MCQ Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 5

Find the median of

$15\dfrac {2}{3}, 15.03, 15, 15\dfrac {1}{3}$ and

$15.3$

0%
$15\dfrac {1}{3}$
0%
$15.3$
0%
$15\dfrac {2}{3}$
0%
$15.03$

Explanation

Arranging in scaling ascending order, we get observations as

$15,15.03,15.30,15.50,15\cfrac{2}{3}$

so, median

$=15.30$

Sam's test scores are History

$76$ , Geography

$74$ , Math

$92$ , English

$81$ and Chemistry

$80$ . If the average (arithmetic mean) score is

$M$ and the median score is

$m$ , what is the value of

$M-m$ ?

0%
0.4
0%
0.5
0%
0.6
0%
0.8

Explanation

Score in History

$76$ , Geography

$74$ , Math

$92$ , English

$81$ , Chemistry

$80$

$\therefore$ average Sore

$M=$

$\dfrac{76+74+92+81+81}{5}=\dfrac{403}{5}=80.6$

Score

$=74,76,80,81,92$

$\therefore$ Median

$(m)=80$

$\Rightarrow M-m=80-80.6=0.6$

For which state the average number of candidates selected over the years is the maximum?

0%
Delhi
0%
H.P
0%
U.P
0%
Punjab

Explanation

The average number of candidates selected over the given period for various states are:

For Delhi

$= \dfrac {94 + 48 + 82 + 90 + 70}{5}$

$= \dfrac {385}{5} = 76.8$

For U.P.

$= \dfrac {78 + 85 + 48 + 70 + 80}{5}$

$= \dfrac {361}{5} = 72.2$

For Punjab

$= \dfrac {85 + 70 + 65 + 84 + 60}{5}$

$= \dfrac {364}{5} = 72.8$

For Haryana

$= \dfrac {75 + 75 + 55 + 60 + 75}{5}$

$= \dfrac {340}{5} = 68$

The average is maximum for Delhi.

What is the mode of

$2, 7, 8, 11, 7, 1, 9, 8, 7$ ?

0%
$9$
0%
$8$
0%
$7.5$
0%
$7$

Explanation

$2,7,8,11,7,1,9,8,7$

As frequency of

$7$ is maximum

$=3$

$\therefore$

$\boxed{Mode\, is\, 7}$

What is the median of the values

$11, 7, 6, 9, 12, 15, 19$ ?

0%
$9$
0%
$11$
0%
$12$
0%
$15$

Explanation

To find median, we write all the scores from smallest to largest and then choose the score in the middle.

Ordered data :

$6,7,9,11,12,15,19$

Therefore, median is

$11$ .

What is the average amount of interest per year which the company had to pay during this period?

0%
$Rs. 32.43$ lakhs
0%
$Rs. 33.72$ lakhs
0%
$Rs. 34.18$ lakhs
0%
$Rs. 36.66$ lakhs

Explanation

Average amount of interest paid by the Company during the given period

$= Rs. \left [\dfrac {23.4 + 32.5 + 41.6 + 36.4 + 49.4}{5}\right ]lakhs$

$= Rs. \left [\dfrac {183.3}{5}\right ]lakhs$

$= Rs. 36.66\ lakhs$

Mean weight of a class is

$50Kg$ . Mean weight of boys and girls are

$52Kg$ and

$42Kg$ respectively.
Find the ratio of number of boys and girls in the school.

0%
$5:1$
0%
$4:1$
0%
$3:2$
0%
$3:5$

Explanation

$u_g=42$ ,

$u_b=52$ ,

$u=50$
Assume the total strength of class to be 100.
So,

$n_b=n$ and

$n_g=100-n$

$u=\dfrac{u_g \times n_g + u_b \times n_b}{n_g+n_b}$

$\therefore 50=\dfrac{42 \times (100-n)+ 52 \times n}{100}$

$\therefore 50=\dfrac{52n+4200-42n}{100}$

$\therefore 800=10n$
Thus,

$n=80$ and

$100-n=20$
Hence the ratio :

$(4:1)$
Ans-Option B.

