MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 2 - MCQExams.com
CBSE
Class 11 Commerce Economics
Measures Of Central Tendency
Quiz 2
Find the median of the data: $$ 35, 48, 92, 76,52, 51, 63$$ and $$71$$
Report Question
0%
$$71$$
0%
$$64$$
0%
$$52$$
0%
$$48$$
Explanation
Arranging the data in ascending order:
$$35,48,51,52,63,64,71,76,92$$
Mean $$=\dfrac {\text { sum of all numbers }}{ \text {count of all numbers }} = \dfrac { 35+48+51+52+63+64+71+76+92 }{ 9 } =\dfrac { 552 }{ 9 } =61.33$$
Since the number of data is odd the median will be the middlemost term.
So, Median $$=63$$ of
the numbers would become:
$$ 35,48,52,63,64,66,71,76,92$$
The new Median is $$64$$.
The marks obtained by twenty students are given as follows.
$$27, 22, 24, 17, 16, 15, 14, 22, 17, 15, 22, 26, 19, 22, 20, 22, 16, 15, 22, 27$$
Find the modal marks.
Report Question
0%
$$20$$
0%
$$22$$
0%
$$24$$
0%
$$26$$
Explanation
Marks obtained by $$20$$ students :
$$27, 22, 24, 17, 16, 15, 14, 22, 17, 15, 22, 26, 19, 22, 20, 22, 16, 15, 22, 27$$
Arranging them in ascending order: $$14, 15, 15,15, 16,16, 17, 17, 19, 20, 22, 22, 22, 22, 22, 22, 24, 26, 27, 27$$
$$22$$ occurs $$6$$ and maximum number of times in the data. Hence, $$22$$ is the modal marks.
Find the median of the following data.
$$150, 147, 140, 120, 135, 139, 148, 148$$
Report Question
0%
$$140$$
0%
$$143.5$$
0%
$$148$$
0%
$$150$$
Explanation
The data points are:
$$150, 147, 140, 120, 135, 139, 148, 148$$
Arranging in ascending order: $$120, 135, 139, 140, 147, 148, 148, 150$$
The number of observations are $$8$$. Since, the observations are even, the median will be the mean of $$\cfrac{n}{2}$$ and $$\cfrac{n+2}{2}$$ observations
median $$=$$ mean of $$\cfrac{8}{2}$$ and $$\cfrac{8+2}{2}$$ observations
median $$=$$ mean of 4th and 5th observations
median $$= \cfrac{140+ 147}{2}$$
median $$= 143.5$$
A cricket player scored runs in different matches as follows.
$$36, 41, 57, 89, 105, 103, 17$$. Find the median.
Report Question
0%
$$41$$
0%
$$57$$
0%
$$89$$
0%
$$103$$
Explanation
The data points are:
$$36, 41, 57, 89, 105, 103, 17$$
Arranging in ascending order: $$17, 36, 41, 57, 89, 103, 105$$
The number of observations are $$7$$. Since, the observations are odd, the median will be the $$\cfrac{n+1}{2}$$ observations
median $$= \cfrac{7+1}{2}$$ observations
median $$= 4^{th}$$ observations
median $$= 57$$
Find the median of the following data:
$$3.6, 9.4, 3.8, 5.6, 6.5, 8.9, 2.7, 10.8, 15.6, 1.9$$ and $$7.6.$$
Report Question
0%
$$6.5$$
0%
$$7.5$$
0%
$$4.5$$
0%
$$5.5$$
Explanation
Given data is $$3.6, 9.4, 3.8, 5.6, 6.5, 8.9, 2.7, 10.8, 15.6, 1.9, 7.6$$
Arranging them in ordered pair,
$$1.9, 2.7, 3.6, 3.8, 5.6, 6.5, 7.6, 8.9, 9.4, 10.8, 15.6$$.
Here, middle score is $$6.5$$.
Hence, median is $$6.5$$.
The weights of students of a certain class are given below.
$$39, 42, 47, 38, 42, 40, 42, 38, 43, 42, 38, 44, 46, 39, 42, 40, 43, 42, 41$$
Find the mode.
