MCQ Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 1

The weight ( in kg ) of

$5$ men are

$62, 65, 69, 66$ and

$61.$ The median is

0%
$45$ kg
0%
$66$ kg
0%
$65$ kg
0%
$55$ kg

Explanation

Arranging in the ascending order we have

$61, 62, 65, 66, 69.$
Hence median

$=65$ kg.

_________ of a distribution is that value of the variant which divides it into two equal parts.

0%
Mode
0%
Mean
0%
Median
0%
Harmonic mean

What is a reasonable explanation for a data set of the test score of 75 students in which the mean score is 81 and the median is 68?

0%
An extremely low score
0%
A number of typical scores
0%
Too many scores
0%
Two extremely high scores

The sum of four consecutive odd integers

$w, x, y,$ and

$z$ is

$24$ . What is the median of the set

$\left \{w, x, y, z, 24\right \}$ ?

0%
$3$
0%
$5$
0%
$7$
0%
$9$
0%
$24$

Explanation

Given that the sum of four consecutive odd integers is

$24$ .

which implies

$x=w+2 , y=w+4 , z=w+6$ .

Therefore

$4w+12=24$ , which implies

$w=3$ .

So, the median of set

${3,5,7,9,24}$ is

$7$ .

Let the sum of the first three terms of an A.P. be

$39$ and the sum of its last four terms be

$178$ . If the first term of this A.P. is

$10$ , then the median of the A.P. is:

0%
$29.5$
0%
$28$
0%
$31$
0%
$26.5$

Explanation

Let the first three terms of the A.P. be

$a-d,a, a+d$
Given

$a-d=10 ...(1)$
Also given

$a-d+a+a+d=39$

$\Rightarrow 3a=39 ...(2)$

Solving

$(1)$ and

$(2)$ , we get

$a=13, d=3$
So, the A.P. is

$10,13,16,...$

Now, let the last four terms be

$a_n, a_{n-1}, a_{n-2},a_{n-3}$

$a+(n-1)d+a+(n-2)d+a+(n-3)d+a+(n-4)d=178$

$\Rightarrow 4a+(4n-10)d=178$

$\Rightarrow n=14$

So, total number of terms in the A.P. are

$14$ (even).
Hence, median of the A.P. will be mean of

$T_7, T_8$

$T_7=10+6(3)=28$

$T_8=10+7(3)=31$
Median

$=\cfrac{28+31}{2}=29.5$

Median of the data:

$1, 9, 4, 3, 7, 6, 8, 8, 12, 15$ is.

0%
$6.5$
0%
$7.5$
0%
$8$
0%
Any number between $7$ and $8$

Explanation

First arrange the set of data in order(from least to highest)

$1,3,4,6,7,8,8,9,12,15$

As there are even number of items,

$\therefore$ Median

$=\dfrac{7+8}{2}=\dfrac{15}{2}=7.5$

The median of

$10,\space 14,\space 11,\space 9,\space 8,\space 12,\space 6$ is

0%
$14$
0%
$11$
0%
$10$
0%
$12$

Explanation

Given numbers can be arranged in ascending order as,

$6, 8, 9, 10,11, 12, 14$ .
Number of term is

$7$ .
Thus median

$\displaystyle =\left(\frac{7+1}{2}\right)^{th}$ term

$= 10$

The mean of

$x, y, z$ is

$y$ , then

$x+z =$

0%
$y$
0%
$2y$
0%
$3y$
0%
$zy$

Explanation

The question tells us that the mean of

$x, y$ and

$z$ is

$y$ .

i.e.

$\dfrac {x+y+z}{3} =y$

i.e.

$x+z=3y-y$

$\rightarrow x+z=2y$

A set of numbers consists of three

$4$ 's, five

$5$ 's, six

$6$ 's, eight

$8$ 's and seven

$10$ 's. The mode of this set of numbers is

0%
$6$
0%
$7$
0%
$8$
0%
$10$

Explanation

Since,

$8$ occurs maximum number of times. It is the mode of the set.

A student obtained

$60, 75$ and

$85$ marks respectively in three monthly examinations in Mathematics and

$95$ marks in the final examination. The three monthly examinations are of equal weightage whereas the final examination is weighted twice as much as a monthly examination. The mean marks of Mathematics are

0%
$80$
0%
$81$
0%
$82$
0%
$85$

Explanation

We have

$\quad x_1 = 60, x_2 = 75, x_3 = 85, x_4 = 95, w_1 = 1, w_2 = 1, w_3 = 1$ and

$w_4 = 2$

$\therefore \quad \overline{X_w} = \displaystyle\frac{w_1x_1 + w_2x_2+ w_3x_3 + w_4x_4 }{w_1 + w_2 + w_3 + w_4}$

$\Rightarrow \quad \overline{X_w} = \displaystyle\frac{1\times 60+1\times75+1\times85+2\times95}{1+1+1+2} = \displaystyle\frac{410}{5} = 82$

The median of

$34,32,48,38,24,30,27,21,35$ is

0%
$32$
0%
$34$
0%
$38$
0%
$30$

Explanation

Arranging the data in ascending order of magnitude, we have

$21,24,27,30,32,34,35,38,48$
Since there are

$9$ , an odd number of items, Therefore the median is the
value of

$\left(\displaystyle\frac{9+1}{2}\right)^{th}$ observation, i.e.

