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CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 3 - MCQExams.com
CBSE
Class 11 Commerce Economics
Measures Of Central Tendency
Quiz 3
The mode of a set of observations is the value which
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occurs most frequently
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is central
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is between maximum and minimum
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none of the foregoing
Explanation
By definition, mode is the value which occurs most frequently.
The arithmetic mean of first five natural numbers is
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$$3$$
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$$4$$
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$$5$$
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$$6$$
Explanation
Formula used:
$$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$$
Apply the above formula, we get
$$\displaystyle AM=\frac{1+2+3+4+5}{5}=3$$
The daily earnings ( in rupees ) of $$10$$ workers in a factory are $$8, 16, 19, 8, 16, 19, 16, 8, 19, 16.$$ The median wage is ,
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Rs. $$17.50$$
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Rs. $$8.00$$
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Rs. $$19.00$$
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Rs. $$16.00$$
Explanation
Arranging the daily earnings in ascending order, we have $$8, 8, 8, 16, 16, 16, 16, 19, 19, 19.$$ The median is average of $$5^{th}$$ and $$6^{th}$$ wages, i.e., $$\displaystyle \frac{16+16}{2}=16$$.
The average weight of 29 students is 28 kg. By the admission of a new student the average weight is reduced to 27.8 kg. The weight of the new student is
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22 kg
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21 kg
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22.4 kg
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21.6 kg
Explanation
Weight of new student
$$\displaystyle = 30\times 27.8-29\times 28$$
$$\displaystyle = 834-812=22\ kg.$$
The marks of 10 students in a certain subject in a class are 20, 19, 50, 48, 50, 36, 35, 50, 40, 40 The mean and mode are respectively as .............
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Mean =38.8, Mode = 50
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Mean = 40, Mode= 35
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Mean = 30, Mode= 33
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Mean = 35, Mode= 25
Explanation
$$\Rightarrow$$ Arranging given numbers in ascending order $$19,\,20,\,35,\,36,\,40,\,40,\,48,\,50,\,50,\,50$$.
$$\Rightarrow$$ Number of observation $$=10$$
$$\Rightarrow$$ $$Mean=\dfrac{19+20+35+36+40+40+48+50+50+50}{9}=\dfrac{388}{10}=38.8$$
$$\therefore$$ $$Mean=38.8$$
$$\Rightarrow$$ We can number $$50$$ occurs maximum times i.e. $$3$$ times.
$$\therefore$$ $$Mode=50$$
Find the inter quartile range for the data: $$9,11,15,19,17,13,7.$$
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$$7$$
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$$6$$
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$$8$$
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None of the above
Explanation
Interquartlie is the difference between upper quartile and lower quartile.
Lower quartile is the median of the lower half of the data set.
Upper quartile is the median of the upper 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 $$Q_1=9$$
Upper quartile is given by $$Q_3=\dfrac 34(n+1)=\dfrac 34(7+1)=\dfrac 34(8)=6^{th} value$$
Therefore, the upper quartile is $$Q_3=17$$
Therefore, the inter quartile is given by $$Q_3-Q_1=17-9=8$$
Which of the following is not central
tendency?
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Mode
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Arithmetic mean
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Median
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None
Explanation
Central tendency of any data is defined by mean or median or mode.
Hence Mean deviation is not a central tendency.