Report Question
0%
$$39$$
0%
$$40$$
0%
$$41$$
0%
$$42$$
Explanation
weight of students :
$$39, 42, 47, 38, 42, 40, 42, 38, 43, 42, 38, 44, 46, 39, 42, 40, 43, 42, 41$$
Arranging them in ascending order: $$38, 38, 38, 39, 39, 40, 40, 41, 42,42,42,42,42,42, 43, 43, 44, 46, 47$$
$$42$$ occurs $$6$$ and maximum number of times in the data. Hence, $$42$$ is the modal weight.
Find the mean and median of the data:
$$35, 48, 92, 76, 64, 52, 51, 63$$ and $$71$$.
If $$51$$ is replaced by $$66$$, what will be the new median?
Report Question
0%
$$65$$
0%
$$64$$
0%
$$66$$
0%
$$67$$
Explanation
Arranging data in ascending order- $$35, 48,51,52, 63, 64, 71,76, 92$$
Count of observations $$=9 $$
Mean $$=$$ Sum of all observations/count $$= \dfrac {552}{9}= 61.33$$
Median for odd number of data is the middle most observation
So median $$=$$ $$5^{th}$$ observation $$= 63$$
If $$51$$ is replaced by $$66$$, then the observations in ascending order will be- $$35, 48,52, 63, 64, 66,71,76, 92$$
Again Median for odd number of data is the middle most observation
So new median $$= 5^{th}$$ observation $$= 64$$
The weights (in $$kg$$) of $$10$$ children are:
$$40,52,34,47,31,35,48,41,44,38$$. Find the median weight
Report Question
0%
$$35.9\:kg$$
0%
$$38.4\:kg$$
0%
$$40.5\:kg$$
0%
$$42.5\:kg$$
Explanation
Arranging the weights in ascending order, we have: $$31$$,$$34$$,$$35$$,$$38$$,$$40$$,$$41$$,$$44$$,$$47$$,$$48$$,$$52$$
Here, $$n = 10$$, which is even
$$\therefore$$ Median weight $$= \cfrac { 1 }{ 2 } \left\{ \begin{pmatrix} \cfrac { 10 }{ 2 } \end{pmatrix}^{th} \text{term}+\begin{pmatrix} \cfrac { 10 }{ 2 } +1 \end{pmatrix}^{th} \text{term} \right\} $$
$$=\cfrac { 1 }{ 2 } \left\{ 5^{th} \text{term}+6^{th} \text{term} \right\} =\cfrac { 1 }{ 2 } \left\{ 40+41 \right\}\: kg$$
$$= \cfrac { 81 }{ 2 }\:kg = 40.5\:kg$$
Hence, median weight $$= 40.5\:kg$$
The following observations have been arranged in ascending order. If median of these observations is $$58$$, find the value of $$x$$.
$$24, 27, 43, 48, x - 1, x + 3, 68, 73, 80, 90$$
Report Question
0%
$$58$$
0%
$$57$$
0%
$$55$$
0%
$$59$$
Explanation
Given series is: $$24,27,43,48, x - 1, x + 3, 68, 73, 80, 90$$
Since, the series has $$10$$ observations, the median will be the mean of $$5^{th}$$ and $$6^{th}$$ observation.
Median $$= \dfrac{x - 1 + x +3}{2}$$
Median $$= x +1$$
$$\Rightarrow 58 = x + 1$$
$$\Rightarrow x = 57$$
Find the median of:
$$233, 173, 189, 208, 194, 204, 194, 185, 200$$ and $$220.$$
Report Question
0%
$$199$$
0%
$$200$$
0%
$$197$$
0%
$$198$$
Explanation
Arrange the series in ascending order:
$$173, 185, 189, 194, 194, 200, 204, 208, 220, 233$$
The series has $$10$$ numbers, even numbers.
Hence, the median will be the mean of the two middle numbers
Median $$=$$ mean of $$5^{th}$$ and $$6^{th}$$ terms
Median $$= \dfrac{194 + 200}{2}$$
Median $$= 197$$
To find mean, we use the formula
Report Question
0%
$$\displaystyle \sum_{i=1}^{n}f_ix_i$$
0%
$$N\displaystyle \sum_{i=1}^{n}f_ix_i$$
0%
$$\frac {1}{N}\displaystyle \sum_{i=1}^{n}f_ix_i$$
0%
$$\displaystyle \sum_{i=1}^{n}\left (\frac {f_i(x_i)^2}{N}\right )$$
Explanation
mean of the data = $$\frac{Sum \quad of \quad obeservations}{Number \quad of \quad observations}$$
If the data is a grouped data, frequency of each observation denoted by $$f_i$$ and observation by $$x_i$$, then
Mean = $$\frac {1}{N}\displaystyle \sum_{i=1}^{n}f_ix_i$$
Find the median of:
$$25, 16, 26, 16, 32, 31, 19, 28$$ and $$35$$.
