MCQ Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 7

If in a data,

$10$ numbers arranged in increasing order. If the

$7$ th entry is increased by

$4,$ then the median increases by

0%
zero
0%
$4$
0%
$6$
0%
$5$

Explanation

The median would be average of the

$5^{th}$ and

$6^{th}$ observations. Since they are unaffected by increase in

$7^{th}$ entry, the median will be unchanged.

The numbers

$3, 5, 7$ and

$9$ have their respective frequencies

$x-2, x+2, x-3$ and

$x+3$ . If the arithmetic mean is

$6.5$ , then the value of

$x$ is

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

Mean

$= 6.5 = \dfrac{3 \times (x-2) + 5 \times (x + 2) + 7 \times (x - 3) + 9 \times (x + 3)}{x-2+x+2+x-3+x+3}$

$\therefore$

$6.5 = \dfrac{3x - 6 + 5x + 10 + 7x - 21 + 9x + 27}{4x}$

$\therefore$

$6.5 \times 4x = 24x + 10$

$\therefore$

$26x - 24x = 10$

$\therefore$

$x = 5$

Mode of the data

$15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15$ is

0%
$14$
0%
$15$
0%
$16$
0%
$17$

Explanation

Given Data:

$15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15$

Arranging it in ascending order:

$14,14,14,14,15$ ,

$15,15,15,15,16,17,18,19,19,20$

$15$ occurs

$5$ times and most number of time.

Hence, it is the mode of the data.

The mean is

0%
$70$
0%
$68$
0%
$71$
0%
$69$

Explanation

Mean of a frequency distribution =

$\displaystyle \dfrac{\sum f_i x_i}{\sum f_i}$
Mean

$= \dfrac {5520}{80}$
Mean

$= 69$

Median of the following numbers:

$4, 4, 5, 7, 6, 7, 7, 12, 3$ is

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

Explanation

Terms are:

$4, 4, 5, 7, 6, 7, 7, 12, 3$ .
Arranging the terms in ascending order:

$3,4,4,5,6,7,7,7,12$ .

Since the number of terms is odd the median will be the middle term i.e.

$5^{th}$ term which is

$6$ .

The number of observations in a group is

$40$ . If the average of first

$10$ is

$4.5$ and that of the remaining

$30$ is

$3.5$ , then the average of the whole group is

0%
$\displaystyle \frac{1}{5}$
0%
$\displaystyle \frac{15}{4}$
0%
$4$
0%
$8$

Explanation

$\textbf{Step -1: Finding the sum of first 10 observations.}$

$\text{Average of first 10 observations}=4.5$

$\mathbf{\because\text{Average}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}.}$

$\therefore4.5=\dfrac{\text{Sum of first 10 observations}}{10}$

$\Rightarrow\text{Sum of first 10 observations}=4.5\times10=45$

$\textbf{Step -2: Finding the sum of remaining 30 observations.}$

$\text{Average of remaining 30 observations}=3.5$

$\therefore3.5=\dfrac{\text{Sum of remaining 30 observations}}{30}$

$\Rightarrow\text{Sum of remaining 30 observations}=3.5\times30=105$

$\textbf{Step -3: Finding the average of the whole group.}$

$\text{Sum of total 40 students}=45+105=150$

$\text{Number of total observations}=40$

$\text{Average of the whole group}=\dfrac{150}{40}$

$=\dfrac{15}{4}$

$\textbf{ Hence, option B is correct.}$

The mean weight of

$20$ students is

$25$ kg and the mean weight of another

$10$ students is

$40$ kg. Find the mean weight of the

$30$ students.

0%
$20$
0%
$30$
0%
$40$
0%
$50$

Explanation

$\text{Given Mean weight of 20 students}=25\\ \Rightarrow \dfrac { \text{Sum of weight of 20 students} }{ 20 } =25\\ \Rightarrow \text{Sum of weight of 20 students}=25\times 20=500kg\\ \text{Given Mean weight of 10 students}=40\\ \Rightarrow \dfrac { \text{Sum of weight of 10 students} }{ 10 } =40\\ \Rightarrow \text{Sum of weight of 10 students}=40\times 10=400kg$
Mean weight of all 30 students