If the mean of 1,7,5,3,4,4 is m and 3,2,4,2,3,3, p is m-1 and median is q, then the value of p and q will be-
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$$p = 3, q = 4$$
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$$p=4, q = 3$$
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$$p = 4, q = 4$$
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$$p = 4 , q = 2$$
Explanation
$$\displaystyle \frac{1+7+5+3+4+4}{6}=m$$
$$\displaystyle m=\frac{24}{6}=4$$
and, $$\displaystyle \frac{3+2+4+2+3+3+p}{7}=m-1$$
$$\displaystyle \frac{17+p}{7}=4-1$$
17+p=21
p=4
Now, median of 3,2,4,2,3,3,p, is q. Then,
For n=7(odd)
Median$$\displaystyle =\frac{n+1}{2}$$th term
$$\displaystyle q=\left ( \frac{7+1}{2} \right )$$ th term
q = 4th term
q = 2
Hence, $$p= 4$$
$$q = 2$$
Find the median of the following data
$$25, 21, 18, 27, 29, 16, 22, 15$$
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$$21.5$$
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$$27.5$$
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$$21$$
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$$16$$
Explanation
Arranging the data in ascending order of magnitude
$$15, 16, 18, 21, 22, 25, 27, 29$$
Since there are $$8$$, i.e. an even number of observation
therefore, median = arithmetic mean of $$\left(\displaystyle\frac{8}{2}\right)^{th}$$ and $$\left(\displaystyle\frac{8}{2} + 1\right)^{th}$$ observation
Median = $$\displaystyle\frac{21+22}{2} = 21.5$$
Which of the following is correct for the given data $$-1, 0, 1, 2, 3, 5, 5, 6, 8, 10, 11, ?$$
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Mean=Mode=Median
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Mean $$=5$$
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Mean=Mode
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Mode=Median
Explanation
$$\Rightarrow$$ The given numbers are $$-1,0,1,2,3,5,5,6,8,10,11$$
$$\Rightarrow$$ $$Mean=\dfrac{-1+0+1+2+3+5+5+6+8+10+11}{11}=4.54$$
$$\Rightarrow$$ Here, middle number is $$6^{th}$$ number which is $$5$$.
$$\therefore$$ $$Median=5$$
$$\Rightarrow$$ In given numbers $$5$$ occurs two times.
$$\therefore$$ $$Mode=5$$
From above we can see that,
$$\therefore$$ $$Mode=Median$$
The median of 3, 5, 6, x, 9,15 isThe value of x is
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$$7$$
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$$4$$
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$$8$$
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none of these
Explanation
$$Median=\dfrac{6+x}{2}$$
$$\Rightarrow 7=\dfrac{6+x}{2}$$
$$\Rightarrow 14=6+x$$
$$\Rightarrow x=14-6$$
$$\Rightarrow x=8$$
Mean of 10 values is 32.If another values is included the mean becomesThe included value is
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$$16$$
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$$14$$
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$$15$$
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$$19$$
Explanation
Included value $$\displaystyle =31\times 11-32.6\times 10=15$$
$$\displaystyle{x}=a+\frac{\Sigma{fd}}{N}$$ is the formula of
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Median
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Mode
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Arithmetic mean
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Mean deviation
Explanation
The given formula is to determine Arithmetic mean using assumed mean method, where $$ a $$ is the assumed mean.
A candidate obtained the following percentage of marks in an examination English 60, Maths 90, Physics 75, ChemistryIf weight 2,4,3,3 are allotted to these subjects respectively, then the weight mean is given by
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$$\displaystyle \frac{60+90+75+66}{2+4+3+3}$$
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$$\displaystyle \frac{60\times 2+90\times 4+75\times 3+66\times3 }{2+4+3+3}$$
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$$\displaystyle \frac{60\times 2+90\times 4+75\times 3+66\times 3}{4}$$
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$$\displaystyle \frac{60+90+75+66}{4}$$
Explanation
Percentage marks in the examination-
English=60
Maths=90
Physics=75
Chemistry=66
Weight allotted to each subjects are-
English=2,Maths=4,English=3 and chemistry=66
Mean=$$\dfrac{sum\ of\ all\ observation\ value}{Sum\ of\ all\ observation}$$
$$\therefore$$ Weight mean=$$\dfrac{60\times 2+90\times 4+75\times 3+66\times 3}{2+4+3+3}$$
The median of the first 12 prime numbers is
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$$12$$
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$$13$$