$32$

Which one of the following is not central tendency ?

0%
Mean deviation
0%
Arithmetic mean
0%
median
0%
Mode

Percentile divides the number of items into ______ equal parts.

0%
$10$
0%
$100$
0%
$20$
0%
$50$

If a variable takes discrete values

$x+4, \space x-\displaystyle\frac{7}{2}, \space x-\displaystyle\frac{5}{2}, \space x-3, \space x-2, \space x+\displaystyle\frac{1}{2}, \space x-\displaystyle\frac{1}{2}, \space x+5, \space (x>0)$ then the median is

0%
$x - \displaystyle\frac{5}{4}$
0%
$x - \displaystyle\frac{1}{2}$
0%
$x-2$
0%
$x + \displaystyle\frac{5}{4}$

Explanation

Given data's are,

$\space x-\displaystyle\frac{7}{2}, \space x-3 ,\space x-\displaystyle\frac{5}{2} , \space x-2, \space x-\displaystyle\frac{1}{2}, \space x+\displaystyle\frac{1}{2}, x+4, \space x+5$
Hence required mean

$=$ mean of

$4^{th}$ and

$5^{th}$ term

$=x-\cfrac{5}{4}$ (Since number of terms are even.)

A set of numbers consists of three 4's, five 5's, six 6's, eight 8's and seven 10's. The mode of this set of numbers is

0%
More than 5
0%
More than 5 but less than 10
0%
8
0%
None

Which of the following is not a measure of central tendency:

0%
Mean
0%
Median
0%
Class interval
0%
Mode

Explanation

$\textbf{Step -1: Describing Measures of central tendency.}$

$\text{As central tendency is a tendency for a set of values to gather around the middle of the set.}$

$\text{And mean, median and mode all three are the measures of central tendency.}$

$\textbf{Step -2: Finding the correct option.}$

$\text{But class interval is the numeric width of any class in a particular distribution.}$

$\text{So, option A, B, D}\text{ are the measures of central tendency.}$

$\text{And option C is not a measure of central tendency.}$

$\textbf{ Hence, the correct option is C.}$

The median of the data

$5,6,7,8,9,10$ is

0%
$7$
0%
$8$
0%
$7.5$
0%
$8.5$

Explanation

There are six observations

$\therefore \quad$ Median =

$AM$ of

$3^{rd}$ and

$4^{th}$ observation

$\Rightarrow \quad \mbox{Median} = \displaystyle\frac{7+8}{2} = 7.5$

Find the mode of:

$5, 6, 9, 13, 6, 5, 6, 7, 6, 6, 3$

0%
$6$
0%
$7$
0%
$9$
0%

Explanation

Given series is

$5,6,9,13,6,5,6,7,6,6,3$

Here digit

$6$ is repeated five times.

Hence, here mode is

$6$ .

The mode of the following data:

$120,110,130,110,120,140,130,120,140,120$ is

0%
$110$
0%
$120$
0%
$130$
0%
$115$

Explanation

The frequency table for the given data is as given below

Value $(x_i)$	$110$	$120$	$130$	$140$
Frequency $(f_i)$	$2$	$4$	$2$	$2$

We observe that the value

$120$ has the maximum frequency. Hence, the mode or modal value is

$120$ .

Find the mode of

$27,12,18,19, 18,24,17,18$

0%
$17$
0%
$18$
0%
$19$
0%
$20$

Explanation

The numbers are:

$27,12,18,19,18,24,17,18$
Arranging in ascending order:

$12, 17, 18, 18, 18, 19, 24, 27$

$18$ occurs

$3$ times and maximum number of times. Hence, it is the mode.

Find the median of the following data:

$9, 3, 20, 13, 0, 7$ and

$10$

0%
$7$
0%
$9$
0%
$10$
0%
None of these

Explanation

Given data is

$9,3,20,13,0,7, 10$

Arranging them in order, we get

$0, 3, 7,9, 10,13,20$

Median is the middle score in ordered data.

Therefore, here median is

$9$ .

