Data Handling - Class 7 Maths - Extra Questions

Find the mode and mean of the data : 3,4,5,4,6,4,7,5,8,4,6,4,7,8,6,4,3

According to the problem 

Mean $$ = \dfrac{{3 + 4 + 5 + 4 + 6 + 4 + 7 + 5 + 8 + 4 + 6 + 4 + 7 + 8 + 6 + 4 + 3}}{{17}}$$

$$ = \dfrac {88}{17}$$

$$ = 5.16$$

The mode of the data is $$4$$ because it appears most frequently.


Find the mean of each of the data. $$3,8,6,17,12,14$$

$$ The $$ $$ observations $$ $$ are : 3,8,6,17,12,14. $$
$$Mean=$$ $$\dfrac { sum\quad of\quad observations }{ no.\quad of\quad observations } $$ $$=$$ $$\dfrac { 3+8+6+17+12+14 }{ 6 } $$ $$=$$ $$\dfrac { 60 }{6 }$$ $$=$$ $$10$$


Find the mean of $$x+3,x+5,x+7,x+9$$ and $$x+11$$.

There are total $$5$$ terms,

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

$$\implies \dfrac{5x+35}{5}$$

$$\implies x+7$$       .......  Required mean


If the mean of $$16, 14, x, 23, 20$$ is $$18$$, find the value of $$x$$.

As we  know, $$\overline{x} = \dfrac{x_1 + x_2 + ....+x_n}{n}$$
here $$\overline{x} = 18$$
$$n = 5$$
So, $$18 = \dfrac{16 + 14 + x + 23 + 20}{5}$$
$$\therefore 90 = 73 + x$$
$$\therefore x = 90 - 73$$
$$x = 17$$


The mean (average) weight of three students is $$40\ kg$$. One of the students Ranga weighs $$46\ kg$$. The other two students, Rahim and Reshma have the same weight. Find Rahims weight

$$\textbf{Step -1: Finding total weight .}$$
                $$\text{Given that the mean of the weights of three students was 40kg, we can calculate }$$
                $$\text{the total sum of their weights}$$
                $$\text{Let the weight of Rahim and Reshma be x}$$
                $$\text{Total weight =mean}\times \text{Number of students}$$
                $$\implies\text{Total weight}=40\times 3$$
                $$\implies\text{Total weight}=120$$
$$\textbf{Step -2: Finding the value of x.}$$
                $$\text{Total weight}=46+2x$$
                $$\implies 120=46+2x$$
                $$\implies 2x=74$$
                $$\implies x=37kg$$
$$\textbf{Hence, Rahim weighs 37kg.}$$


The average of $$2,4,6$$ and $$8?$$ 

$$2+4+6+8=20 \div 4=5$$
The statement , that the Average of $$2, 4, 6, 8 $$ is $$5$$ coorect.


The mean of the numbers $$p+5,25,32,11,3p-1$$ and $$16$$ is 22.Find the value of $$p$$.

Mean$$=\dfrac{p+5+25+32+11+3p-1+16}{6}=22$$
$$\Rightarrow \dfrac{4p+88}{6}=22$$
$$\Rightarrow 4p+88=22\times 6=132$$
$$\Rightarrow 4p=132-88=44$$
$$\Rightarrow p=\dfrac{44}{4}=11$$


Find the mean of each of the data $$23,36,29,31,43$$

$$ The $$ $$ observations $$ $$ are : 23,36,29,31,43 . $$

$$Mean=$$ $$\dfrac { sum\quad of\quad observations }{ no.\quad of\quad observations } $$ $$=$$ $$\dfrac { 23+36+29+31+43 }{ 5 } $$ $$=$$ $$\dfrac { 162 }{ 5 }$$ $$=$$ $$32.4$$


The marks (out of $$100$$) obtained by group of students in a science test are $$85, 76, 90, 85, 39, 48, 56, 95, 81$$ and $$75.$$ Find the:

1)Highest and lowest marks obtained by students.
2)Mean marks obtained.
3)No. of students who obtained marks more than mean marks.

Arranging given marks in ascending order,
$$39,48,56,75,76,81,85,85,90,95$$

$$\therefore$$  Highest and lowest marks obtained by students are $$39$$ and $$95$$.

Number of students $$=10$$

$$\Rightarrow$$  Mean marks obtained 

$$=\dfrac{39+48+56+75+76+81+85+85+90+95}{10}=\dfrac{730}{10}=73$$

$$\Rightarrow$$  Number of students securing more than mean marks $$73$$ $$=7$$ students.                  {$$75,76,81,85,85,90,95$$}


The weight of $$11$$ members of the school football team given in kg is as follows $$38, 41, 47, 32, 33, 36, 38, 40, 41, 49, 44 $$
What is the average weight?

$$\dfrac{38+41+47+32+33+36+38+40+41+49+44}{11}$$

$$=\dfrac{439}{11}$$

$$=39.9\approx 40 $$kg 


A football player $$3,\ 4,\ 3,\ 5,\ 4,\ 5,\ 4$$ goals in $$7$$ matches. Find the mean of the goals.

Mean goal or average goal $$=\dfrac {\text{Total no. of goals}}{\text{Total no. of matches}}$$

                                              $$=\dfrac {28}{7}$$

                                              $$= 4$$


The height (in cm) of 8 girls of a class are 140, 142, 135, 137,150,148 and 138 respectively. Find the mean height of these girls and their median height.

Arranged data $$\rightarrow 135,137,138,140,142,148,150$$
$$N=7$$, So, Medium $$={ \left( \dfrac { 7+1 }{ 2 }  \right)  }^{ th }$$ then $$={ 4 }^{ th }$$ then $$=140cm$$
$$\overline { x } =$$ Mean $$=\dfrac { \sum { x }  }{ N } =\dfrac { 990 }{ 7 } =141.42cm$$


Find the average of each of the following sets.
$$(i)$$ $$48,\;23,\;13,\;54,\;7$$
$$(ii)$$ $$3\text{ kg},\;47\text{ kg},\;92\text{ kg},\;28\text{ kg}$$

Average of raw data is equal to sum of data divided by number of data.
$$\text{Average} = \dfrac{\text{Sum of data}}{\text{Number of data}}$$

$$(i)$$
$$\begin{aligned}{}\text{Average}& = \frac{{48 + 23 + 13 + 54 + 7}}{5}\\ &= \frac{{145}}{5}\\ &= 29\end{aligned}$$

$$(ii)$$
$$\begin{aligned}{}\text{Average}& = \frac{{3+47+92+28}}{4}\\ &= \frac{{170}}{4}\\ &= 42.5\text{ kg}\end{aligned}$$


Find the mean of numbers $$25, 30, 27, 23$$ and $$25$$. 

25,30,27,23,& 25
mean of given numbers
is $$ =\frac{25+30+27+23+25}{5}$$
$$ \bar{x} =\frac{130}{5} =26$$
$$\therefore $$   mean   is 26.

1201940_1389060_ans_a1f4499d30e04e85b94d58b02bf8f4af.JPG


The daily rainfall for each day of a week in a certain city is given in millimetres. Find the average rainfall during the week.
$$9,12,11,8,20,16,18$$

Average =$$\dfrac{sum \,of \,terms}{numbers\, of \,terms}=\dfrac{9+12+11+8+20+16+18}{7}=\dfrac{94}{7}=13.43mm$$ 


Find the mean of $$73,96,80,88,78$$

Given data is $$73,96,80,88,78$$
No .of observations is $$5$$
Mean of data is given as $$\dfrac{73+96+80+88+78}{5}=\dfrac{415}{5}=83$$


The mean of $$13,98,90,31,15$$ is.