Class	0-20	20-40	40-60	60-80	80-100
Frequency	19	$f_1$	32	$f_2$	19	Total 120

Mean of the above frequency table is given as

$50$ . Find

$f_1$ and

$f_2$ .

0%
$28$ and $22$
0%
$25$ and $25$
0%
$24$ and $26$
0%
$20$ and $30$

Explanation

Class Marks will be

$10,30,50,70,90$ (

$x_i$ ).
Mean,

$u=\dfrac{\sum x\times f}{\sum f}$

$u=50=\dfrac{3500+30 \times f_1+70 \times f_2}{120}$

$30 \times f_1+70 \times f_2=2500$
Also,

$f_1+f_2=120$
Solving the above equation we get,

$f_1=25$ and

$f_2=25$ .
Ans- Option B

Mean of marks obtained by

$10$ students is

$30$ .
Marks obtained are

$25,30,21,55,47,10,15,x,45,35$ .
Find the value of

$x$ .

0%
$25$
0%
$37$
0%
$69$
0%
$17$

Explanation

Let

$u$ be the average marks of the class.

$\therefore u=\dfrac{25+30+21+55+47+10+15+x+45+35}{10}=\dfrac{283+x}{10}$
But

$u=30$ .
So,

$30=\dfrac{283+x}{10}$ .

$\therefore x=17$
Ans-Option D

There are

$60$ students out of which

$25$ are girls. The average weight of girls is

$40Kg$ and that of boys is

$53Kg$ . Mean weight of the entire class?

0%
$44.285$
0%
$47.54$
0%
$47.583$
0%
$49.695$

Explanation

$u_g=40$ ,

$u_b=53$

$n_g=25$ ,

$n_b=35$

$u=\dfrac{u_g \times n_g + u_b \times n_b}{n_g+n_b}$

$\therefore u=\dfrac{25 \times 40+ 35 \times 53}{25 + 35}$

$\therefore u=\dfrac{2855}{60}=47.583$
Ans-Option C.

Find AM of divisors of

$100$ .

0%
$24$
0%
$25.5$
0%
$24.11$
0%
$21.9$

Explanation

Divisors of 100 are:

$1, 2, 4, 5, 10, 20, 25, 50, 100$

$\therefore n=9$

$\therefore S=1+2+4+5+10+20+25+50+100=217$

$\therefore A.M.=\dfrac{S}{n}=\dfrac{217}{9}=24.11$

Mean of

$40$ observations was given as

$160$ .It was detected that

$125$ was misread as

$160$ . Find the correct mean.

0%
$158$
0%
$169$
0%
$159$
0%
$161$

Explanation

Formula used:

$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$

Sine, number of observations

$,n=40$
Mean

$=160$ (initial wrong mean)
Incorrected sum

$=n \times u=40 \times 160=6400$
Now

$125$ was read as

$165$ , so
Corrected sum

$=6400-165+125=6360$

$\therefore$ Corrected mean

$=\dfrac{6360}{40} =159$

If the mode of the following data

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

$3$ , then value of

$P$ is ___________.

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

Given data is

$4,3,2,5,P,4,5,1,7,3,2,1$ and mode is

$3$

If mode of the data given is

$3$ , then

$3$ should appear most

But from the data

$1,2,3,4,5$ all are appearing twice for making

$3$ as mode,

$P$ should be

$3$ .

Find the mode of the data.

$14, 6, 9, 15, 14, 9, 21, 21, 25, 21, 27, 29, 21, 8, 6,$

$15, 25, 14, 21, 9, 21, 25, 27, 29, 6, 14, 21, 21, 27, 25, 27, 9, 15, 14, 9$ .

0%
$25$
0%
$21$
0%
$14$
0%
$9$

Explanation

Given data is

$14,6,9,15,14,9,21,25,21,27,29,21,8,6,15,15,25,14,21,9,21,25,27,29,6,14,21,21,27, 25,27,9,15,14,9$

Mode of the data is the number that appears most in the data, i.e.

$21$ in the given data.

