$8$	$4$

0%
$-12$
0%
$12$
0%
$12.36$
0%
$152.77$

Explanation

Marks	Mid-values(X)	f	dxi $=X-25/10$	$fdx_if$	$dx_i^2$
$0-10$	$5$	$10$	$-2$	$-20$	$40$
$10-20$	$15$	$8$	$-1$	$-8$	$8$
$20-30$	$25$	$15$	$0$	$0$	$0$
$30-40$	$35$	$8$	$1$	$8$	$8$
$40-50$	$45$	$4$	$2$	$8$	$16$
Total		$45$		$-12$	$72$

$\sqrt{\displaystyle\frac{72}{45}-\frac{144}{45\times 45}}\times 10$

$=12.36$

Calculate standard deviation of the following data.

X	$10$	$11$	$12$	$13$	$14$	$15$	$16$	$17$	$18$
f	$2$	$7$	$10$	$12$	$15$	$11$	$10$	$6$	$3$

0%
$3.95$
0%
$1.99$
0%
$14$
0%
None of the above

Explanation

X	f	fx	X-Mean	$f(X-x)^2$
$10$	$2$	$20$	$-4$	$32$
$11$	$7$	$77$	$-3$	$63$
$12$	$10$	$120$	$-2$	$40$
$13$	$12$	$156$	$-11$	$2$
$14$	$15$	$210$	$0$	$0$
$15$	$11$	$165$	$1$	$11$
$16$	$10$	$160$	$2$	$40$
$17$	$6$	$102$	$3$	$54$
$18$	$3$	$54$	$4$	$48$
Total	$76$	$fX=1064$		$f(X-X)^2=300$

$^-{X}=\displaystyle\frac{1064}{76}=14$

$\sigma x=\sqrt{\displaystyle\frac{300}{76}}$

$=\sqrt{3.95}$

$=1.99$ .

Y bought a CD Player costing Rs.

$13,000$ by making a down payment of Rs.

$3,000$ and agreeing to make equal annual payment for four years. How much would be each payment if the interest on unpaid amount be

$14\%$ compounded annually?

$[P(4, 0.14)2.91371]$

0%
Rs. $3,432.05$
0%
Rs. $3,423.50$
0%
Rs. $3,342.05$
0%
Rs. $3,234.50$

Explanation

$=$ A.P(n, i)
Here

$n=4$ and

$i=0.14$

$=\displaystyle\frac{10,000}{P(4, 0.14)}$

$[P(20, 0.10)=8.51356$ from table

$2$ (a)]

$=\displaystyle\frac{10,000}{2.91371}$

$=$ Rs.

$3,432.05$ . Therefore each payment would be Rs.

$3,432.05$ .

The mean and standard deviation of

$10$ observations are

$35$ and

$2$ respectively. Find the changed mean and standard deviation if each observation is increased by

$4$ .

0%
$39$ & $6$
0%
$2$ & $39$
0%
$39$ & $2$
0%
$6$ & $39$

Explanation

When each observation is increased by

$4$ , the mean of the changed observations will also increase and it will become equal to

$35+4=39$ .
Since increment in all observations by a constant does not change the standard deviation. Thus, the mean and standard deviation when each observation is increased by

$4$ would be

$39$ and

$2$ respectively.

Paasche Price Index = ?

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

Explanation

Paasche Price Index Number

$P^{P}_{01} =\dfrac{\Sigma P_{1}Q_{1}}{\Sigma P_{0}Q_{1}} \times {100} = \dfrac{22}{14}\times{100} = 157.14$ .

Fisher Price Index Number = ?

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

Explanation

Fisher Price Index Number

$P^{p}_{01} = \sqrt{P^{l}_{01}\times P^{p}_{01}} = \sqrt{(221.98\times216.3)\times{100}} = 219.12$ .

Find the mean and standard deviation of the following observations: X

$=2, 5, 7, 8, 13$ .

0%
$7$ & $3.63$
0%
$3.63$ & $7$
0%
$7.63$ & $3$
0%
$3$ & $7.63$

Explanation

$\displaystyle\frac{2+5+7+8+13}{5}=7$

$\sqrt{\displaystyle\frac{4+25+49+64+169}{5}}-49=3.63$ .

Let

${ x }_{ 1 },{ x }_{ 2 },.....{ x }_{ n }$ be

$n$ observations with mean

$m$ and standard deviation

$s$ . Then the standard deviation of the observations

$a{ x }_{ 1 },a{ x }_{ 2 },.....a{ x }_{ n }$ , is

0%
$a+s$
0%
$s/a$
0%
$\left| a \right| s$
0%
$as$

Explanation

$S.D(ax) = \sqrt{E[ax-a\bar{x}]^2}=\sqrt{E[a(x-\bar{x})^2]}=\sqrt{a^2.E[(x-\bar{x})^2]}=|a|.\sqrt{E[(x-\bar{x})^2]}=|a|.S.D(x)=|a| s$

The Mean deviation about A.M. of the numbers

$3, 4, 5, 6, 7$ is :

0%
$25$
0%
$5$
0%
$1.2$
0%
$0$

Explanation

Mean deviation is defined as the average of the absolute deviation of the observations from their mean.
Here, the arithmetic mean of the given observations is

$\dfrac{3 + 4 + 5 + 6 + 7}{5} = 5$
Thus, mean deviation =

$\dfrac{|3 - 5| + |4 - 5| + |5 - 5| + |6 - 5| + |7 - 5|}{5} = \dfrac{6}{5} = 1.2$
Hence, option 'C' is correct.