Given data is $$13,98,90,31,15$$
No .of observations $$5$$
Sum of observatins $$13+98+90+31+15=247$$
Mean of data is $$\dfrac{247}{5}=49.4$$


The mean of first 20 natural numbers is 

Sum of n natural number is given by $$=\dfrac{n(+1)}{2}$$
The sum of first $$20$$ natural numbers is 
$$\dfrac {20\times 21}{2}=210$$
The mean is given as $$\dfrac {210}{20}=10.5$$


Find the mean of $$12,15,18,21,24$$

Given series is $$12,15,18,21,24$$
No. of observations is $$5$$
Sum of observations $$12+15+18+21+24=54$$
Mean $$\dfrac{54}{3}=18$$


find the mean of $$ 37,51,42,96,84$$

The data is $$37,51,42,96,84$$
The sum of observations is $$37+51+42+96+84=310$$
The mean of observation is $$\dfrac {310}{5}=62$$


The mean of $$42,61,53,80,x$$ is  $$55$$ find x.

The no .of observations $$5$$
The mean of data $$55$$
Sum of observations $$5\times 55=275$$
The sum of observations is $$42+61+53+80+x=236+x\\275=236+x\\x=275-236=39$$


Find the mean of $$1,6,7,3,8$$

Given data $$1,6,7,3,8$$
No. of observations $$5$$
Mean of data $$\dfrac{1+6+7+3+8}{5}=\dfrac{25}5=5$$


If the mean of $$8, 12, 21, 42, x$$ is $$20$$ find the value of $$x$$.

Given 
Mean of $$8,12,21,42,x$$ is $$20$$
$$\implies \dfrac{8+12+21+42+x}{5}=20$$
$$\implies x+83=100$$
$$\implies x=17$$


8kg of tea consisting Rs240 per kg is mixed with 9kg of tea costing Rs25o per kg. The average price per kg of the mixed tea is

Total quantity of tea = $$8+9$$=$$17$$kg
Total amount of tea = $$8\times240+9\times250$$=$$4170$$
Average price per kg = $$\dfrac{4170}{17}$$=$$245.29$$


Find the arithmetic mean of the following distribution:
Marks obtained1015202530
No. of students246810

Marks obtained
$$({ x }_{ i })$$		No. of students
$$({ f }_{ i })$$
$$ { x }_{ i }{ f }_{ i }$$
10220
15460
206120
258200
3010300
Total$$\sum { { f }_{ i } } =30$$
$$\sum { { f }_{ i } } { x }_{ i }=700$$

Now, $$\overline { x } =\cfrac { \sum { { f }_{ i } } { x }_{ i } }{ \sum { { f }_{ i } }  } =\quad \cfrac { 700 }{ 30 } =23.33$$


Find the mean of the following distribution:
ClassNumber of students
4-82
8-1212
12-1615
16-2025
20-2418
24-2812
28-3213
32-363

ClassNo. of students
$$({ f }_{ i })$$		Class Mark $$({ x }_{ i })$$
$${ x }_{ i }=\frac {\text{ Upper class limit+ Lower class limit} }{ 2 } $$

$${ x }_{ i }{ f }_{ i }$$
4-8  2612
8-12  1210120
12-16  1514210
16-20  2518450
20-24  1822396
24-28  1226312
28-32  1330390
32-36  334102
Total$$\sum { { f }_{ i } } =100$$


$$\sum { { f }_{ i } } { x }_{ i }=1992$$
Now, $$\overline { x } =\cfrac { \sum { { f }_{ i } } { x }_{ i } }{ \sum { { f }_{ i } }  } = \cfrac { 1992 }{ 100 } =19.92$$


Find the mean of the following data:
Class interval0-1010-2020-3030-4040-50
Frequency12118109

Class Frequency
$$(\sum { { f }_{ i } })$$		  Class Mark $$({ x }_{ i })$$
$${ x }_{ i }=\frac { \text{ Upper class limit+ Lower class limit} }{ 2 } $$		$${ x }_{ i }{ f }_{ i }$$
0-10  12560
10-20  1115165
20-30  825200
30-40  1035350
40-50  945405
Total$$\sum { { f }_{ i } } =50$$


$$\sum { { f }_{ i } } { x }_{ i }=1180$$
Now, $$\overline { x } =\cfrac { \sum { { f }_{ i } } { x }_{ i } }{ \sum { { f }_{ i } }  } = \cfrac { 1180 }{ 50 }= 23.6$$


Find the mean of the first $$10$$ natural numbers.

Given data 1, 2, 3, .......10
$$\therefore$$ Arithmetic mean (A. M)
$$=\, \displaystyle \frac{The\, sum\, of\, observations}{The\, total\, number\, of\, observations}$$
$$=\, \displaystyle \frac{1\, +\, 2\, +\, 3\, +\, ...\, +\, 10}{10}\, =\, \displaystyle \frac{55}{10}\, =\, 5.5$$


During the twelve months of 1998, an executive charged $$4,1,5,6,3,5,1,0,5,6,4$$ and $$3$$ business luncheons at the Wardlaw Club. What was the mean monthly number of luncheons charged by the executives? If the mean is $$x$$ then find $$100x$$.

The mean of given $$12$$ numbers is $$\dfrac {(4+1+5+6+3+5+1+0+5+6+4+3)}{12}$$
$$ = \dfrac {43}{12} = 3.58$$ 
The mean monthly income charged by executives is $$3.58$$.
$$x = 3.58$$
$$\therefore 100x = 358$$


The marks obtained by $$10$$ students in a test are $$15, 75, 33, 67, 76, 54, 39, 12, 78, 11$$. Find the arithmetic mean.

Marks obtained by $$10$$ students in a test
$$=15,75,33,67,76,54,39,12,78,11$$

Arithmetic Mean = $$\dfrac{sum\  of\  all\  terms\  in\  data}{total \ number\  of\  terms\  in\  data}        \ =     $$$$\cfrac { 15+75+33+67+76+54+39+12+78+11 }{ 10 } =46$$


Find the arithmetic mean of the marks $$72, 73, 75, 82, 74$$ obtained by a student in $$5$$ subjects in an annual examination.

Here $$n = 5$$

$$Mean=\dfrac { {\text {Sum of all observations}}}{\text{number of observations}} =\dfrac { 72+73+75+82+74 }{ 5 } $$

$$=\dfrac { 376 }{ 5 }$$

$$ =75.2$$

$$\therefore $$ Mean $$= 75.2$$


Rice consumed in a school under the mid-day meal program for 5 consecutive days is 15.750 kg, 14.850 kg, 16.500 kg, 14.700 kg, and 17.700 kg. Find the average rice consumption for the 5 days.
704628_4c8189208ef3435196735457b9e9dd12.png

$$\Rightarrow$$  Rice consumed in a school for $$5$$ consecutive days is $$15.750\,kg,\,14.850\,kg,\,16.500\,kg,\,14.700\,kg,\,17.700\,kg$$
$$\Rightarrow$$  Required average $$=\dfrac{Total\,rice\,consumption\,in\,5\,days}{5}$$
                                     $$=\dfrac{15.750+14.850+16.500+14.700+17.700}{5}$$
                                     $$=\dfrac{79.500}{5}=15.900\,kg$$
$$\therefore$$  The average rice consumption for $$5$$ days are $$15.900\,kg$$


Write at least two questions to find the mean, giving suitable data.

$$Question 1$$.A survey was conducted by group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
                  
Number of plants        0-2  2-4  4-6  6-8  8-10  10-12  12-14
Number of houses        1      2      1      5      6        2        3
                 
 Which method did you use for finding the mean, why?

$$Question 2$$.Consider the following distribution of daily wages of 50 workers a factory.
                 
