MCQ Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 6

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

0%
10%
0%
40%
0%
50%
0%
none of these

Explanation

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

$=\cfrac{\sigma}{\bar{x}}\times 100 =\cfrac{5}{50}\times 100$ %

$= 10$ %

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

0%
$\displaystyle \beta=5\alpha$
0%
$\displaystyle \beta=4\alpha+3$
0%
$\displaystyle \beta=4\alpha$
0%
None

Explanation

$A$

$B$

$3$

$15$

$4$

$19$

$5$

$23$

$.$

$n$

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

$\dfrac{\sum B}{n}=\beta$

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

$B$

$a=15$

$d=4$

$\sum B=\dfrac{n}{2}[2a+(n-1)d]$

$\dfrac{n}{2}[30+(n-1)4]$

$\dfrac{n}{2}[26+4n]$

$\beta=\dfrac{\sum B}{n}=\dfrac{26+4n}{2}$

$\beta=\dfrac{26+4(2\alpha-5)}{2}$

$2\beta=6+8\alpha$

$\beta=4\alpha+3$

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

0%
$4$
0%
$5$
0%
$6$
0%
$7$

Explanation

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

Which of the following are true or false?

a) T-distribution varies from+infinity to-infinity.
b)The variance of t distribution and the variance of normal distribution become closer and closer as the size of the sample increases.

0%
both (a) and (b) are true
0%
both (a) and (b) are false
0%
(a) is true but (b) is false
0%
(a) is false but (b) is true

Standard deviation is ________.

0%
most widely used measure of dispersion
0%
independent of origin
0%
not independent of scale
0%
all of these

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

0%
$9$
0%
$10$
0%
$11$
0%
$12$

Explanation

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

The mean of a finite set X of numbers is

$14$ , the median of this set of numbers is

$12$ , and the standard deviation is

$1.8$ . A new set Y is formed by multiplying each member of the set S by

$3$ . Determine the correct statements w.r.t. set Y:
I. The mean of the numbers is set Y is

$42$
II. The median of the numbers in set Y is

$36$
III. The standard deviation of the numbers is set Y is

$5.4$

0%
I only
0%
II only
0%
I and II only
0%
I and III only
0%
I, II, and III

Explanation

For set

$x$

Mean

$=14$

Median

$=12$

S.D.

$=\sqrt { \sigma }$

$=1.8$ .

For set

$y$

Mean

$=14\times 3= 42$ .

Median

$=12\times 3=36$ .

$=1.8\times 3= 5.4$ .

Hence all three are correct.

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

0%
$43$
0%
$44$
0%
$45$
0%
$46$

Explanation

Answer:- By shortcut Method

Class interval width (i)= 24-12 = 12

X	F	$d=\cfrac{x-A}{i}$	Fd
12=A	1	0	0
24	2	1	2
36	3	2	6
48	4	3	12
60	5	4	20
	$\Sigma f = 15$		$\Sigma fd = 40$

Mean =

$A+\cfrac{\Sigma fd}{\Sigma f} \times i=12+\left(\cfrac{40}{15} \times 12\right)={12+32}=44$

B)44

Find the coefficient of variation.

0%
$72.66$
0%
$81.24$
0%
$264$
0%
$330$
0%
None

Explanation

$\quad X\\ \quad 03\\ \quad 05\\ \quad 12\\ \quad 03\\ \quad 02\\ \_ \_ \_ \_ \_ \\ \quad 25$

$x-\overline { x } \\ -2\\ \quad 00\\ \quad 07\\ -2\\ -3$

${ \quad \quad (x-\overline { x } ) }^{ 2 }\\ \quad \quad \quad 04\\ \quad \quad \quad 00\\ \quad \quad \quad 49\\ \quad \quad \quad 04\\ \quad \quad \quad 09\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \sum { { (x-\overline { x } ) }^{ 2 }=66 }$

Variance

$= \cfrac { \sum { { (x-\overline { x } ) }^{ 2 } } }{ n-1 } \\ =\cfrac { 66 }{ 4 } \\ =16.5$ .

$\sigma =S.D.=\sqrt { 16.5 }$ .

Coefficient of variation

$= \cfrac { \sigma }{ x } \\ =\cfrac { \sqrt { 16.5 } }{ 5 } \\ =0.8124$ .

% of coefficient of variation

$= 0.8124*100\\ =81.24$ .

Which of the following can be used as measures of dispersions?

0%
Range
0%
Percentiles
0%
Standard Deviation
0%
All

The mean of a distribution is

$14$ and standard deviation is

$5$ . What is the value of the coefficient of variation?

