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CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 5 - MCQExams.com
CBSE
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 $$15.3$$
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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 $$M-m$$?
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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?
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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 $$2, 7, 8, 11, 7, 1, 9, 8, 7$$?
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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 $$11, 7, 6, 9, 12, 15, 19$$?
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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?
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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.
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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 $$f_2$$.
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$$28$$ 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 $$x$$.
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$$25$$
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?
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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 $$100$$.
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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.
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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 ___________.
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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.
$$14, 6, 9, 15, 14, 9, 21, 21, 25, 21, 27, 29, 21, 8, 6,$$
$$ 15, 25, 14, 21, 9, 21, 25, 27, 29, 6, 14, 21, 21, 27, 25, 27, 9, 15, 14, 9$$.
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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 $$x =$$
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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.
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$$1$$ only
0%
$$2$$ only
0%
$$1$$ and $$2$$ both
0%
Neither $$1$$ nor $$2$$
Explanation
Statement 1: is correct because positional measure of dispersion describes about the position that a particular data value has within a data set.
Statement 2:is correct because quartiles and percentiles are positional measure of dispersion.
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 ____________.
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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 __________.
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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-
$$15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15$$.
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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.
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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:
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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 $$9$$.
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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:
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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
Explanation
It is not uncommon in a median
To locate it graphically
to use it then data is quantitive
to use it when data is rigidly defined
If $$1, 2, 3, 4, 5, 6, 7,$$ are numbers then $$Q_1$$ & $$Q_3$$ are respectively.
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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 ____________.
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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.
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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
$$0$$
$$230$$
$$20$$
$$218$$
$$30$$
$$200$$
$$40$$
$$165$$
$$50$$
$$123$$
$$60$$
$$73$$
$$70$$
$$288$$
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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 _______.
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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:
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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?
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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 $$80$$
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