$\Rightarrow \dfrac { \text{Sum of weight of 30 students} }{ 30 } \\ \Rightarrow \dfrac { \text{Sum of weight of 20 students+Sum of weight of 10 students} }{ 30 } \\ =\dfrac { 500+400 }{ 30 } =\dfrac { 900 }{ 30 } =30kg$

If three sets of data had means of

$15, 22.5$ and

$24$ based on

$6, 4,$ and

$5$ observations respectively, then the mean of these three sets combined is

0%
$20.0$
0%
$20.5$
0%
$22.5$
0%
$24.0$

Explanation

Mean of combined three sets

$=\dfrac{Total \space Sum}{Total \space Count}$

$=\dfrac{15\times6 + 22.5\times4 + 24\times5}{6+4+5}$

$=\dfrac{300}{15}$

$= 20$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

Arranging the terms in ascending order, 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52
median value

$=\displaystyle \left ( \frac{11+1}{2} \right )^{th}=6^{th} value = 27$

The mean of the following distribution is

class	Frequency
0-5	4
5-10	5
10-15	7
15-20	12
20-25	7
25-30	5

0%
$15$
0%
$16$
0%
$17$
0%
$18$

Explanation

Mean

$=$

$\dfrac{sum \ of \ (Class mark \times frequency)}{Sum \ of \ frequency}$

Mean

$=$

$\dfrac{2.5\times 4 + 7.5\times5 + 12.5\times 7+ 17.5\times12 + 22.5\times 7 + 27.5\times 5}{4 +5+7+12+7 +5}$
Mean =

$\dfrac{640}{40} = 16$

The numbers

$4, 6\ and\ 8$ are the frequencies of

$(x + 2), x \ and\ (x - 1)$ respectively . If their arithmetic mean is

$8,$ the value of

$x$ is

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

$8 =$

$\dfrac{4\times \left ( x+2 \right )+6x+8\times \left ( x-1 \right )} {4+6+8}$

$=\dfrac{4x+8+6x+8x-8}{18}$

$\therefore$

$x = 8$

A contractor employed

$18$ labourers at Rs.

$12$ per day,

$10$ labourers at Rs.

$13.50$ per day,

$5$ labourers at Rs.

$25$ per day and

$2$ labourers at Rs.

$42$ per day. The average wage of a labourer per day is

0%
Rs. $16$
0%
Rs. $20$
0%
Rs. $24$
0%
Rs. $28$

Explanation

Mean =

$\dfrac{Total Sum}{Total Count}$

Mean =

$\dfrac{18\times12 + 10\times13.50 + 5\times25 + 2\times42}{18+10+5+2}$
=

$\dfrac{560}{35}$

$= 16$

In a class test in English

$10$ students scored

$75$ marks,

$12$ students scored

$60$ marks,

$8$ scored

$40$ marks and

$3$ scored

$30$ marks, the mode for their score is

0%
$75$
0%
$30$
0%
$60$
0%
$25$

Explanation

$\textbf{Step -1: Defining mode.}$

$\text{In a given data set, the mode is the value that appears most often among every other values.}$

$\text{Given data set says, }$

$\text{10 Students scored 75 marks.}$

$\text{12 Students scored 60 marks.}$

$\text{8 Students scored 40 marks.}$

$\text{3 Students scored 30 marks.}$

$\textbf{Step -2: Finding the mode of their score.}$

$\text{The mode of their score will be the marks occurring most often.}$

$\therefore \text{12 Students scored 60 marks.}$

$\text{Hence 60 is the mode of their score.}$

$\textbf{Hence,The mode of their score is 60. (Option C)}$

The median of the following numbers

$25, 27, 21, 23, 24$ is ___________.

0%
$21$
0%
$23$
0%
$25$
0%
$24$

Explanation

Arranging the numbers in ascending order:

$21,23,24,25,27$

Median of the numbers is

$\dfrac{n+1}{2}th$ number from starting,

where

$n$ is the no of total numbers.

Here

$n=5$

$\text{median}=3^{rd}$ term

$=24$

The examination marks of

$250$ pupils are recorded below. Find the mean.