 Daily wages(in Rs)        100-120  120-140  140-160  160-180  180-200  
                   
Number of houses            12            14           8              6            10
                   
Find the mean daily wages of the workers of the factory by using appropriate method         
 


Find the mean of the following distribution
x:1030507089
f:78101510

Calculation of Mean
 $$x_i$$ $$f_i$$$$f_ix_i$$
 $$10$$ $$7$$ $$70$$
 $$30$$ $$8$$ $$240$$
 $$50$$ $$10$$ $$500$$
 $$70$$ $$15$$ $$1050$$
 $$89$$ $$10$$ $$890$$

$$\displaystyle N= \sum f_i= 50$$       $$\displaystyle \sum f_ix_i=2750$$
$$\therefore $$ Mean = $$\dfrac{\sum f_ix_i}{N}= \dfrac{2750}{50}= 55$$


The average temperature on Saturn is $$288^oF$$, while that on Jupiter is $$162^oF$$. Which planet has the average temperature?

Average temperature $$ = \dfrac{{{{288}^0}F + {{162}^0}F}}{2}$$
$$ = \dfrac{{{{450}^0}}}{2}F$$
$$ = {225^0}F$$


The following are the numbers of books issued in a school library during a week:$$105, 216, 322, 167, 273, 405$$ and $$346$$.Find the average number of books issued per day.

$$\quad Average   = \cfrac { total\quad number\quad of\quad books\quad issued\quad in\quad the\quad week }{ total\quad number\quad of\quad days\quad in\quad a\quad week } \\ \therefore \quad Total\quad number\quad of\quad books\quad issued\quad in\quad the\quad week\quad =\quad 216+105+322+167+273+405+346\\ =1834\quad books\\ Number\quad of\quad days\quad in\quad the\quad week\quad =\quad 7\\ Average\quad =\quad \cfrac { 1834 }{ 7 } \quad =\quad 262$$

Therefore, the average number of books issued per day is $$262$$.


The mean of three numbers $$a, b$$ and $$c$$ is $$8$$ and the mean of five numbers $$a, b, c, d$$ and $$e$$ is $$15$$. Find the mean of $$d$$ and $$e$$?

Mean of $$a,b$$ and $$c$$ is $$8,$$
$$\dfrac{a+b+c}{3}$$$$=8$$
$$\Rightarrow a+b+c=24\quad\quad\quad\dots(i)$$

Mean of $$a,b,c,d$$ and $$e$$ is $$15,$$
$$\dfrac{a+b+c+d+e}{5}=15$$
$$\Rightarrow a+b+c+d+e=75\quad\quad\quad\dots(ii)$$

Subtract equation $$(i)$$ from equation $$(ii),$$
$$d+e = 51$$
$$\Rightarrow \dfrac{d+e}{2} =25.5$$

So, the mean of $$d$$ and $$e$$ is $$25.5.$$


The mean of $$x, x+2, x+4, x+6$$ and $$x+8$$ is $$11$$, find the mean of the first three observations.


$$\textbf{Step -1: Finding the value of x.}$$

                $$\text{Here, number of observations }=5$$

                $$\text{Sum of given observations }=x+x+2+x+4+x+6+x+8$$

                $$=5x+20$$

                $$\text{Mean of given observations}=11$$

                $$\mathbf{\because \text{Mean}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}}$$

                $$\Rightarrow 11=\dfrac{5x+20}{5}$$

                $$\Rightarrow 5x+20=55$$

                $$\Rightarrow 5x=35$$

                $$\Rightarrow x=7$$

$$\textbf{Step -2: Finding the mean of the first three observations.}$$

                $$\text{First three observations are }x,x+2,x+4$$

                $$\text{Number of observations }=3$$

                $$\text{Sum of first three observations }=x+x+2+x+4$$

                $$=3x+6$$

                $$=3\times 7+6$$

                $$=27$$

                $$\text{Mean of the first three observations }=\dfrac{27}{3}=9$$

$$\textbf{Hence,the mean of the first three observations is }\mathbf{9}\textbf{.}$$



The mean of $$18,24,15,2x+1$$ and $$12$$ is $$21$$. Find the value of $$x$$.

The mean of 5 numbers $$18,24,15,2x + 1,12$$ is given as 21.

$${\rm{Mean}} = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{observations}}}}{{{\rm{Total}}\;{\rm{observations}}}}$$

$$21 = \frac{{18 + 24 + 15 + 2x + 1 + 12}}{5}$$

$$105 = 70 + 2x$$

$$2x = 35$$

$$x = 17.5$$



$$34, 66, 30, 38, 44, 50, 40, 60, 42, 51$$.

The given data is $$34,66,30,38,44,50,40,60,42,51$$.
Number of values $$=10$$
sum of the values $$=34+66+30+38+44+50+40+60+42+51$$
                               $$=455$$
Arithmetic mean$$=\dfrac{\text{sum of values}}{\text{No. of values}}$$
                            $$=\dfrac{455}{10}$$
                            $$=45.5$$


Find arithmetic mean of following data:
$$38, 70, 48, 34, 42, 55, 63, 46, 54, 44$$.

given raw data is $$38,70,48,34,42,55,63,46,54,44.$$
so the number of values $$=10$$
Sum of the values $$=38+70+48+34+42+55+63+46+54+44$$
                               $$=494$$
Arithmetic mean of the raw data 
$$=\dfrac{\text{sum of values}}{\text{number of values}}=\dfrac{494}{10}$$
$$=49.4$$


$$22, 24, 30, 27, 29, 31, 25, 28, 41, 42$$.

The given data is $$22,24,30,27,29,31,25,28,51,42.$$
Number of values given $$=10$$
Sum of the values $$=22+24+30+27+29+31+25+28+41+42$$
                               $$=299$$
Arithmetic mean$$=\dfrac{\text{sum of values}}{\text{no. of values}}$$
                           $$=\dfrac{299}{10}$$
                           $$=29.9$$


The mean of the numbers $$6, y, 7, x, 14$$ is $$8$$. Express $$y$$ in terms of $$x$$

The mean of the $$5$$ numbers $$6,y,7,x,14$$ is $$8$$.

$${\rm{Mean}} = \cfrac{{{\rm{number}}\;{\rm{of}}\;{\rm{observations}}}}{{{\rm{total}}\;{\rm{observations}}}}$$

$$8 = \cfrac{{6 + y + 7 + x + 14}}{5}$$

$$40 = 27 + y + x$$

$$y + x = 13$$

$$y = 13 - x$$



Show that the set of letters needed to spell "CATARACT" and the set of letters needed to spell "TRACT" are equal.

$$A=A= Set \,  of \,  letters \,  of \,  the \,  word$$ "$$CATARACT$$"$$ =$${$$A,C,R,T$$}
={A,C,R,T}
$$B=B= Set \,  of \,  letters \,  of \,  the \,  word$$ "$$TRACT$$"$$ =$${$$A,C,R,T$$}

Clearly, $$ A=B$$


Find the mean and variance for the given data
First $$10$$ multiplies of $$3$$

$$\Rightarrow$$  First $$10$$ multiple of $$3$$ are, $$3,6,9,12,15,18,21,24,27,30.$$
$$\Rightarrow$$  $$Mean(\overline{x})=\dfrac{3+6+9+12+15+18+21+24+27+30}{10}$$
                         $$=\dfrac{165}{10}$$
                         $$=16.5$$
$$\therefore$$  $$\overline{x}=16.5$$

$$x_i$$ $$x_i-\overline{x}$$ $$(x_i-\overline{x})^2$$ 
$$3$$ $$-13.5$$ $$182.25$$ 
$$6$$ $$-10.5$$ $$110.25$$ 
$$9$$ $$-7.5$$ $$56.25$$ 
$$12$$ $$-4.5$$ $$20.25$$ 
$$15$$ $$-1.5$$ $$2.25$$ 
$$18$$ $$1.5$$ $$2.25$$ 
$$21$$ $$4.5$$ $$20.25$$ 
$$24$$ $$7.5$$ $$56.25$ 
$$27$$ $$10.5$$ $$110.25$$ 
$$30$$ $$13.5$$ $$182.25$$ 
  $$\sum(x_i-\overline{x})^2=742.5$$
$$\Rightarrow$$  $$Variance(\sigma)^2=\dfrac{x_i\overline{x})^2}{n}=\dfrac{742.5}{10}=74.25$$



Out of $$25$$ numbers, the mean of $$15$$ numbers is $$36$$ and the mean of the remaining number is $$26$$; find the mean of all the $$25$$ numbers.