If the Arithmetic mean of

$8, 6, 4, x, 3, 6, 0$ is

$4$ ; then the value of

$x =$

0%
$7$
0%
$6$
0%
$1$
0%
$4$

Explanation

Arithmetic mean

$= \cfrac{\text{sum of all observations}}{\text{no. of observations}}$

$\Rightarrow 4=\cfrac { 8+6+4+x+3+6+0 }{ 7 } \\ \Rightarrow 28=27+x\\ \Rightarrow x=1$

Which one is correct?
Statement 1:Positional measure of dispersion describes about the position that a particular data value has within a data set.
Statement 2:Quartiles and percentiles are positional measure of dispersion.

0%
$1$ only
0%
$2$ only
0%
$1$ and $2$ both
0%
Neither $1$ nor $2$

Mean of

$9$ observations was founded to be

$35$ . Later on, it was detected that an observation

$81$ was misread as

$18$ , then the correct mean of the observations is ____________.

0%
$40$
0%
$41$
0%
$42$
0%
$43$

Explanation

Mean of

$9$ observations

$= 35$

Sum of

$9$ observations

$= 35\times9 = 315$

Since

$81$ was misread as

$18$

New sum

$= 315-18+81= 378$

The correct mean

$= \dfrac{378}{9} = 42$

The median of the observations

$30, 91, 0, 64, 42, 80, 30, 5, 117, 71$ is __________.

0%
$52$
0%
$51$
0%
$53$
0%
$52.5$

Explanation

Arranging the given numbers in ascending order

$0, 5,30,30,42,64,71,80,91,117$

Median of the above numbers is:

$=\dfrac{42+64}{2}$

$=\dfrac{106}{2}=53$

The mode of the following numbers is-

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

0%
$14$
0%
$16$
0%
$19$
0%
$15$

Explanation

Number	Frequency
14	4
15	5
16	1
17	1
18	1
19	2
20	1

Mode is the number in a set that occurred mostly or present in maximum number in that set of data

Frequency of the number

$15$ is maximum

so mode is

$15$

The mean salary paid per week to

$1000$ employees of an establishment was found to be Rs.

$900$ . Later on, it was discovered that the salaries of two employees were wrongly recorded as Rs.

$750$ and Rs.

$365$ instead of Rs.

$570$ and Rs.

$635$ . Find the corrected mean salary.

0%
$900.90$
0%
$1,115$
0%
$1,225$
0%
$900.09$

Explanation

We first find the corrected sum of observation.
Corrected sum of observations

$=$ (Sum of total incorrect observation)

$-$ (Sum of incorrect data)

$+$ (Sum of correct data).

$=900\times 1,000-(750+365)+(570+635)=9,00,090$

$\therefore$ Corrected Mean

$=Rs.\dfrac{9,00,090}{1,000}=Rs.900.09$

Mode of

$2,4,6,3,4,3,3,4,4,2$ will be:

0%
$2$
0%
$3$
0%
$4$
0%
$6$

Explanation

The mode is the value which appears most often in the data.

In this case,

$4$ is appearing most often, therefore

$4$ is the mode.

Find the arithmetic mean of numbers from

$1$ to

$9$ .

0%
$9$
0%
$5$
0%
$8$
0%
$3$

Explanation

The numbers from

$1$ to

$9$ are

$1,2,3,4,5,6,7,8,9$

$A.M. =\dfrac{\text{Sum of observations}}{\text{Number of observations}}= \dfrac {1+2+3+4+5+6+7+8+9}{9} = \dfrac {45}{9} = 5$

It is not uncommon in a median:

0%
To locate it graphically
0%
To use it then data is quantitive
0%
To use it when data is rigidly defined
0%
All of the above

$1, 2, 3, 4, 5, 6, 7,$ are numbers then

$Q_1$ &

$Q_3$ are respectively.

0%
$2, 4$
0%
$2, 6$
0%
$4, 6$
0%
$3, 5$

Explanation

The given numbers are,

$1,2,3,4,5,6,7$

Number of observation

$(N)=7$

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

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

$2^{nd}term=2$

$\therefore$

$Q_1=2$

Now,

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

$6^{th}term=6$

$\therefore$

$Q_3=6$

The mean of

$\displaystyle\frac{1}{3},\frac{3}{4},\frac{5}{6},\frac{1}{2}$ and

$\displaystyle\frac{7}{12}$ is ____________.