Class interval	0-9	10-19	20-29	30-39	40-49
Frequency	0	2	6	24	36
Class interval	50-59	60-69	70-79	80-89	90-99
Frequency	47	55	40	27	13

0%
$50.12$
0%
$90.31$
0%
$60.82$
0%
$75.12$

Explanation

$Mean = \displaystyle \frac{Total (mid value \times frequency)}{Total frequency}$

Class interval	Mid- value	Fequency.	Mid-value $\times$ frequency
0-9	4.5	0	0
10-19	14.5	2	29
20-29	24.5	6	147
30-39	34.5	24	828
40-49	44.5	36	1602
50-59	54.5	47	2561.5
60-69	64.5	55	3547.5
70-79	74.5	40	2980
80-89	84.5	27	2281.5
99-99	94.5	13	1228.5
		Total - 250	Total 15205

$Mean = \displaystyle \frac{Total (mid value \times frequency)}{Total frequenct} = \frac{15205}{250}= 60.82$

The marks of

$20$ students in a test were as follows:

$5, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20$

The median is

0%
$13.5$
0%
$12.5$
0%
$14.5$
0%
$15.5$

Explanation

Here,

$n$

$=$

$20$ which is an even number.

$\therefore$ Median

$= \displaystyle \cfrac{\left ( \dfrac{n}{2} \right )^{th} \text{term} + \left ( \dfrac{n}{2}+1\right )^{th} \text{term}}{2}$

$\therefore$ Median

$= \displaystyle \cfrac{\left ( \dfrac{20}{2} \right )^{th} \text{term} + \left ( \dfrac{20}{2}+1\right )^{th} \text{term}}{2}$

$\displaystyle =\cfrac{10^{th}\: \text{term} + 11^{th}\: \text{term}}{2}$

$\displaystyle =\cfrac{13+14}{2}=13.5$

The median of variables 5, 8, 10, 13, 17, 21, 27, 30, 33, 42 is

0%
21
0%
17
0%
38
0%
19

Arithmetic mean for the given data is

Marks	No. of students
0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60	5 10 25 30 20 10

0%
$55$
0%
$35$
0%
$2$
0%
$33$

Explanation

Mean

$=$

$\dfrac{\text {Sum of} (\text {class mark} \times \text {frequency})}{\text {Sum of frequency}}$

Mean

$=$

$\dfrac{5\times 5 + 15\times10 + 25\times 25+ 35\times30 + 45\times 20 + 55\times 10}{5 +10+25+30+20 +10}$

Mean

$=$

$\dfrac{3300}{100}$ =

$33$

The average weight of

$10$ men is decreased by

$3 \space kg$ when one of them whose weight is

$80 kg$ is replaced by a new person. The weight of the new person is

0%
$70$
0%
$60$
0%
$50$
0%
$73$

Explanation

Let the mean of

$10$ persons =

$M$
Sum of weight of the persons

$= 10M$

After replacing a person weighing 80 Kg with another weighing say

$x$ Kg,

$sum = 10M - 80 +x$
New Mean

$= M-3$
hence,

$10(M-3) = 10M - 80 +x$

$10M- 30= 10M -80 +x$

$x = 50 Kg$

In a family, there are

$8$ men,

$7$ women and

$5$ children whose mean ages separately are respectively

$24,20$ and

$6$ years. The mean age of the family is

0%
$17.1$ years
0%
$18.1$ years
0%
$19.1$ years
0%
none of these

Explanation

Here we have three collections for which

${ A }_{ 1 }=24,{ n }_{ 1 }=8;{ A }_{ 2 }=20,{ n }_{ 2 }=7$ and

${ A }_{ 3 }=6,{ n }_{ 3 }=5$ .
Their combined mean is the required mean.
By the formula

$\displaystyle A=\frac { { n }_{ 1 }{ A }_{ 1 }+{ n }_{ 2 }{ A }_{ 2 }+{ n }_{ 3 }{ A }_{ 3 } }{ { n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 } } =\frac { 8\times 24+7\times 20+5\times 6 }{ 8+7+5 }$

$\displaystyle =\frac { 192+140+30 }{ 20 } =\frac { 362 }{ 20 } =18.1$

$\therefore$ The mean age of the family

$=18.1$ years

Marks obtained by 60 students of a class are shown as

Marks	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Number of students	8	7	$f_1$	15	5	$f_2$	7

the mean marks are 64, then

$f_1$ :

$f_2$ is

0%
$7 : 4$
0%
$4 : 5$
0%
$5 : 4$
0%
None of these

Explanation

Total number of student is 28.