Report Question
0%
$$25$$
0%
$$24$$
0%
$$26$$
0%
$$28$$
Explanation
Arranging the series in ascending order:
$$16, 16, 19, 25, 26, 28, 31, 32, 35$$
The series has $$9$$ terms, the median is the middle term.Hence, $$26$$ is the median of the series.
Find the median of:
$$241, 243, 347, 350, 327, 299, 261, 292, 271, 258$$ and $$257.$$
Report Question
0%
$$271$$
0%
$$275$$
0%
$$276$$
0%
$$285$$
Explanation
Arrange the series in ascending order:
$$241, 243, 257, 258, 261, 271, 292, 299, 327, 347, 350$$
The series has $$11$$ terms. Hence, the median will be the middle term, i.e. $$6^{th}$$ term.
Hence, the median is $$271$$
While computing mean of grouped data, we assume that the frequencies are:
Report Question
0%
evenly distributed over all the class
0%
centred at the classmarks of the class
0%
centred at the upper limits of the class
0%
centred at the lower limits of the class
Explanation
While computing mean of grouped data, we assume that the frequencies are centred at the class marks of the class.
Hence, option B is correct.
The mode of the distribution $$3, 5, 7, 4, 2, 1, 4, 3, 4$$ is
Report Question
0%
$$7$$
0%
$$4$$
0%
$$3$$
0%
$$1$$
Explanation
Observation
Frequency
$$3$$
$$2$$
$$5$$
$$1$$
$$7$$
$$1$$
$$4$$
$$3$$
$$2$$
$$1$$
$$1$$
$$1$$
$$4$$ has the highest frequency of $$3$$. So $$4$$ is the mode
What is the average of squares of consecutive odd numbers between $$1$$ and $$13 $$?
Report Question
0%
$$49$$
0%
$$51$$
0%
$$53$$
0%
$$57$$
Explanation
The consecutive odd numbers from $$1$$ to $$13 = 3, 5, 7, 9, 11$$
Thus required average = $$\displaystyle \frac{3^{2}+5^{2}+7^{2}+9^{2}+11^{2}}{5}=\frac{9+25+49+81+121}{5}=\frac{285}{5}=57$$
Find the mean of the following data, by using the assumed mean method.
Class Interval
$$0-10$$
$$10-20$$
$$20-30$$
$$30-40$$
$$40-50$$
Frequency
$$7$$
$$8$$
$$12$$
$$13$$
$$10$$
Report Question
0%
$$10$$
0%
$$30$$
0%
$$27.2$$
0%
None of these
Explanation
Answer:-
Class interval
Frequency ($$f_i$$)
$$ x_i = \cfrac{\text{lower limit + upper limit}}{2} $$
$$ f_ix_i $$
$$0-10$$
$$ 7$$
$$5$$
$$35$$
$$10-20$$
$$8$$
$$15$$
$$120$$
$$20-30$$
$$12$$
$$25$$
$$300$$
$$30-40$$
$$13$$
$$35$$
$$455$$
$$40-50$$
$$10$$
$$45$$
$$450$$
$$ \Sigma f_i = 50 $$
$$ \Sigma f_ix_i = 1360 $$
Mean = $$ \cfrac{\Sigma f_ix_i}{\Sigma f_i}= \cfrac{1360}{50} = 27.2 $$
The weights (in $$kg$$) of $$13$$ students are given below:
$$44$$, $$42$$, $$43$$, $$46$$, $$52$$, $$39$$, $$40$$, $$42$$, $$41$$, $$47$$, $$49$$, $$38$$, $$50$$ Calculate the median weight
Report Question
0%
$$58\: kg$$
0%
$$43\: kg$$
0%
$$36\: kg$$
0%
$$31\:kg$$
Explanation
Weight of 13 students is:
$$44$$, $$42$$, $$43$$, $$46$$, $$52$$, $$39$$, $$40$$, $$42$$, $$41$$, $$47$$, $$49$$, $$38$$, $$50$$
