MCQ Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 2

Find the median of the data:

$35, 48, 92, 76,52, 51, 63$ and

$71$

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.

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.

$150, 147, 140, 120, 135, 139, 148, 148$

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.

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

$7.6.$

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.

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?

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

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

$x$ .

$24, 27, 43, 48, x - 1, x + 3, 68, 73, 80, 90$

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

$220.$

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

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

$35$ .

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

$257.$

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:

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

The mode of the distribution

$3, 5, 7, 4, 2, 1, 4, 3, 4$ is

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

$13$ ?

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	$7$	$8$	$12$	$13$	$10$

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

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

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:

$20$ ,

$18$ ,

$20$ ,

$16$ ,

$15$ ,

$23$ ,

$16$ ,

$17$ ,

$15$ ,

$16$

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

0%
$7$
0%
$6$
0%
$6.5$
0%
$5$

Explanation

For the given data:

$5$ ,

$7$ ,

$2$ ,

$8$ ,

$5$ ,

$8$ ,

$2$ ,

$5$ ,

$8$ ,

$7$ ,

$6$

$5$ occurs most frequently. hence, the mode is

$5$

Mean of the first six even numbers is

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:

$(i)$

$7$ ,

$10$ ,

$5$ ,

$20$ ,

$18$ ,

$25$ ,

$17$

$(ii)$

$3$ ,

$5$ ,

$9$ ,

$2$ ,

$8$ ,

$7$ ,

$6$ ,

$7$ ,

$4$

$(iii)$

$37$ ,

$42$ ,

$31$ ,

$46$ ,

$25$ ,

$27$ ,

$30$ ,

$32$ ,

$41$

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

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:

$36$ ,

$32$ ,

$31$ ,

$29$ ,

$31$ ,

$30$ ,

$32$ ,

$40$ ,

$46$ ,

$31$ ,

$32$ ,

$35$ ,

$37$ ,

$40$ . If one observation

$31$ is replaced by

$32$ , will the modal value change? If yes, find it

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

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

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:

$9,11,15,19,17,13,7$ .

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

0%
mean
0%
median
0%
mode
0%
all the above three measures

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?

0%
$4$
0%
$6$
0%
$8$
0%
$10$

Explanation

Since there are

$11$ values
median = middle value =

$6^{th}$ value =

$6$

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 2 - 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 2 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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