No of observations is $$25$$

Mean of $$15$$ observations is $$36$$

Sum of observations is $$36\times 15=540$$

Mean of $$10$$ observations is $$26$$

Sum of observations is $$26\times 10=260$$

Total sum is $$540+260=800$$

Mean of the dta is $$\dfrac{800}{25}=32$$


Sum of deviation of variate fron their A.M is always zero that is $$\sum  ({ x }_{ i }-\overline { x } )=0$$ sum of square of deviation of variate from their A.M is minimum that is $$\sum  ({ x }_{ i }-\overline { x } )=0^2$$ is minimum

consider 3 number
3,4,5
$$\overline { x } =\frac { 3+4+5 }{ 3 } =4$$
$$\sum (x_i-\overline { x } )=(3-4)+(4-4)+(5-4)$$
                                       $$=-1+0+1$$
                                        $$=0$$ (always correct)
$$\sum (x_i-\overline { x } )^2=(-1)^2+o^2+1^2$$
                                          $$=2$$(minimum)



Find the mean for the data.
1224517_7647c87c2d2e4dc799badaade22646b5.png

$$Mean = \dfrac{{\sum {xifi} }}{{\sum {fi} }} = \frac{{350}}{{25}} = 14$$
1380944_1224517_ans_42efda7bbb7141fca06a0921262be306.png


The average of $$6$$ numbers is $$9$$. If $$2$$ is added to each number, what will be the new average?

$$\begin{array}{l} Sum\, \, \, of\, 6\, \, numbers=6\times 9=54 \\ New\, \, average=\dfrac { { 6\times 2+54 } }{ 6 }  \\ =\dfrac { { 12+54 } }{ 6 } =\dfrac { { 66 } }{ 6 }  \\ =11 \end{array}$$


Find the mean and median of following data :
$$38, 70, 48, 40, 42, 46, 66, 68, 46, 54, 44$$


1484999_1328072_ans_09899aa491f840a3861cb953aae9c306.jpeg


The mean age of $$5$$ children in a family is $$12$$ years, if four of them are respectively $$6,11,13$$ and $$16$$ years find the age of the fifth child.


1209939_1274006_ans_7d2b49c4ebe94a719d99d2806c4db5ea.jpg


The table shows the points of each player scored in four games.
 Players Game I Game II Game III  Game IV
 X 1416 10 10 
 Y 08  6 4
 Z 811 Did not play   13
$$Y$$ played in all the four games, how would you find mean?

$$Y'$$s points 
Game $$I=0,$$  Game $$II-8$$ 
Game $$III=6$$, Game$$IV-4$$

$$Y's \ \  mean=\dfrac{sum \ of\ \ observation}{No. \ of \  observation}$$

                     $$=\dfrac{0+8+6+4}{4}=\dfrac{18}{4}=4.5$$

$$\therefore$$ The mean of $$Y=4.5$$


The yield of soyaber per acre in the form of mode for $$7$$ years was $$10, 7, 5, 3, 9, 6, 9$$ quintal. Find the mean of yeild per acre .

The yeild of soyaber per acre in the form of mode for 7 years
was $$10, 7, 5, 3, 9, 6, 9$$ quintals

we know that

mean $$ = \dfrac{sum \ of \ the \ observations}{ No \ of \ observations}$$

$$= \dfrac{10 + 7 + 5 + 3 + 9 + 6 + 9}{7}$$

$$= \dfrac{49}{7}$$

Mean $$= 7$$

Hence
Mean of yeild per acre $$= 7 $$ quintal 


Following table shown the points of each player scored in from games.
PlayerGame $$1$$Game $$2$$Game $$3$$Game $$4$$
$$A$$$$14$$$$16$$$$10$$$$10$$
$$B$$$$0$$$$8$$$$6$$$$4$$
$$C$$$$8$$$$11$$$$did\ not\ play$$$$13$$
Who is the best performer?

Average points of player $$A$$ in all the games he played $$= \cfrac{14 + 16 + 10 + 10}{4} = \cfrac{50}{4} = 12.5$$
Average points of player $$B$$ in all the games he played $$= \cfrac{0 + 8 + 6 + 4}{4} = \cfrac{18}{4} = 4.5$$
Average points of player $$C$$ in all the games he played $$= \cfrac{8 + 11 + 13}{3} = \cfrac{32}{3} = 10.67$$
Since playes $$A$$ has the best average, thus player $$A$$ is the best performer.


The ages of $$5$$ girls in a group are : $$16$$ years $$4$$ months; $$15$$ years $$8$$ months ; $$14$$ years $$9$$ months; $$15$$ years $$3$$ months and $$14$$ years $$8$$ months. Find their mean.

Given that 
The ages of $$5$$ girls in a group are

we know $$12$$ months $$ = 1$$ year

$$8$$ month $$= \dfrac{1}{12} $$ years

$$16 \, years \,\,\, 4 months$$
$$15 \, years \,\,\, 8 months$$
$$14 \, years \,\,\, 9 months$$
$$15 \, years \,\,\, 3 months$$
$$\underline{14 \, years \,\,\, 8 months}$$
$$74 years \,\,\, 32 months$$

$$\dfrac{32}{12} $$ years

$$74 \ years + 2 \ years \, 8 months$$                $$8 \ months = \dfrac{8}{12} \ years$$

$$76 years \, 8 months$$                                                       $$= 0.7 \ years$$

$$76.7 years$$

$$\therefore $$ mean $$= \dfrac{76.7 \ years}{5}$$

mean $$= 15.34 \ years$$


Find the mean of the numbers $$8, 12, 5, 6, 10, 7$$.

Given numbers $$8, 12, 5 , 6, 10, 7$$ no of number $$= 6$$

we know $$mean = \dfrac{sum \ of \ numbers}{no \ of \ numbers}$$

$$mean = \dfrac{8 + 12 + 5 + 6 + 10 + 7}{6}$$

$$= \dfrac{48}{6} = 8$$

$$mean = 8$$


The mode of students in maths are $$10,11,13,12$$, find average marks of students. 

Given the (marks of students) in maths are $$10, 11, 13, 12$$ we have average $$= \dfrac{sum \ of \ observations}{no \, of \ observations}$$

$$\therefore$$ average marks of students

$$= \dfrac{10 + 11 + 13 + 12}{4} = \dfrac{46}{4}$$

$$= 11.5$$


If the mean of five observation $$x,x+2,x+4,x+6$$ and $$x+8$$ is $$13$$, then find the value of $$x$$.