$\Rightarrow f_1+f_2+42=60\Rightarrow f_1+f_2 = 18$ (i)

And mean

$=\cfrac{35\times 8+45\times 7+55f_1+65\times 15+75\times 5+85f_2+95\times 7}{60} = 64$ (given)

$\Rightarrow 11f_1+17f_2 = 246$ (ii)

solving these two equation we get,

$f_1 = 10, f_2 = 8$

A group of

$10$ items has arithmetic mean

$6$ . If the arithmetic mean of

$4$ of these items is

$7.5$ , then the mean of the remaining items is

0%
$6.5$
0%
$5.5$
0%
$4.5$
0%
$5.0$

Explanation

We have,

$\bar{X}=6,\bar{X_1} = 7.5, \bar{X_2}=?, w_1=4, w_2=6$

$\therefore \displaystyle \bar{X} = \frac{\bar{X_1}w_1+ \bar{X_2}w_1 }{w_1+w_2}=\frac{7.5\times 4+\bar{X_2}\times 6}{10}=6\Rightarrow \bar{X_2} = 5$

The daily wages (in Rs.) of ten workers are

$20,25,17,18,8,15,22,11,9,14$ . The median of wages is

0%
$15$
0%
$16$
0%
$17$
0%
$14$

Explanation

Arranging the wages in ascending order of magnitude, we have

$\quad 8,9,11,14,15,17,18,20,22,25$ .
Since there are

$10$ observations. Therefore, Median is arithmetic mean of

$\left(\displaystyle\frac{10}{2}\right)^{th}$ and

$\left(\displaystyle\frac{10}{2} + 1\right)^{th}$ observations.
So, the Median =

$\displaystyle\frac{15+17}{2} = 16$

The number of observations in a group is

$40$ . If the average of first

$10$ is

$4.5$ and that of the remaining

$30$ is

$3.5$ , then the average of the whole group is:

0%
$\displaystyle\frac{15}{4}$
0%
$\displaystyle\frac{1}{5}$
0%
$8$
0%
$4$

Explanation

$\textbf{Step -1: Finding the sum of first 10 observations.}$

$\text{Average of first 10 observations}=4.5$

$\mathbf{\because\text{Average}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}.}$

$\therefore4.5=\dfrac{\text{Sum of first 10 observations}}{10}$

$\Rightarrow\text{Sum of first 10 observations}=4.5\times10=45$

$\textbf{Step -2: Finding the sum of remaining 30 observations.}$

$\text{Average of remaining 30 observations}=3.5$

$\therefore3.5=\dfrac{\text{Sum of remaining 30 observations}}{30}$

$\Rightarrow\text{Sum of remaining 30 observations}=3.5\times30=105$

$\textbf{Step -3: Finding the average of the whole group.}$

$\text{Sum of total 40 students}=45+105=150$

$\text{Number of total observations}=40$

$\text{Average of the whole group}=\dfrac{150}{40}$

$=\dfrac{15}{4}$

$\textbf{ Hence, option A is correct.}$

Obtain the median for the following frequency distribution

x:	1	2	3	4	5	6	7	8	9
f :	8	10	11	16	20	25	15	9	6

0%
$6$
0%
$4$
0%
$3$
0%
$5$

Explanation

Calculate of Median
Here N

$= 120 \Rightarrow \displaystyle \frac{N}{2}=60$
we find that the cumulative frequency just greater than N/2, is 65 and the value of x corresponding to 65 is 5. Therefore, median is 5.

x	f	c.f
1	8	8
2	10	18
3	11	29
4	16	45
5	20	65
6	25	90
7	15	105
8	9	114
9	6	120
		N=120

$x_i$ 's are the mid points of the class intervals of grouped data,

$f_1$ 's are the corresponding frequencies and

$\bar x$ is the mean, then

$\sum (f_i x_i - \bar x)$ is equal to

0%
$0$
0%
$-1$
0%
$1$
0%
$2$

Explanation

Given that,

$x_i$ are the mid points of the class intervals.

$f_i$ are the corresponding frequencies.