Mean deviation for

$n$ observations

${ x }_{ 1 },{ x }_{ 2 },.....{ x }_{ n }$ from their mean

$\bar { X }$ is given by:

0%
$\sum _{ i=1 }^{ n }{ ( { x }_{ i }-\bar { X }) }$
0%
$\cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-\bar { X } \right| }$
0%
$\sum _{ i=1 }^{ n }{ { { (x }_{ i }-\bar { X } ) }^{ 2 } }$
0%
$\cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ { { (x }_{ i }-\bar { X } ) }^{ 2 } }$

Explanation

It is fundamental concept that mean deviation of

$n$ observations

$x_1, x_2, x_3, ......x_n$ about mean

$\bar{X}$ is given by,

$\displaystyle \frac{1}{n} \sum_{i=1}^{i=n} |x_i-\bar{X}|$

If a variable

$X$ takes values

$0,1,2,....,n$ with frequencies

${ _{ }^{ n }{ C } }_{ 0 },{ _{ }^{ n }{ C } }_{ 1 },{ _{ }^{ n }{ C } }_{ 2 },......{ _{ }^{ n }{ C } }_{ n }\quad$ respectively, then S.D. is equal to :

0%
$\cfrac { n }{ 4 }$
0%
$\cfrac { n }{ 2 }$
0%
$\cfrac { \sqrt { n } }{ 2 }$
0%
$\cfrac { \sqrt { n } }{ 4 }$

Explanation

Here,

$\displaystyle \mu1'$

$=\frac{\sum r\frac{n}{r}^{n-1}C_{r-1}}{\sum ^{n}C_{r}}$

$\displaystyle$

and

$\displaystyle \mu 2'$

$\displaystyle=\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )^{n}C_{r}+\frac{n}{2}$

$\displaystyle =\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )\frac{n\left ( n-1 \right )}{r\left ( r-1 \right )}^{n-2}C_{r-2}+\frac{n}{2}$

$\displaystyle =\frac{n\left ( n-1 \right)}{2^{n}}.2^{n-2}+\frac{n}{2}$

$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}$

$\therefore$ Variance

$\displaystyle \sigma^2=\mu 2'-\left (\mu 1' \right )^{2}$

$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}-\left ( \frac{n}{2} \right )^{2}=\frac{n}{4} \mbox{or} S.D = \sigma = \sqrt{\sigma^2} = \frac{\sqrt{n}}{2}$

If the mean deviation about mean

$1,1+d,1+2d,....1+100d$ from their mean is

$255$ , then the

$d$ is equal to

0%
$10$
0%
$20$
0%
$10.1$
0%
$20.2$

Explanation

Clearly given observation is in A.P, and have

$101$ terms.

$\therefore$ Mean

$\displaystyle =\frac{1}{2}\left \{ 1+1+100d\right \}=1+50d$
And thus mean deviation about mean

$\displaystyle =\frac{1}{101}\sum \left | x-\bar{x} \right |$

$\displaystyle =\frac{1}{101}.2d\left ( 1+2+3\cdots +50 \right )$

$\displaystyle =\frac{50\left ( 51 \right )d}{101}=255$ (given)

$\Rightarrow d = 10.1$

The mean deviation of the series

$a,a+d,a+2d+......,a+(2n-1)d,a+2nd$ about the mean is

0%
$\cfrac { n+1 }{ 2n+1 }$
0%
$\cfrac { n(n+1)d }{ 2n+1 }$
0%
$\cfrac { (n+2)d }{ 2n }$
0%
$\cfrac { (n-1)d }{ 2n+1 }$

Explanation

Mean

$\displaystyle\overline{x}=\frac{(a)+(a+d)+(a+2d)+...+(a+2nd)}{2n+1}$
where

$\displaystyle\overline{x}=\frac{\sum x_i}{n}$
Then,

$\displaystyle\frac{\frac{2n+1}{2}(a+a+2nd)}{2n+1}$

From an A.P.

$\displaystyle S_n = \frac{n}{2} (a+l)$ which gives

$\overline{x} = a+nd$
The series being a,a+d,a+2d,...,a+(n-1)d,a+nd,a+(n+1)d,...,a+2nd
Mean deviation from the mean
=

$\displaystyle\frac{1}{N} \sum f_i [x_i- \overline{x}]$
=

$\displaystyle\frac{1}{2n+1} \sum [x_i -a-nd]$
=

$\displaystyle\frac{1}{2n+1} [nd+(n-1)d+(n-2)d+...+d+0+d+...+nd]$
=

$\displaystyle\frac{2d}{2n+1} [n+(n-1)+(n-2)+...1]$
=

$\displaystyle\frac{2d}{2n+1} .\frac{n(n+1)}{2} = \frac{n(n+1)d}{2n+1}$

CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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