Given that the mean of five observation $$x, x + 2, x + 4, x + 6$$ and $$x + 8 $$ is $$13$$.
we know that 

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

$$\Rightarrow 13 = \dfrac{x + (x + 2) +(x + 4) + (x + 6) + (x + 8)}{5}$$

$$ 13 = \dfrac{5x + 20}{5}$$

$$13 = \dfrac{5 (x + 4)}{5}$$

$$13 = x + 4$$

$$x + 4 = 13$$            value of $$x = 9$$

$$x = 13 - 4$$

$$x = 9$$


Find the mean of the following data.
$$\begin{array} { | c | c | c | c | c | c | c | } \hline \text { Class Intervals } & { 0 - 10 } & { 11- 20 } & { 21 - 30 } & { 31 - 40 } & { 41 - 50 } & { 51 - 60 } \\ \hline \text { Frequency } & { 4 } & { 7 } & { 9 } & { 10 } & { 7 } & { 3 } \\ \hline \end{array}$$

$$x_i=5,15,25,35,45,55\sum f_i=4+7+9+10+7+3=40\sum x_if_i=20+105+225+350+315+165=1180\bar x=\dfrac{1180}{40}=29.5$$'




Find the mean of given data.
$$x + 77 , x + 7 , x + 5 , x + 3 , x + 2.$$

Given observations are $$x + 77,\  x + 7,\  x + 5,\  x + 3,\  x + 2.$$.

We know that, arithmetic mean $$=\dfrac{\text{Sum of all observations}}{\text{Total number of observations}}$$

Hence, the arithmetic mean of the given $$5$$ observations $$=\dfrac{x+77+x+7+x+5+x+3+x+2}{5}$$

$$\Rightarrow \dfrac{5x+94}{5}=\dfrac{5x}{5}+\dfrac{94}{5}$$

$$\Rightarrow x+18.8$$

Hence, the mean of the given data is $$x+18.8$$.


The average of a set of 5 numbers isA number 14 is deleted from the set. What is the average of the remaining four numbers.

Given the average of set of $$5$$ numbers is $$10$$

mean (Average) $$= 10$$

no. of numbers(observations) $$= 5$$

Here A number $$14$$ is deleted.

we know average $$= \dfrac{sum \ of \ numbers}{No. \ of \ numbers(observations)}$$

Sum of $$5$$ numbers $$= 10 \times 5$$

$$= 50$$

Here remaining sum of no"
i.e. 4 numbers sum

$$ = 50 - 14$$

$$= 36$$

$$\therefore$$ new mean (Average) $$= \dfrac{sum \ of \ 4 \ numbers}{No. \ of \ numbers(observations)}$$

$$= \dfrac{36}{4}$$

Average of the Remaining four numbers $$= 9$$


Find the sum of $$-59,-41,73,-92,81,-(-41),-3$$.

Find sum of $$-59 , -41, 73 , -92, 81 , -(-41), - 3$$

$$= -59 + (-41) + 73 + (-92) + 81 + -(-41), - 31$$

$$= -59 - 41 + 73 - 92 + 81 + (41 - 3)$$

$$= -59 - 41 + 73 - 92 + 81 + 41 - 3$$

$$ = -195 + 195$$

$$= 0$$


Average of $$9$$ numbers is $$14$$. If one number $$12$$ is excluded, find the new average.

Average of $$9$$ numbers$$=14$$
Let sum of $$9$$ no.s$$=x$$
$$\cfrac { x }{ 9 } =14$$
$$x=14\times 9=126$$
If $$12$$ is excluded, sum$$=126-12=114$$
New average$$=\cfrac { new\quad sum }{ 8 } $$
$$=\cfrac { 114 }{ 8 } =14.25$$ 


Find the A.M. between $$7$$ and $$13$$.

$$\textbf{Step-1: Apply relevant property of Arithmetic Mean.}$$
                 $$\text{We have,}$$ $$7$$ $$\text{and}$$ $$13$$

                 $$\therefore$$  $$\text{A.M. =}$$ $$\dfrac{7+13}2$$  $$[\textbf{If a & b are two numbers then}$$ $$\textbf{A.M.}$$ $$\boldsymbol{=\dfrac{a+b}2]}$$
                 $$=\dfrac {20}2$$ 
                 $$=10$$
$$\textbf{Hence, answer is 10}$$


Find the mean of the first five even whole numbers.

First five even whole numbers are $$0, 2, 4, 6, 8$$
Number of data $$= 5$$
$$\therefore Mean = \dfrac{0 + 2 + 4 + 6 + 8}{5} = \dfrac{20}{5} = 4$$


Find the A.M. of $$12$$ and $$-8$$.

$${\textbf{Step 1: Apply the formula of A.M. }}$$

                $$\text{Let  }{a = 12}{\text{ and }}{b = - 8}$$ 

                $${\text{Now, A}}{\text{.M}}{\text{.}} = \dfrac{{12 + ( - 8)}}{2}$$

                                     $$ = \dfrac{4}{2} = 2$$ 

$${\textbf{The A.M. of }}\mathbf{12}{\textbf{ and }}\mathbf{ - 8}{\textbf{ is }}\mathbf{2.}$$

 



Find the A.M. of $$(x-y)$$ and $$(x+y)$$.

Let $$a=x-y$$ and $$b=x+y$$

A.M. of two numbers $$a$$ and $$b$$ is $$=\dfrac{a+b}2$$

$$=\dfrac{x-y+x+y}2=\dfrac {2x}2=x$$
$$\therefore$$  A.M. of $$x-y$$ and $$x+y$$ is $$x$$


A cricketer scores the following runs in 8 innings : 58, 76, 40, 35, 48, 45, 0,Find the mean score.

Given A cricketer Scores the following runs in $$8$$ innings
$$58, 76, 40, 35, 48, 45, 0, 100$$

No. of innings(n)$$=8$$

We know that Mean $$=\dfrac{sum \,of \, observations(scores)}{No. \,of \, observations(innings)}$$

                                   $$=\dfrac{58+76+40+35+48+45+0+100}{8}$$

                                   $$=\dfrac{402}{8}$$

                                   $$=50.25$$

$$\therefore$$ Mean score$$=50.25$$ Runs.


Calculate the arithmetic mean of $$5.7, 6.6, 7.2, 9.3, 6.2$$.

Given observations are $$5.7, 6.6, 7.2. 9.3, 6.2$$.
Number of observations = $$5$$
Mean = sum of observations / number of observations
Mean = $$(5.7+6.6+7.2+9,3+6.2)/5$$
$$=35/5=7$$
Hence the mean of the given observations is $$7$$.


The weight (in kg) of $$8 $$ new born babies are $$3,3.2,3.4,3.5,4,3.6,4.1,3.2 $$. Find the mean weight of the babies. 

Given weight of babies are $$3, 3.2, 3.4, 3.5, 4, 3.6, 4.1, 3.2$$
Number of observations =$$ 8$$
Mean = sum of observations / number of observations
$$\therefore$$ Mean = $$(3+3.2+3.4+3.5+4+3.6+4.1+3.2)/8$$
$$= 28/8 = 3.5 \,kg$$
Hence the mean of the weight of babies is $$3.5\, kg$$.


In a class test, the mean of marks scored by a class of $$46$$ students was calculated as  $$18.2$$. Later on, it was detected that the marks of one student were wrongly copied as  $$21$$ instead of $$29$$. Find the correct mean.

Mean of marks $$= 18.2$$
Number of students $$= 40$$
Total marks of  $$40$$ students $$= 40 \times 18.2 = 728$$
Difference of marks when copied wrongly $$= 29-21 = 8$$
So total marks $$= 728+8 = 736$$
$$\therefore$$mean $$= 736/40$$
$$= 18.4$$
Hence the correct mean is $$18.4$$.


The marks obtained by $$15$$ students in a class test are $$12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20$$.find the mean of their marks when $$2$$ marks are deducted from the marks of each student.

If mark of each student is deducted by $$2$$, total deducted marks$$ = 2\times15 = 30$$
Total decrease in sum of marks $$= 207-30 = 177$$
$$\therefore$$Mean = sum of marks/number of students
$$\therefore$$mean $$= 177/15 = 11.8$$
Hence the mean is $$27.6$$


The marks obtained by $$15$$ students in a class test are $$12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20$$.find the mean of their marks when the marks of each student are increased by $$4$$.