0%
$57.7\%$
0%
$45.7\%$
0%
$35.7\%$
0%
None of these

Explanation

Coefficient of variation is given by

$CV = \dfrac{SD}{Mean}\times 100$

$\Rightarrow \dfrac{5}{14}\times 100 = 35.7\%$

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

0%
$15.833$
0%
$16.833$
0%
$17.833$
0%
$18.833$

Explanation

Answer:- By shortcut Method

Class interval width (i) =

$5-0 = 5$

X	C. I. mid point	F	$d=\cfrac{X-A}{i}$	Fd
0-5	2.5	4	-1	-4
5-10	7.5=A	8	0	0
10-15	12.5	12	1	12
15-20	17.5	16	2	32
20-25	22.5	20	3	60
		$\Sigma F = 60$		$\Sigma Fd=100$

Mean =

$A+\cfrac{\Sigma fd}{\Sigma f} \times i=7.5+\left(\cfrac{100}{60} \times 5\right)={7.5+8.33}=15.83$

$15.83$

A random variable

$X$ has the probability distribution given below. Its variance is

X	1	2	3	4	5
P(X=x)	k	2k	3k	2k	k

0%
$\dfrac{16}{3}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{5}{3}$
0%
$\dfrac{10}{3}$

Explanation

We know that

$\sum P(x) = 1$

$k+2k+3k+2k+k=1$

$9k=1$

$k=\frac{1}{9}$

Now, var

$\sigma^2 = \sum xi^2P(xi)-(\sum\,xiP(xi))^2$

$=1(k)+2^2(2k)+3^2(3k)+4^2(2k)+5^2(k)-[1(k)+2(2k)+3(3k)+4(2k)+5(k)]^2$

$=93k-(27k)^2$

$=93(\cfrac{1}{9})-[27(\frac{1}{9})]^2$

$=\cfrac{31}{3}-3^2$

$=\cfrac{31-27}{3}$

$=\cfrac{4}{3}$

If the standard deviation of a set of scores is

$1.2$ and their mean is

$10$ , then the coefficient of variation of the scores is

0%
$12$
0%
$0.12$
0%
$20$
0%
$120$

Explanation

Given : standard deviation

$(\sigma)=1.2,$ mean

$(\overline {X})=10$ .

Coefficient of variation(C.V.)

$=\dfrac{\sigma}{\overline {X}}\times 100=\dfrac{1.2}{10}\times 100=12$

$\therefore$ C.V.

$=12$

Hence, option

$A$ is correct.

Standard Deviation of n observations

$a_1, a_2, a_3 .....a_n$ is

$\sigma$ Then the standard deviation of the observations

$\lambda a_1, \lambda a_2, ..., \lambda a_n$ is

0%
$\lambda \sigma$
0%
$-\lambda \sigma$
0%
$|\lambda| \sigma$
0%
${\lambda}^n \sigma$

Explanation

Let the mean of original data is

$\mu$

Then mean of observation if all the terms are multiplied by the constant

$\lambda$ will be

$\lambda \mu$

$\sigma$ be the standard deviation of original observation , then

$\sigma=\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n (x_i-\mu)^2 }$

Now let the standard deviation of new observation is

$\sigma'$ , then

$\sigma'=\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n (\lambda x_i-\lambda\mu)^2 }$

$=\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n \lambda^2( x_i-\mu)^2 }$

$=|\lambda|\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n( x_i-\mu)^2 }=|\lambda |\sigma$

Note that: standard deviation can never be negative

Find the variance of the following distribution.

Class interval	$20-24$	$25-29$	$30-34$	$35-39$	$40-44$	$45-49$
Frequency	$15$	$25$	$28$	$12$	$12$	$8$

0%
$416$
0%
$46$
0%
$16$
0%
None of these

Explanation

$Class$	$Mid-point$ $(x)$	$Frequency$ $(f)$	$xf$	$(x-\overline{x})$	$(x-\overline{x})^2$	$(x-\overline{x})^2f$
$20-24$	$22$	$15$	$330$	$-10.25$	$105.0625$	$1575.9375$
$25-29$	$27$	$25$	$675$	$-5.25$	$27.5625$	$689.0625$
$30-34$	$32$	$28$	$896$	$-0.25$	$0.0625$	$1.75$
$35-39$	$37$	$12$	$444$	$4.75$	$22.5625$	$270.75$
$40-44$	$42$	$12$	$504$	$9.75$	$95.0625$	$1140.75$
$45-49$	$47$	$8$	$376$	$14.75$	$217.5625$	$1740.5$
		$\sum f=100$	$\sum xf=3225$			$\sum(x-\overline{x})^2f=5418.75$