0%
$\displaystyle\frac{2}{5}$
0%
$\displaystyle\frac{3}{5}$
0%
$\displaystyle\frac{1}{5}$
0%
None of these

Explanation

Given data is

$\dfrac{1}{3},\dfrac{3}{4},\dfrac{5}{6},\dfrac{1}{2},\dfrac{7}{12}$

Mean

$=$ Sum of observations

$\div$ Total number of observations

Sum of observations

$=$

$\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{5}{6}+\dfrac{1}{2}+\dfrac{7}{12}$

Mean

$=$

$\dfrac{\dfrac{4+9+10+6+7}{12}}{5}$

$= \dfrac{3}{5}$

The ______ is the middle element when the data set is arranged in order of the magnitude.

0%
median
0%
mean
0%
mode
0%
quartile

Explanation

The median is the middle value of the data set (that is arranged in ascending order). It is the value that divides a data set into two equal parts.

In an individual data set it is calculated by using the formula:

(i) If the data has an odd number of figures

$\text{Median} = \dfrac{N+1}{2}$ th value

(ii) If the data has an even number of figures

$\text{Median} = \dfrac{N}{2}$ th value

Find the median of the following data.

Age greater than (Years)	No. of Persons
$0$	$230$
$20$	$218$
$30$	$200$
$40$	$165$
$50$	$123$
$60$	$73$
$70$	$288$

0%
$31.5$
0%
$41.5$
0%
$41.6$
0%
$31.6$

Explanation

Note that it is a greater than type cumulative frequency distribution. First we convert it into a less than type form.

Age greater than(Years)	Greater than Cumulative frequency	Frequency	Less than Cumulative frequency
$0-10$	$230$	$12$	$12$
$10-20$	$218$	$18$	$30$
$20-30$	$200$	$35$	$65$
$30-40$	$165$	$42$	$107$
$40-50$	$123$	$50$	$157$
$50-60$	$73$	$45$	$202$
$60-70$	$28$	$20$	$222$
$70-80$	$8$	$8$	$230$

$N/2=230/2=115$ , therefore median class is

$40-50$
Also

$L_m=40, f_m=50, h=10, C=107$

$\therefore M_d=40+\displaystyle\frac{115-107}{50}\times 10=41.6$ .

The number of trees in different parks of a city are

$33, 38, 48, 33, 34, 33$ and

$24$ . The mode of this data is _______.

0%
$24$
0%
$34$
0%
$33$
0%
$48$

Explanation

We have,

$33,\,38,\,48,\,33,\,34,\,34,\,33$ and

$24$ .

On arranging the data in ascending order, we get

$24,\,33,\,33,\,33,\,34,\,34,\,38$ and

$48$ .

Here,

$33$ occurs more frequently, i.e.

$3$ times.

$\therefore$ Mode of data is

$33$ .

The mean age of

$25$ students in a class is

$12$ . If the teacher's age is included, the mean age increases by

$1$ . Then teacher's age (in years) is:

0%
$24$
0%
$38$
0%
$30$
0%
$48$

Explanation

Sum of

$25$ students ages be

$x$

$\Rightarrow$ Mean

$=\dfrac {x}{25}=12\Rightarrow x=300$ equation - (1)

Let the teacher's age be

$y$

$\Rightarrow$ New mean

$=\dfrac {x+y} {26}=13$

$\Rightarrow 300+y=338$

$\Rightarrow y=38$

Teacher's age

$=y=38$

The average life expectancy in three countries

$A,B$ and

$C$ was

$75,90$ and

$80$ years respectively. The ratio of population in the three countries wa

$4:2:3$ respectively. What was the average life expectancy of the three countries combined?

0%
$75$
0%
$80$
0%
$81$
0%
$92$

Explanation

given avergare life expectency are

$75,90,80$ respectively

ratio of population

$=4:2:3$

let the populations be

$4x,2x,3x$

therefore the total population combined

$=9x$

average life expectency

$=\dfrac{75.4x+90.2x+80.3x}{9x}=80$

average life expectency combined is

$80$

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 5 - 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 5 - MCQExams.com

0:0:1

Practice Class 11 Commerce Economics Quiz Questions and Answers

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