If mark of cach student is increased by 4, total increased marks $$=4\times15 = 60$$
Total increase in sum of marks $$= 207+60 = 267$$
$$\therefore$$ Mean $$=\dfrac{sum\ of\ marks}{number\ of\ students}$$

$$\therefore$$ mean $$= \dfrac{267}{15} = 17.8$$
Hence the mean is $$17.8$$.


The marks obtained by $$15$$ students in a class test are $$12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20$$.find the mean of their marks.

Marks obtained by students are $$12, 14, 07, 09, 23, 11, 08, 13. 11, 19. 16, 24, 17, 03. 20$$.
Number of students = $$15$$
Mean = sum of observations / number of observations
$$= 124+14+07+09+23+11+08+ 13+ 11+19+16+24+17+03+20$$
$$= 207/15$$
$$=13.8$$
Hence the mean of their marks is $$13.8$$.


Find the mean of $$25$$ given numbers when the mean of $$16$$ of them is $$13$$ and the mean of the remaining numbers is $$18$$.

Mean of $$10$$ numbers $$= 13$$
$$\therefore$$Sum of numbers $$= 13\times10 = 130$$
Mean of remaining $$15$$ numbers $$= 18$$
$$\therefore$$Sum of numbers $$= 15\times18 = 270$$
$$\therefore$$Sum of all numbers $$= 130+270 = 400$$
$$\therefore$$Mean = sum of numbers/$$25 = 400/25 = 16$$
Hence the mean of $$25$$ numbers is $$16$$.


The marks obtained by $$15$$ students in a class test are $$12, 14, 07, 09, 23, 11, 08, 13, 11, 19, 16, 24, 17, 03, 20$$.find the mean of their marks when the marks of each student are doubled.

If the mark of each student is doubled, then the new sum of marks $$= 2 \times 207 = 414$$
$$\therefore$$mean = new sum of marks/number of students
$$\therefore$$mean $$= 414/15 = 27.6$$
Hence the mean is $$27.6$$.


Find the mean of the following data:
$$40, 30, 30, 0, 26, 60$$

Mean of $$40, 30, 30, 0, 26, 60$$
No. of data (n) $$= 6$$
$$\therefore Mean = \dfrac{40 + 30 + 30 + 0 + 26 + 60}{6} = \dfrac{186}{6} = 31$$


Fill in the blanks:
The mean of the first $$8$$ natural numbers is ............

The mean of the first $$8$$ natural numbers is $$4.5$$
Ex- First $$8$$ natural number $$=1,2,3,4,5,6,7,8$$
$$mean=\dfrac{1+2+3+4+5+6+7+8}{8}=\dfrac{36}{8}=4.5$$


Find the mean of the following distributions:
Number5101520253035
Frequency1 2 5 6 3 2 1

Number (x)frequency (f)fx
       5          1$$5 \times 1=5$$
      10          2$$10 \times 2=50$$
      15          5$$15 \times 5=75$$
      20          6$$20 \times 6=120$$
      25          3$$25 \times 3=75$$
      30          2$$30 \times 2=60$$
      35          1$$35 \times 1=35$$
    Total$$\Sigma f=20$$$$\Sigma fx=390$$

$$\therefore Mean=\Sigma fx/\Sigma f$$
$$=390/20=19.5$$
Hence the mean is 19.5.


At a shooting competition, the scores of a competitor were as given below :

Score

0

1

2

3

4

5

No. of shots

0

3

6

4

7

5

What was his mean?

We write the marks in cumulative frequency table.


Score x

   No. of shots (f)             

fx

Cumulative frequency

0

       0

0

0

1

       3

3

3

2

       6

12

9

3

       4

12

13

4

       7

28

20

5

       5

25

25

Total

$$\Sigma f = 25$$

$$\Sigma fx = 80$$

$$Mean = \Sigma fx/\Sigma f$$

$$= 80/25$$

$$= 3.2$$

Hence his mean score is $$3.2$$.



A boy scored the following marks in various class tests during a term each test being marked out of $$20$$: $$15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16$$. What are his mean marks?

$$Mean = \Sigma x_i/n$$

$$= 7+ 10+12+12+14+ 15+16+16+16+ 17+ 19$$

$$= 154/11$$

$$= 14$$

Hence the mean is $$14$$.



If the arithmetic mean of $$8,4,x,6,2,7$$ is $$5$$, then the value of $$x$$ is _____

Given
Mean of $$8,4,x,6,2,7,$$ is $$5$$.
We know that,
$$Mean= \dfrac{Sum\, of\,all\, observations}{Number\, of\, observations}\\$$
$$\Rightarrow 5=\dfrac{8+4+x+6+2+7}{6}$$
$$\Rightarrow x+27=30$$
$$\Rightarrow x=3$$


A car seller collects the following data of cars sold in his shop.
Colour of CarNumber of Cars Sold
        Red         15
      Black
         20
      White
         17
     Silver         12
     Others          9
(a) Which colour of the car is most liked?
(b) Which measure of central tendency was used in (a)?

(a) The black color of the car is most liked because number of black cars sold is the greatest.
(b) Mode is used in (a) part as mode is the data that occur most frequently in mention set. In the given data black cars are sold the most, therefore, black car csn be considered as the mode.



Find the mean of first six multiples of 4.

The first six multiples of $$4$$ are : $$4,8,12,16,20,24$$
We know that 
$$\text{Mean}=\dfrac{\text{sum of observations}}{\text{number of observations}}\\$$
$$\therefore\text{ Mean}=\dfrac{4+8+12+16+20+24}6=\dfrac{84}{6}= 14$$

Hence, the mean of the first six multiples of $$4$$ is $$14$$.


The following are weights (in kg) of 12 people.
$$70,62,54,57,62,84,75,59,62,65,78,60$$
(a) Find the mean of the weights of the people.
(b) How many people weigh above the mean weight ?
(c) Find the range of the given data.

Rearranging the data into ascending order, we get 
$$ 54,57,59,60,62,62,62,65,70,75,84$$

Here,
(a) Mean $$=$$ Sum of observations $$\div$$ number of observations
$$=\dfrac{54+54+57+59+60+62+62+62+65+70+75+84}{12}$$
$$=\dfrac{788}{12}= 65.6$$ kg

(b) $$4$$ people have a weight above the mean weight.

(c) Range $$=$$ greatest observation $$-$$ lowest observation  
$$=84-54= 30$$ kg


The mean of three numbers isThe mean of other four numbers is $$12.$$ Find the mean of all the numbers.

Given that 
Mean of $$3$$ numbers $$=10$$.
$$\Rightarrow 10=\dfrac{\text{Sum of 3 observations}}{3}$$
$$\Rightarrow$$ Sum of 3 observation $$= 30\quad...(1)$$

Also given that
Means of other$$ 4$$ numbers $$= 12$$
$$\Rightarrow 12=\dfrac{\text{Sum of 4 observations}}{4}$$
$$\Rightarrow$$ Sum of 4 observation $$= 48\quad...(2)$$

From equation $$(1)$$ and $$(2)$$, we get 
Sum of 7 observation $$= 30 + 48=78$$
$$\therefore$$  Mean $$=\dfrac{78}{7}= 11.14$$


Find the mean of the following data:
$$3, 5, 7, 9, 11, 13, 15$$

Mean of $$3, 5, 7, 9, 11, 13, 15$$
No. of data $$= 7$$
$$\therefore Mean = \dfrac{3 + 5 + 7 + 9 + 11 + 13 + 15}{7} = \dfrac{63}{7} = 9$$


A batsman scored the following number of runs in six innings:
$$36, 35, 50, 46, 60, 55$$
Calculate the mean runs scored by him in an inning.