$\Rightarrow$

$Mean(\overline{x})=\dfrac{\sum xf}{\sum f}=\dfrac{3225}{100}=32.25$

$\Rightarrow$

$Variance(\sigma^2)=\dfrac{\sum(x-\overline{x})^2f}{\sum f}=\dfrac{5418.75}{100}=54.18$

If the median of the data 6,7,x-2,x,18,21 written in ascending order is 16, then the variance of that data is

0%
$30\dfrac{1}{5}$
0%
$31\dfrac{1}{3}$
0%
$32\dfrac{1}{2}$
0%
$33\dfrac{1}{3}$

Explanation

$M=\dfrac{x-2+x}{2}=16$

Variance =

$\dfrac{1}{n}\sum (x_i-\overline{x})^{2}$

Required variance=

$31\dfrac{1}{3}$

If the mean of the numbers

$a,b,8,5,10$ is

$6$ and their variance is

$68$ , then

$ab$ is equal to

0%
$6$
0%
$7$
0%
$12$
0%
$14$
0%
$25$

Explanation

Given,

$\cfrac { a+b+8+5+10 }{ 5 } =6$

$\Rightarrow a+b+c+23=30$

$\Rightarrow a+b=7....(i)$
and

$\quad \cfrac { \sum { { \left( { x }_{ 1 }-\overline { x } \right) }^{ 2 } } }{ n } ={ \sigma }^{ 2 }\quad$

$\cfrac { { (a-6) }^{ 2 }+{ (b-6) }^{ 2 }+4+1+16 }{ 5 } =\cfrac { 34 }{ 5 } \quad$

$\Rightarrow { (a-6) }^{ 2 }+{ (b-6) }^{ 2 }+21=34\quad$

$\Rightarrow { a }^{ 2 }+{ b }^{ 2 }-84+93=34\quad \left[ \because a+b=7 \right]$

$\quad \Rightarrow { a }^{ 2 }+{ b }^{ 2 }=25...(i)\quad$
From Eq.(i)

$\Rightarrow { (a+b) }^{ 2 }={ 7 }^{ 2 }$

$\Rightarrow { a }^{ 2 }+{ b }^{ 2 }+2ab=49$

$\Rightarrow 25+2ab=49$ (From Eq.(ii))

$\Rightarrow 2ab=24\Rightarrow ab=12\quad$

$n=10, \bar{x}=12$ and

$\sum x^2=1530$ , then calculate the coefficient of variation.

0%
$20$
0%
$25$
0%
$30$
0%
$35$

Explanation

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

$=\sqrt{\dfrac{1530}{10}-(12)^2}$

$=\sqrt{153-144}$

$=\sqrt{9}$

$=3$

Coefficient of variation

$=\dfrac{\sigma}{\overline{x}}\times 100$

$=\dfrac{3}{12}\times 100$

$=\dfrac{1}{4}\times 100$

$=25$

Calculate the standard deviation of the following data.

x	$3$	$8$	$13$	$18$	$23$
f	$7$	$10$	$15$	$10$	$8$

0%
$6.32$ .
0%
$2.32$ .
0%
$3.32$ .
0%
None of these

Explanation

$x$	$f$	$xf$	$(x-\overline{x})$	$(x-\overline{x})^2$	$(x-\overline{x})^2f$
$3$	$7$	$21$	$-10.2$	$104.04$	$728.28$
$8$	$10$	$80$	$-5.2$	$27.04$	$270.4$
$13$	$15$	$195$	$-0.2$	$0.04$	$0.6$
$18$	$10$	$180$	$4.8$	$23.04$	$230.4$
$23$	$8$	$184$	$9.8$	$96.04$	$768.32$
	$\sum f=50$	$\sum xf=660$			$\sum(x-\overline{x})^2=1998$

$\Rightarrow$

$\overline{x}=\dfrac{\sum xf}{\sum f}=\dfrac{660}{50}=13.2$

Now, calculate standard deviation :

$S=\sqrt{\dfrac{(x-\overline{x})^2}{\sum f}}=\sqrt{\dfrac{1998}{50}}=\sqrt{39.96}=6.32$

Calculate the standard deviation of the following data.

$10, 20, 15, 8, 3, 4$

0%
$5.97$
0%
$59.7$
0%
$4.97$
0%
None of these

Explanation

$\Rightarrow$ The given numbers are

$3,4,8,10,15,20$ .