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$$14$$
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$$15$$
Explanation
First 12 prime numbers are 2,3,5,7,11,13,17,19,23,29,31,37
Median$$=\dfrac{13+17}{2}=\dfrac{30}{2}=15$$
The numbers 3,5,6 and 4 have frequencies of x x+2, x-8 and x+6 respectively If their mean is 4 then the value of x is
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$$5$$
0%
$$6$$
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$$7$$
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$$4$$
Explanation
x
f
fx
3
x
3x
5
x+2
5x+10
6
x-8
6x-48
4
x+6
4x+24
$$\bar x=\dfrac{\sum fx}{n}$$
where n=The sum of the frequency
Given $$\bar x=4$$
$$\therefore \bar 4=\dfrac{18x-14}{4x}$$
$$\Rightarrow 16x=18x-14$$
$$\Rightarrow 16x-14x=-14$$
$$\Rightarrow -2x=-14$$
$$\Rightarrow x=\dfrac{-14}{-2}=7$$
A contractor employed 18 labourers at Rs 12 per day 10 labourers at Rs 13.50 per day 5 labourers at Rs 25 per day and 2 labourers at Rs 42 per day The average wage of a laborer per day is
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$$Rs\ 16$$
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$$Rs\ 20$$
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$$Rs\ 24$$
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$$Rs\ 28$$
Explanation
The average wage$$=\dfrac{18\times 12+10\times 13.50+5\times 25+42\times 2}{35}$$
$$\Rightarrow \dfrac {216+135+125+84}{35}=\dfrac{560}{35}=Rs.16$$
The weighted arithmetic mean of the first n natural numbers whose weights are equal to the corresponding numbers is given by
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$$\displaystyle \frac{1}{2}\left ( n+1 \right )$$
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$$\displaystyle \frac{1}{2}\left ( n+2 \right )$$
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$$\displaystyle \frac{1}{3}\left ( 2n+1 \right )$$
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$$\displaystyle \frac{1}{3}n\left ( 2n+1 \right )$$
Explanation
The first 'n' natural numbers are 1,2,3,......,n
and their corresponding weights are 1,2,3,...,n
$$\therefore Weighted\ Mean=\bar{x}_w=\dfrac{\sum_{i=1}^{n}w_ix_i}{\sum_{i=1}^{n}w_i}$$
$$\Rightarrow \dfrac{1\times 1+2\times 2+3\times 3+.....+n\times n}{1+2+3+.....+n}$$
$$\Rightarrow \dfrac{1^2+2^2+3^3+.......+n^2}{1+2+3+...+n}$$
As sum of square of n natural numbers is $$=\dfrac{n(n+1)(2n+1)}{6}$$
$$\therefore mean= \dfrac{\dfrac{1}{6}n(n+1)(2n+1)}{\dfrac{1}{2}n(n+1)}$$
$$\Rightarrow \dfrac{1}{3}(2n+1)$$
A student obtained the following marks percentage in an examination English - 50, Accounts - 75, Economic - 60, B. Std. -Hindi -if weights are 2,3,3,2,1 respectively allotted to the subjects, his weighted mean is
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$$\displaystyle \frac{50+75+60+80+55}{2+3+3+2+1}$$
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$$\displaystyle \frac{\left ( 50\times 2 \right )+\left ( 75\times 3 \right )+\left ( 60\times 3 \right )+\left ( 80\times 2 \right )+\left ( 55\times 1 \right )}{5}$$
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$$\displaystyle \frac{50\times 2+75\times 3+60\times 3+80\times 2+55\times 1}{2+3+3+2+1}$$
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none
Explanation
Percentage marks in the examination-
English=50
Accounts=75
Economics=60
B.Std=80
Hindi=55
Weight allotted to each subjects are-
English=2,Accounts=3,Economics=3,B.std=2 and Hindi=1
Mean=$$\dfrac{sum\ of\ all\ observation\ value}{Sum\ of \ all\ observation}$$
$$\therefore$$ Weight mean=$$\dfrac{50\times 2+75\times 3+60\times 3+80\times 2+55\times 1}{2+3+3+2+1}$$
In an examination a candidate scores the following percentage of marks English-44; Hindi-58; Maths-74; Physics-61; Chemistry-62 If weights 2,4,4,5,3 respectively are allotted to these subjects then the candidate's weighted mean percentage is
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$$61$$
0%
$$61.5$$
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$$62$$
0%
$$62.5$$
Explanation
Percentage marks in English, Hindi,Maths,Physics,Chemistry=44,58,74,61,62
Their weight=2,4,4,5,3
$$\therefore Weighted mean=\dfrac{2\times 44+4\times 58+4\times 74+61\times 5+62\times 3}{2+4+4+5+3}$$
$$\Rightarrow \dfrac{88+232+296+305+186}{18}=\dfrac{1107}{18}=61.5$$
The mean of the following is given by
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$$10$$
0%
$$12$$
0%
$$18$$
0%
$$20$$
Explanation
x
f
$$f_i.x_i$$
10
3
30
12
10
120
20
15
300
25
7
175
35
5
175
mean$$=\dfrac{\sum f_i.x_i}{\sum f_i}$$
$$\Rightarrow \dfrac{30+120+300+175+175}{3+10+15+7+5}=\dfrac{800}{40}=20$$
Which of the following cannot be determined graphically?