Arranging the weights in ascending order:
$$38, 39, 40, 41,42, 42, 43, 44, 46,47, 49, 50, 52$$
Since, the number of students is $$13$$, the median will be the middle most observation,
here, it is $$7^{th}$$ term i.e. $$43$$
The median of the following data $$46$$, $$64$$, $$87$$, $$41$$, $$58$$, $$77$$, $$35$$, $$90$$, $$55$$, $$33$$, $$92$$ is
Report Question
0%
$$87$$
0%
$$77$$
0%
$$58$$
0%
$$60.2$$
Explanation
Given observations are: $$46,64,87,41,58,77,35,90,55,33,92$$
Arranging the observations in ascending order:
$$33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92$$
The series has $$11$$ observations, the median will the middle observation i.e. $$6^{th}$$ observation,
hence, the median $$= 58$$
Calculate the mode for the following data:
$$20$$, $$18$$, $$20$$, $$16$$, $$15$$, $$23$$, $$16$$, $$17$$, $$15$$, $$16$$
Report Question
0%
$$23$$
0%
$$15$$
0%
$$20$$
0%
$$16$$
Explanation
From the given observations:
$$20$$, $$18$$, $$20$$, $$16$$, $$15$$, $$23$$, $$16$$, $$17$$, $$15$$, $$16$$
Here, $$16$$ is the most frequent observation, hence it is the mode of the series.
Mode of the observations $$5, 7, 2, 8, 5, 5, 8, 2, 5, 8, 7, 6$$ is
Report Question
0%
$$7$$
0%
$$6$$
0%
$$6.5$$
0%
$$5$$
Explanation
For the given data: $$5$$, $$7$$, $$2$$, $$8$$, $$5$$, $$5$$, $$8$$, $$2$$, $$5$$, $$8$$, $$7$$, $$6$$
$$5$$ occurs most frequently. hence, the mode is $$5$$
Mean of the first six even numbers is
Report Question
0%
$$7$$
0%
$$5$$
0%
$$3$$
0%
$$1$$
Explanation
Formula used:
$$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$$
Apply the above formula, we get
Mean$$=\displaystyle \frac{2+4+6+8+10+12}{6}$$
= $$\displaystyle \frac{42}{6}=7$$
Find the median of each of the following data:
$$(i)$$ $$7$$, $$10$$, $$5$$, $$20$$, $$18$$, $$25$$, $$17$$
$$(ii)$$ $$3$$, $$5$$, $$9$$, $$2$$, $$8$$, $$7$$, $$6$$, $$7$$, $$4$$
$$(iii)$$ $$37$$, $$42$$, $$31$$, $$46$$, $$25$$, $$27$$, $$30$$, $$32$$, $$41$$
Report Question
0%
$$(i)$$ $$18$$
$$(ii)$$ $$4$$
$$(iii)$$ $$20$$
0%
$$(i)$$ $$5$$
$$(ii)$$ $$36$$
$$(iii)$$ $$13$$
0%
$$(i)$$ $$17$$
$$(ii)$$ $$6$$
$$(iii)$$ $$32$$
0%
$$(i)$$ $$9$$
$$(ii)$$ $$29$$
$$(iii)$$ $$30$$
Explanation
$$(i)$$ By arranging the given no's in increasing order, we get:
$$ 5,7,10,17,18,20,25 $$
Total terms $$= 7$$
since, the middle most term is $$ 4^{th} $$.
$$\therefore $$ value of median is $$17$$
$$(ii)$$ By arranging the given no's in increasing order, we get:
$$ 2,3,4,5,6,7,7,8,9 $$
Total terms $$= 9$$
since, the middle most term is $$ 5^{th} $$.
$$\therefore $$ value of median is $$6$$
$$(iii)$$ By arranging the given no's in increasing order, we get:
$$ 25,27,30,31,32,37,41,42,46 $$
Total terms $$= 9$$
since, the middle most term is $$ 5^{th} $$.