The mean

$\bar x$ for the grouped data is

$\bar x=\dfrac{\sum f_ix_i}{n}$

$\implies \bar xn=\sum f_ix_i$ ------(1)

$\sum (f_ix_i-\bar x)=\sum f_ix_i-\sum \bar x$

$\implies \sum (f_ix_i-\bar x)=\bar xn-\sum \bar x$ (from (1))

$\implies \sum (f_ix_i-\bar x)=\bar xn-\bar xn=0$

$u_i \displaystyle = \frac{x_i - 25}{10} , \sum f_i u_i=20$ and

$\sum f_i =100$ , then

$\overline {x}$ is equal to

0%
$27$
0%
$25$
0%
$30$
0%
$35$

Explanation

$\text{Mean} = \displaystyle a + \frac{\sum f_i u_i}{\sum f_i} \times h , u_i = \frac{x_i -a}{h}$

$a = 25 , h = 10, \sum f_i u_i = 20, \sum f_i = 100$
Mean

$= 25 + \displaystyle \frac{20}{100} \times 10 = 27$

A firm of readymade garments make both men's and women's shirt. Its profit average is

$6$ % of sales. Its profits in men's shirt average

$8$ % of sales and women's shirts comprise

$60$ % of output. The average profit per sales rupee in women's shirts is :

0%
$0.0466$
0%
$0.0166$
0%
$0.0666$
0%
None of these

Explanation

Here

$\overline { x } =6,\overline { { x }_{ 1 } } =8,{ n }_{ 1 }=40,{ n }_{ 2 }=60$ .
Assuming that the total output is

$100,$ we are required to find out

$\overline { { x }_{ 2 } }$ , we know that

$\displaystyle \overline { x } =\frac { { n }_{ 1 }\overline { { x }_{ 1 } } +{ n }_{ 2 }\overline { { x }_{ 2 } } }{ { n }_{ 1 }+{ n }_{ 2 } } =\frac { 40\times 8+60\overline { { x }_{ 2 } } }{ 40+60 }$

$\displaystyle \Rightarrow 6=\frac { 320+60\overline { { x }_{ 2 } } }{ 100 } \Rightarrow \overline { { x }_{ 2 } } =\frac { 600-320 }{ 60 } =\frac { 280 }{ 60 } =4.66$
Thus, the average profit in womens shirt is

$4.66$ % of sales or

$Rs \space 0.0466$ per sale rupee.

Following table gives frequency distribution of trees planted by different housing societies in a particular locality.
Find mean number of trees planted by housing society by using 'step deviation method'.

No. of tress	10 - 15	15 - 20	20 - 25	25 - 30	30 - 35	35 - 40
No. of Societies	2	7	9	8	6	4

0%
$25.42$ trees
0%
$27.42$ trees
0%
$29.42$ trees
0%
$31.42$ trees

Explanation

Consider the following table, to calculate mean by "step deviation method":

$x_i$ =mid value of class interval

Assumed mean

$a=27.5$

class interval

$c=5$

$ci$	$f_i$	$x_i$	$u_i=\dfrac{x_i-a}{c}$	$f_iu_i$
10-15	2	12.5	$\dfrac{12.5-27.5}{5}=$ -3	-6
15-20	7	17.5	-2	-14
20-25	9	22.5	-1	-9
25-30	8	27.5	0	0
30-35	6	32.5	1	6
35-40	4	37.5	2	8

$N=\Sigma f_i=36$

$\Sigma f_iu_i=-15$

Mean

$\overline x=a +\dfrac {\Sigma f_iu_i}{N}\times c$

$\therefore \overline x=27.5 + \dfrac{-15}{36} \times 5=25.42$

Hence, option

$A$ is corect.

The results of two colleges are as follows: then

		M.A	M.Sc.	B.A.	B.Sc.	Total
A	Appeared	100	60	120	200	480
	passed	90	45	75	150	360
B	Appeared	240	200	160	200	800
	Passed	200	160	100	140	600

0%
A is better than B
0%
None is better than the other
0%
A is inferior to B
0%
can not say

Explanation

Total number of students of College A appeared

$=480$

Total number of students of College A passed

$=360$

$\therefore$ passing percentage

$=\dfrac{360}{480}\times 100=75$ %

Total number of students of College B appeared

$=800$

Total number of students of College A passed

$=600$

$\therefore$ passing percentage

$=\dfrac{600}{800}\times 100=75%$ %

Since Both colleges have passing percentage as same,none is better than other.

Hence, option

$B$ is correct.

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 7 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 7 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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