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mean
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mode
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median
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none of these
Explanation
$$Mean =\dfrac{Sum of all number}{Total no of numbers}$$
hence mean cannot be determined graphically.
If the mean of the following is 15 the value of 'a' is
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$$4$$
0%
$$6$$
0%
$$8$$
0%
$$10$$
Explanation
x
f
$$f_i.x_i$$
5
6
30
10
a
10a
15
6
90
20
10
200
25
5
125
mean$$=\frac{\sum f_i.x_i}{\sum f_i}$$
$$\Rightarrow15= \dfrac{30+10a+90+200+125}{6+a+6+10+5}$$
$$\Rightarrow 15=\dfrac{445+10a}{27+a}$$
$$\Rightarrow 445+10a=405+a$$
$$\Rightarrow 5a=40$$
$$\Rightarrow a=\dfrac{40}{5}=8$$
The median of 1, 3, 6, 8, 4, 2, 7, 10 is
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0%
6
0%
5
0%
7
0%
5.5
Explanation
Arranging in order 1, 2, 3, 4, 6, 7, 8, 10
Median=$$\displaystyle \frac{4+6}{2}=\frac{10}{2}=5$$
The mode of the series 2,3,1,2,5,3,2,2,3,5 is
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0%
5
0%
1
0%
2
0%
3
Explanation
2 has highest frequency
A data has 25 observations (arranged in descending order) which observations represents the median?
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$$\displaystyle 13^{th}$$
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$$\displaystyle 14^{th}$$
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$$\displaystyle 15^{th}$$
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none of these
Explanation
Number of observations =25
Number of observation are odd then
Median=value of $$\left(\dfrac{n+1}{2}\right)^{th}$$observation
$$\Rightarrow \dfrac{25+1}{2}=\dfrac{26}{2}=13^{th} $$observation
The arithmetic mean of first five natural number is
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0%
$$2$$
0%
$$3$$
0%
$$4$$
0%
$$8$$
Explanation
Formula used:
$$\text{Arithmetic mean}= \dfrac{\text{Sum of given numbers}}{\text{Total numbers}}$$
Apply the above formula, we get
$$\text{Arithmetic mean}=\cfrac{1+2+3+4+5}{5}$$
$$=3$$
The median of 9, 5, 7, 11, 13, 3 is
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0%
8
0%
9
0%
7
0%
11
Explanation
Arranging data in ascending order: 3, 5, 7, 9, 11, 13
Middle terms: 7, 9
Median=$$\displaystyle \frac{7+9}{2}=8$$
The weight (in kg) of $$7$$ women are $$40, 64, 65, 30, 80, 67$$ and $$70$$. The median is-
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$$80$$ kg
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$$30$$ kg
0%
$$65$$ kg
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$$70$$ kg
Explanation
Arranging in the ascending order we have
$$30, 40, 64, 65, 67, 70, 80$$
Hence median $$=65$$
A candidate obtained the following percentage of marks in an examination English 60, Maths 90, Physics 75, ChemistryIf weighs 2, 4, 3, 3 are allotted to these subjects respectively then the weight mean is given by
Report Question
0%
$$\displaystyle \frac{60+90+75+66}{2+4+3+3}$$
0%
$$\displaystyle \frac{60\times 2+90\times 4+75\times 3+66\times 3}{2+4+3+3}$$
0%
$$\displaystyle \frac{60\times 2+90\times 4+75\times 3+66\times 3}{4}$$
0%
$$\displaystyle \frac{60+90+75+66}{4}$$
Explanation
A candidate obtained the following percentage of marks in an examination English 60, Maths 90, Physics 75, Chemistry 66. If weighs 2, 4, 3, 3 are allotted to these subjects
Then total number English =$$60\times 2$$
And
total number Maths =$$90\times 4$$
And
total number Physics =$$75\times 3$$
And
total number Chemistry =$$66\times 3$$
So weight of subjects mean =
$$\frac{60\times 2+90\times 4+75\times 3+66\times 3}{2+4+3+3}$$
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