In six innings, run scored was
$$36, 35, 50, 46, 60, 55$$
$$\therefore Mean = \dfrac{36 + 35 + 50 + 46 + 60 + 55}{6} = \dfrac{282}{6} = 47$$


The mean of six numbers:$$x-5, x- 1, x, x + 2, x + 4$$ and $$x + 12$$ is $$15$$. Find the mean of first four numbers.

Given  six numbers are $$x-5, x- 1, x, x + 2, x + 4$$ and $$x + 12$$

$$Mean =\dfrac{ [(x – 5) + (x – 1) + x + (x + 2) + (x + 4) + (x + 12)]}{ 6}$$

Given,

$$15 =\dfrac {(x – 5 + x – 1 + x + x + 2 + x + 4 + x + 12)}{6}$$

So we get

$$15 =\dfrac{ (12 + 6x)}{ 6}$$

By cross multiplication

$$12 + 6x = 90$$

$$6x = 90 – 12 = 78$$

$$x = \dfrac{78}{6}= 13$$

So the six numbers are $$(13 – 5), (13 – 1), 13, (13 + 2), (13 + 4), (13 + 12) = 8, 12, 13, 15, 17, 25$$

Here the mean of first four numbers

$$= \dfrac{(8 + 12 + 13 + 15)}{ 4} \\ = \dfrac{48}{4} = 12$$



Find the mean of first six multiples of $$5$$.

Here the first six multiples of $$5$$ are $$5, 10, 15, 20, 25$$ and $$30$$

We know that $$n = 6$$

So $$Mean = (5 + 10 + 15 + 20 + 25 + 30)/ 6$$

By further calculation

$$= 105/6$$

$$= 17.5$$



Find the mean of first six natural numbers.

$${\textbf{Step -1: Write first six natural numbers.}}$$

               $${\text{First six natural numbers are}}$$ $$1,2,3,4,5,6$$

               $${\text{Total number }}$$ $$ = 6$$

$${\textbf{Step -2: Substitute the known values in arithmetic mean formula to find mean.}}$$

               $${\text{As we know that, formula for finding arithmetic mean of n numbers is given by,}}$$

               $${\text{Mean  =  }}\dfrac{{{\text{sum of all observations}}}}{{{\text{Total observations}}}}$$

               $$ \Rightarrow {\text{Mean  =  }}\dfrac{{1 + 2 + 3 + 4 + 5 + 6}}{6}$$

               $$ \Rightarrow {\text{Mean  =  }}\dfrac{{21}}{6}$$  

               $$ \Rightarrow {\text{Mean  =  }}\dfrac{7}{2}$$

               $$ \Rightarrow {\text{Mean  =  3}}{\text{.5}}$$

$${\textbf{Hence, mean of first six natural number is 3.5.}}$$



Find the mean of first six multiples of $$3$$.

We know that the first six multiples of $$3$$ are $$3, 6, 9, 12, 15, 18$$

Here $$n = 6$$

So $$mean = (3 + 6 + 9 + 12 + 15 + 18)/ 6$$

By further calculation

$$= 63/ 6$$

$$= 10.5$$



Find the mean of $$53, 61, 60, 67$$ and $$64$$.

The given numbers are $$53, 61, 60, 67$$ and $$64$$

We know that n$$ = 5$$

So Mean $$= (53 + 61 + 60 + 67 + 64)/ 5$$

By further calculation

$$= 305/5$$

$$= 61$$


Find the mean of the following set of numbers: $$6, 9, 11, 12$$ and $$7$$

By definition, we know
Mean $$= \sum x/ n$$
Here, $$n = 5$$
Thus,
Mean $$= (6 + 9 + 11 + 12 + 7)/ 5 = 45/5 = 9$$


If different values of variable x are $$19.8, 15.4, 13.7, 11.71, 11.8, 12.6, 12.8, 18.6, 20.5\ and\ 21.1$$, find the mean.

The given values are $$19.8, 15.4, 13.7, 11.71, 11.8, 12.6, 12.8, 18.6, 20.5$$  and $$21.1$$

We know that n $$= 10$$

So Mean $$= (19.8 + 15.4 + 13.7 + 11.71 + 11.8 + 12.6 + 12.8 + 18.6 + 20.5 + 21.1)/ 10$$

By further calculation

$$= 158.01/10$$

$$= 15.801$$


Find the mean of squares of first five whole numbers.

Here the first five whole numbers are $$0, 1, 2, 3, 4$$

So the square of whole numbers $$= 0^2, 1^2, 2^2, 3^2, 4^2 = 0, 1, 4, 9, 16$$

We know that $$n = 5$$

So $$Mean = (0 + 1 + 4 + 9 + 16)/ 5$$

By further calculation

$$= 30/5$$

$$= 6$$



The pocket expenses (per day) of Anuj, during a certain week, from Monday to Saturday were $$85.40, 88.00, 86.50, 84.75, 82.60$$ and $$87.25$$. Find the mean pocket expenses per day.

It is given that the pocket expenses (per day) during a certain week are

$$₹85.40, ₹88.00, ₹86.50, ₹84.75, ₹82.60$$ and $$₹87.25$$

We know that $$n = 6$$

So the mean pocket expenses per day $$= (85.40 + 88.00 + 86.50 + 84.75 + 82.60 + 87.25)/ 6$$

By further calculation

$$= 514.5/6$$

$$= ₹85.75$$

Hence, the mean pocket expenses per day are $$₹85.75$$.



Find the mean of the following set of numbers: $$11, 14, 23, 26, 10, 12, 18$$ and $$6$$

$$11, 14, 23, 26, 10, 12, 18$$ and $$6$$
By definition, we know
Mean $$= \sum x/ n$$
Here, $$n = 8$$
Thus,
Mean $$= (11 + 14 + 23 + 26 + 10 + 12 + 18 + 6)/ 8 = 120/8 = 15$$


Marks obtained (in mathematics) by $$9$$ student are given below:
$$60, 67, 52, 76, 50, 51, 74, 45$$ and $$56$$
(a) find the arithmetic mean
(b) if marks of each student be increased by $$4;$$ what will be the new value of arithmetic mean.

$$(a)$$ Mean $$= \sum x/ n$$
Here, $$n = 9$$
Thus,
Mean $$= (60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56)/ 9 = 531/9 = 59$$
$$(b)$$ If the marks of each student be increased by 4 then new arithmetic mean will be $$= 59 + 4 = 63$$


Find the mean of the natural numbers from $$3$$ to $$12.$$

The numbers between $$3$$ to $$12$$ are $$3, 4, 5, 6, 7, 8, 9, 10, 11$$ and $$12.$$
Here $$n = 10$$
Mean $$= \sum x/ n$$
$$= (3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)/ 10 = 75/10 = 7.5$$


$$(a)$$Find the mean of $$7, 11, 6, 5,$$ and $$6$$
$$(b)$$ If each number given in $$(a)$$ is diminished by $$2,$$ find the new value of mean.

$$(a)$$ Mean $$= \sum x/ n$$
here $$n = 5$$
$$= (7 + 11 + 6 + 5 + 6)/ 5 = 35/5 = 7$$
$$(b)$$ If $$2$$ is subtracted from each number, then the mean will he changed as $$7 – 2 = 5$$


The following table gives the ages of $$50$$ students of a class. Find the arithmetic mean of their ages.
1834232_a8cf13b40b694a0b8889c15cb47a1906.PNG

Mean $$= \dfrac {\sum fi xi}{\sum fi} =\dfrac{1074}{50} = 21.48$$
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The following table shows the income of farmers in a grape season. Find the mean of their income.
Income
(Thousand Rupees)		20 - 3030 - 4040 - 5050 - 6060 - 7070 - 80
Farmers101115161814