$\Rightarrow$

$\overline{X}=\dfrac{3+4+8+10+15+20}{6}=10$

$X$	$X-\overline{X}$	$(X-\overline{X})^2$
$3$	$-7$	$49$
$4$	$-6$	$36$
$8$	$-2$	$4$
$10$	$0$	$0$
$15$	$5$	$25$
$20$	$10$	$100$
		$\sum (X-\overline{X})^2=214$

$\Rightarrow$

$S=\sqrt{\dfrac{(X-\overline{X})^2}{N}}$

$=\sqrt{\dfrac{214}{6}}$

$=\sqrt{35.66}$

$=5.97$

Let X denote the number of scores which exceed 4 in 1 toss of a symmetrical die. Consider the following statements :
The arithmetic mean of X is
The standard deviation of X is
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

The only possible favorable outcomes are

$5$ and

$6$ . Hence, the arithmetic mean of

$X$ is 5.5.

The standard deviation of

$X=\dfrac{1}{2}((6-5.5)^2+(5-55)^2)=0.25$

The arithmetic mean of the observations 10,8,5,a,b is 6 and their variance 6.Then ab=

0%
6
0%
4
0%
3
0%
12

Explanation

$\dfrac{10+8+5+a+b}{5}=6,$

$\dfrac{1}{n}\sum \left ( x_i-\overline{x} \right )^{2}=6.8$

solving above two equations we get values of a and b.

If the standard deviation is small, we define a new variable known as ________.

0%
student's F-distribution
0%
student's T-variable
0%
chi-square distribution
0%
student's G-variable

Coefficient of quartile deviation of numbers

$6, 8, 9, 10, 11, 12, 14$ is?

0%
$1/5$
0%
$1/10$
0%
$2/5$
0%
$1/4$

Explanation

Arranging given numbers in ascending order,

$6,8,9,10,11,12,14$

Number of observation

$(N)=7$

$Q_i=\left\{i\left(\dfrac{N+1}{4}\right)\right\}^{th}term$

$Q_1=\left\{1.\left(\dfrac{7+1}{4}\right)\right\}^{th}term=2^{nd}term$

$2^{nd}term=8$

$\therefore$

$Q_1=8$

$Q_3=\left\{3.\left(\dfrac{7+1}{4}\right)\right\}^{th}term=6^{th}term$

$6^{th}term=12$

$\therefore$

$Q_3=12$

$\Rightarrow$ Coefficient of quartile

$=\dfrac{Q_3-Q_1}{Q_3+Q_1}$

$=\dfrac{12-8}{12+8}$

$=\dfrac{4}{20}$

$=\dfrac{1}{5}$

Which of the following are measures of dispersion?

0%
Standard Deviation,Median,Range
0%
Standard Deviation,Mode,Range
0%
Standard Deviation,Variance,Range
0%
Mean,Mode,Median

What are the objectives of measure of dispersion?

0%
Helpful in use of further statistical analysis as in regression, correlation etc.
0%
Reliability of measure of central tendency
0%
Control of variability
0%
All of the above

Which of the following are measures of central tendency.

0%
Percentile, Quartile, Median
0%
Median,Mode, Percentile
0%
Percentile, Quartile, Mode
0%
Mean,Mode, Median

The runs scored by Virat in 7 different matches are given as:

$89,91,54,66,13,97,06$

Calculate the mean deviation about the median for this data.

0%
$31.26$
0%
$30.06$
0%
$29.142$
0%
$27.85$

Explanation

Arranging data in ascending order,

$06,13,54,66,89,91,97$

Here, number of observations

$(n)=7(odd)$

$\therefore$

$Median=\left(\dfrac{7+1}{2}\right)^{th}observation=4^{th}observation.$

$\therefore$

$Median(M) =66$

$x$	$\|x-M\|$
$06$	$60$
$13$	$53$
$54$	$12$
$66$	$0$
$89$	$23$
$91$	$25$
$97$	$31$
	$\sum \|x-M\|=204$

$\Rightarrow$ Mean deviation about median

$=\dfrac{\sum|x-M|}{7}=\dfrac{204}{7}=29.142$

The formula of standard error of sample mean is ________.

0%
$\dfrac{\sqrt{PQ}}{\sqrt{n}}$
0%
$\dfrac{\sigma}{\sqrt{n}}$
0%
$\dfrac{\sqrt{N-n}}{\sqrt{N-1}}$
0%
$\dfrac{\sqrt{N}}{\sigma}$

CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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