Ages of the employees of a company are given below. Find the mode.

Ages (in years)	22	23	25	19	21	27
Number of employees	14	19	17	27	12	15

0%
$18$ years
0%
$19$ years
0%
$20$ years
0%
$21$ years

The ages of teachers of a school are

$53, 37, 39, 51, 46, 42, 44, 47, 55, 48$ . Find the median

0%
$46.5$
0%
$47.5$
0%
$48.5$
0%
$49.5$

Explanation

Age of teachers are:

$53, 37, 39, 51, 46, 42, 44, 47, 55, 48$
Arranging in ascending order:

$37, 39, 42, 44, 46, 47, 48, 51, 53, 55$
There are

$10$ numbers, hence, the median will be the mean of

$\cfrac{n}{2}$ and

$\cfrac{n}{2}+1$ terms
Median

$=$ mean of

$5^{th}$ and

$6^{th}$ terms
Median

$=$ mean of

$46$ and

$47$
Median

$= \cfrac{46 + 47}{2}$
Median

$= \cfrac{93}{2} = 46.5$

The average of daily wages for the workers of the two factories combined is

No.of wage earners	Factory A 250	Factory B 200
Average daily wage	Rs.2.00	Rs.2.50

0%
$2.22$
0%
$3.15$
0%
$4.10$
0%
None of these

Explanation

Average wage

$= \cfrac{x_1w_1+x_2w_2}{w_1+w_2}$

$=\cfrac{250\times 2+200\times 2.5}{250+200}$

$=\cfrac{1000}{450}=2.22$

Which of the following is different from others

0%
$5^{th}Decile$
0%
Median
0%
$50^{th}Percentile$
0%
Upper quartile

The measure of dispersion is

0%
Mean deviation
0%
S.D.
0%
quartile deviation
0%
all of the above

Find the median of the given data.

$30, 32, 24, 34, 26, 28, 30, 35, 33, 25$

0%
$32$
0%
$26$
0%
$30$
0%
$25$

Explanation

Arranging the data in ascending order:

$24,25,26,28,30,30,32,33,34,35$
Mean

$=\dfrac {\text { sum of all numbers} }{ \text {count of all numbers }} = \dfrac { 24+25+26+28+30+30+32+33+34+35}{10} =\dfrac { 297 }{10 } =29.7$
Median of even number of data is the average of the middle two terms that is here the average of

$5^{th}$ and

$6^{th}$ terms.
Median

$=\dfrac { 30+30 }{ 2 } =30$

Frequency distribution of daily commission received by 100 salesmen is given.

Daily commission (in Rs.)	$100 - 120$	$120 - 140$	$140 - 160$	$160 - 180$	$180 - 200$
No. of salesman	$20$	$45$	$22$	$9$	$4$

Find mean daily commission received by a salesman.

0%
$136.4$
0%
$140$
0%
$130$
0%
None of these

Explanation

Rs No.of salesman

$x$

${ f }_{ x }$

$100-120$

$20$

$110$

$2200$

$120-140$

$45$

$130$

$5850$

$140-160$

$22$

$150$

$3300$

$160-180$

$9$

$170$

$1530$

$180-200$

$4$

$190$

$760$

$100$

$13640$

$\bar { x }$ =

$\cfrac { \sum { fx } }{ \sum { f } } =\cfrac { 13640 }{ 100 } =Rs136.4$

The mean of

$x + 3, x + 5, x + 7, x + 9$ and

$x + 11$ is

0%
$2x+7$
0%
$x+8$
0%
$x+7$
0%
None of these

Explanation

The observations are :

$x+3,x+5,x+7,x+9, x+11$
Mean =

$\dfrac{sum}{number \quad of \quad observations}$ =

$\dfrac{x+3,x+5,x+7,x+9, x+11}{5}$ =

$\dfrac{5x + 35}{5} = x + 7$

Find the mean of first ten odd natural numbers.

0%
$10$
0%
$11$
0%
$12$
0%
None of these

Explanation

${\textbf{Step -1: Listing first 10 odd natural numbers .}}$

${\text{First 10 odd natural numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19}}$

${\textbf{Step -2: Calculating mean}}$

${\text{We know that, mean of n numbers = }}\dfrac{{{\text{Sum of n numbers}}}}{{\text{n}}}$

$\therefore {\text{ Mean of first 10 odd natural numbers,}}$

${\text{M = }}\dfrac{{{\text{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}}}}{{{\text{10}}}}$

$\Rightarrow {\text{ M = }}\dfrac{{{\text{100}}}}{{{\text{10}}}}{\text{ = 10}}$

${\textbf{Thus, the mean of first 10 odd natural numbers is 10 .}}$

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 1 - 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 1 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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