Class
(Income in thousand rupees) Class Mark
$$x_{i}$$ Frequency
(No. of farmers
$$f_{i}$$ Class Mark * Frequency
$$x_{i} f_{i}$$
20 - 30               25           10             250
30 - 40               35           11             385
40 - 50               45           15             675
50 - 60                55          16             880
60 - 70                65          18             1170
70 - 80                 75          14             1050
$$\sum f_{i} = 84$$ $$\sum x_{i} f_{i} = 4410$$

Mean - $$\dfrac{\sum x_{i} f_{i}}{\sum f_{i}}$$
$$= \dfrac{4410}{84}$$
= 52.5 thousand rupees
=52500


The loans sanctioned by a bank for construction of farm ponds are shown in the following table. Find the mean of the loans.
Loan
(Thousand Rupees)		40 - 5050 - 6060 - 7070 - 8080 - 90
No. of farm ponds132024
36
7

Mean $$= \dfrac{\sum x_{i} f_{i}}{\sum f_{i}} $$
$$= \dfrac{6540}{100}$$
$$= 65 . 4$$ thousand rupees
= 65400
Hence, the means of the loans is Rs 65400.
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The following frequency distribution table shows the amount of od aid given to 50 flood-affected families. Find the mean of the amount of aid.
Amount of Aid
(thousand rupee)		50-6060-7070-8080-9090-100
No. of families7132064


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The weekly wages of 120 workers in a factory are shown in the following frequency distribution table. Find the mean of the weekly wages.
Weekly wages
(Rupees)		0 - 20002000 - 40004000 - 6000 6000 - 8000
No. of workers15355020


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Find the mean weight from the following table :
Weight (kg)No.of children
2902
3001
3104
3203
3305

  
Weight (kg)     No.of children    'f'
29    02   58
30    01   30
31    04   124
32    03   96
33    05  164
   N = 15$$\Sigma fx  =  473$$
Mean  $$  = \dfrac{\Sigma f_{x}}{N} = \dfrac{473}{15} =  31. 53 $$


Calculate the mean for the following frequency 
Mark10 - 2020 - 30 30 - 4040 - 5050 -6060-  70
Frequency3710682

Mean  = $$  = \dfrac{\Sigma f_{x}}{N} = \dfrac{1710}{40} =  42 . 75$$
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Runs scored by 10 batsman in a one day cricket match are given . Find the average run scored.
23 , 54, 08 ,94 , 60 ,18 , 29 , 44 , 05 , 86.

$$ \Sigma  x =  23 +  54 + 08 + 94 +  60 + 18 +  29 + 44  + 05 +  86  = 421 $$
$$ N = 10 $$
Average = Mean $$ = \overline{X} = \dfrac{\Sigma  x}{N} = \dfrac{421}{10} = 42 . 1 $$


x10151720223035
f51028366

Arithmetic mean $$(\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} = \frac{850}{40} = 21.25$$
Thus, arithmetic mean = 21.25
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x19212325272931
f13151618161513

Arithmetic mean $$(\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} = \frac{2650}{106} = 25$$
Thus, arithmetic mean = 25
1864544_1878101_ans_bdac5acffb5e4b20aa02fd966a5aad22.PNG


x123456
f452519821

Arithmetic mean $$(\bar{x}) = \frac{\sum f_i x_i}{\sum f_i} = \frac{200}{100} = 2$$
Thus, arithmetic mean = 2
1864547_1878103_ans_74f5fe4eb4814c1082720cb139e3b116.PNG


Consider the following distribution of daily wages of 40 workers of a factory and by using suitable method, find mean daily wages of the workers of this factory:
Daily wages (in Rs.)25-3535-4545-5555-6565-75
No. of workers6108124

Arithmetic mean $$ = \frac{\sum f_{i}x_{i}}{\sum f_{i}} = \frac{1980}{40} = 49.50 $$
Thus, mean of daily wages = rs. 49.50 
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Find mean of the following distribution :
Class1-33-55-77-10
Frequency9222717

Thus, arithmetic mean $$ (\bar{x}) = \frac{\sum f_{i}x_{i}}{\sum f_{i}} = \frac{412.5}{75} = 5.5 $$
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To find concentration (part per million) of $$ (SO-{2}) $$ Sulphur dioxide in air data is collected from 30 colonies of a city which are given below :
Concentration
of $$ SO_{2} $$ (ppm)		0.00-0.040.04-0.080.08-0.120.12-0.160.16-0.200.20-0.24
Frequency499242
Find mean of concentration of $$ (SO_{2}) $$ in Air.

Arithmetic mean $$(\bar{x}) = \frac{\sum f_{i}x_{i}}{\sum f_{i}} = \frac{2.96}{30} = 0.0986 or 0.099(approx) $$
Thus, mean of concentration of $$ SO_{2} $$ in Air = 0.099 part per million.
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Weight of 30 students of a class at the time of medical test is recorded as per table given below :
Weight (in kg)40-4545-5050-5555-6060-6565-7070-75
No. of students2386632
Find mean weight of students.

Let assumed mean (A) = 57.5

Arithmetic mean $$ (\bar{x}) = A + \frac{\sum f_{i}x_{i}}{\sum f_{i}} = 57.5 + \frac{-10}{30} = 57.167 = 57.17 kg $$
Thus, mean weight of students = 57.17 kg.
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If arithmetic mean of following data is 15 then find p.
Variable $$ x_{i} $$510152025
Frequency $$ f_{i} $$6P6105

Table of calculating A.M.

Thus,$$ A.M. (\bar{X}) = \frac{\sum fx}{\sum f} $$
$$ 15 = \frac{695 + 10P}{27 + P} $$
$$ \Rightarrow (27 + P)15 = 695 + 10P $$
$$ \Rightarrow 405 + 15P = 695 + 10P $$
$$ \Rightarrow 15P - 10P = 695 + 405 $$
$$ \Rightarrow 5P = 290 $$
$$ \Rightarrow P = \frac{290}{5} = 58 $$
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Find the mean salary of $$60$$ workers of a factory from the following table :


1880154_35681b0e4e92475bbc1c469ba04cef56.PNG

Here, we will find mean salary of $$60$$ workers of a factory by direct method.
Mean $$\bar{(x)} = \dfrac{\sum f_ix_i}{\sum f_i} = \dfrac{305000}{60} = 5083.33$$
Hence, mean salary of $$60$$ workers of a factory is $$Rs 5083.33$$
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Find the average of five numbers 40, 35, 30, 50, 45.

Average = Sum of 5 numbers/Total numbers = (40 + 35 + 30 + 50 + 45)/5 = 200/5 = 40.


If the mean of 1,2,3,.......x is $$\dfrac { 6x }{ 11 } $$, find the value of x


1293616_1537566_ans_77e0c694437a486baec39e62fc9b058b.jpg


A die was thrown $$20$$ times and the following outcomes were recorded :-
     $$5, 2, 1, 3, 4, 4, 5, 6, 2, 2, 4, 5, 5, 6, 2, 2, 4, 5, 5, 1$$
Find the mean and mode of the above data

 Outcome frequency
 $$1$$
$$2$$
$$3$$
$$4$$
$$5$$
$$6$$		$$ 2$$
$$5$$
$$1$$
$$4$$
$$6$$
$$2$$
 $$ 20$$
we know mode means which observation most frequency occured
Here observe that $$5$$  is most frequently occured
So mode is 5

we know 
$$mean = \dfrac{sum \ of \ out \ comes}{No \ of \ out \ comes}$$

$$= \dfrac{5 + 2 + 1+ 3+4+4+5+6+2+2+4+5+5+6+2+2+4+5+5+1}{20}$$

$$= \dfrac{73}{20}$$

$$= 3.65$$

$$mean = 3.65$$


Coefficient of variation of two distributions are 60% and 80% respectively, and their standard deviations are 21 and 16 respectively. Find their arithmetic means.


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Class 7 Maths Extra Questions