$$\therefore $$ value of median is $$32$$
The mean of first six multiples of $$4$$ is
Report Question
0%
$$13.5$$
0%
$$14.5$$
0%
$$14$$
0%
$$16$$
Explanation
$$\textbf{Step 1: Write all 6 multiples of 4}$$
$$\text{First six multiples of 4 are: }4, 8, 12, 16, 20, 24$$
$$\textbf{Step 2: Find Mean of all 6 observations}$$
$$\text{Mean }= \cfrac{4 + 8 + 12 + 16 + 20+ 24}{6}$$
$$\left[\because \boldsymbol{\textbf{Mean }= \cfrac{\textbf{Sum of observations}}{\textbf{Total observations}}}\right]$$
$$\text{Mean }= \cfrac{112}{6}$$
$$\text{Mean }= 14$$
$$\textbf{Hence, Option C is correct.}$$
Find the mode for the following: $$36$$, $$32$$, $$31$$, $$31$$, $$29$$, $$31$$, $$30$$, $$32$$, $$40$$, $$46$$, $$31$$, $$32$$, $$35$$, $$37$$, $$40$$. If one observation $$31$$ is replaced by $$32$$, will the modal value change? If yes, find it
Report Question
0%
$$31$$, Yes, $$32$$
0%
$$33$$, No
0%
$$37$$, Yes, $$30$$
0%
$$32$$, No
Explanation
By arranging the given data in increasing order, we get,
$$ 29,30,31,31,31,31,32,32,32,35,36,37,40,40,46 $$
value that occurs most frequently is $$31$$
$$\therefore $$ Mode $$= 31$$
if one $$31$$ is replaced by $$32$$ then the arranging data is:
$$ 29,30,32,32,32,32,32,32,32,35,36,37,40,40,46 $$
Now the value that occurs most frequently is $$32$$
so, now modal value is change. i.e $$32$$
The runs scored by some players of a cricket team in a one day match are given below:
$$83$$, $$40$$, $$36$$, $$0$$, $$69$$, $$105$$, $$73$$, $$21$$, $$8$$ Find the median runs
Report Question
0%
$$39$$ runs
0%
$$52$$ runs
0%
$$40$$ runs
0%
None of these
Explanation
Runs scored by players are:
$$83$$, $$40$$, $$36$$, $$0$$, $$69$$, $$105$$, $$73$$, $$21$$, $$8$$
Arranging them in ascending order:
$$0, 8, 21, 36, 40, 69, 73, 83, 105$$
The number of players are $$9$$, the median will be the middle most number, i.e. $$5^{th}$$ term.
Hence, median is $$40$$ runs
The marks obtained by $$10$$ students in a test are:
$$12$$, $$22$$, $$32$$, $$41$$, $$26$$, $$30$$, $$14$$, $$11$$, $$18$$, $$35$$. Find the median marks
Report Question
0%
$$24$$ marks
0%
$$28$$ marks
0%
$$32$$ marks
0%
None of these
Explanation
Marks obtained by $$10$$ students in the test:
$$12$$, $$22$$, $$32$$, $$41$$, $$26$$, $$30$$, $$14$$, $$11$$, $$18$$, $$35$$
Arranging the marks in ascending order:
$$11, 12, 14, 18, 22, 26, 30, 32, 35, 41$$
Since, the number of students are $$10$$, median will be the mean of two middle terms
Median $$=$$ mean of $$5^{th}$$ term and $$6^{th}$$ term
Median $$= \dfrac{22 + 26}{2}$$
Median $$= 24$$ marks
Find the lower quartile for the data: $$9,11,15,19,17,13,7$$.
Report Question
0%
$$7$$
0%
$$9$$
0%
$$13$$
0%
None of the above
Explanation
Lower quartile is the median of the lower half of the data set.
Given data set is $$9,11,15,19,17,13,7$$
Arranging the data set in ascending order we get $$7,9,11,13,15,17,19$$
Number of values in the data set is $$n=7$$
Lower quartile is given by $$Q_1=\dfrac{n+1}{4}=\dfrac{7+1}{4}=\dfrac{8}{4}=2^{nd} value$$
Therefore, the lower quartile is $$9$$
Construction of a cumulative frequency table is useful in determining the
Report Question
0%
mean
0%
median
0%
mode
0%
all the above three measures
Explanation
A cumulative frequency table helps to determine the median of the series.
A variate takes $$11$$ values which are arranged in ascending order of their magnitudes It is found that $$4$$th, $$6$$th and $$8$$th observations are $$8$$, $$6$$ and $$4$$ respectively What is the median of the distribution?
Report Question
0%
$$4$$
0%
$$6$$
0%
$$8$$
0%
$$10$$
Explanation
Since there are $$11$$ values
median = middle value = $$6^{th}$$ value = $$6$$
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 11 Commerce Economics Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
