Exercise 9.2

1.Weights of parcels in a transport office are given below.

Weight  (kg)	50	65	75	90	110	120
No.of parcels	25	34	38	40	47	16

Find the mean weight of the parcels.

Sol.

Weight in kg $$x_{i}$$	No.of parcels $$ f_{i}$$	$$f_{i} x_{i}$$
50 65 75 90 110 120	25 34 38 40 47 16	1250 2210 2850 3600 5170 1920

                                                                          $$\sum f=200$$                $$\sum f_{i}x_{i}=17000$$

$$\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}$$ = $$\frac{17000}{200}=\frac{170}{2}$$

$$\bar{x}=85$$

mean = 85

2.Number of families in a village in correspondence with the number of children are given below.

No.of children	0	1	2	3	4	5
No.of families	11	25	32	10	5	1

Find the mean number of children per family.

A.

No.of children $$x_{i}$$	No.of families f_{i}$$	$$f_{i} x_{i}$$
0 1 2 3 4 5	11 25 32 10 5 1	0 25 64 30 20 5

                                                                        $$\sum f_{i}=84$$                 $$\sum f_{i}x_{i}=144$$

$$\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{144}{84}$$

Mean=1.714285

3.If the mean of the following frequency distribution is 7.2 find value of ‘K’.

x	2	4	6	8	10	12
f	4	7	10	16	k	3

Sol.

x	2	4	6	8	10	12
f	4	7	10	16	k	3	40+k
$$\sum f_{i}x_{i}	8	28	60	128	10k	36	260+10k

$$\sum f_{i}=40+k$$;   $$\sum f_{i}x_{i}$$ =260+10k

Given that $$\bar{x}=7.2$$

But $$\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}$$

7.2= $$\frac{260+10k}{40+k}$$

288.0+7.2k=260+10k

10k-7.2k=288-260

2.8k=28

k= $$\frac{28}{2.8}$$=10

4.Number of villages with respect to their population as per India census 2011 are given below.

Population (in thousands)	12	5	30	20	15	8
villages	20	15	32	35	36	7

 

Find the average population in each village.

Sol.

Population(in thousands) $$x_{i}$$	Villages $$f_{i}$$	$$f_{i}x_{i}$$
12 5 30 20 15 8	20 15 32 35 36 7	240 75 960 700 540 56

                                                                   $$\sum f_{i}=145 $$   

$$\sum f_{i}x_{i}=2571$$ thousands

$$\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}$$

Mean = $$\frac{2571}{145}$$=17.731 thousands 

5.AFLATOUN social and financial educational program initiated savings program among the high school children in Hyderabad district.  Mandal wise savings in a month are given in the following table.

Mandal	No.of schools	Total amount saved (in rupees)
Amberpet	6	2154
Thirumalgiri	6	2478
Saidabad	5	975
Khairathabad	4	912
Secunderabad	3	600
Bahadurpura	9	7533

Find arithmetic mean of school wise savings in each mandal. Also find the arithmetic mean of saving of all schools.

Sol.

Mandal	No.of schools $$f_{i}$$	Total amount saved (in rupees) $$f_{i}x_{i}$$	School average in rupees =$$\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}$$
Amberpet	6	2154	$$\frac{2154}{6}=359$$
Thirumalgiri	6	2478	$$\frac{2478}{6}=413$$
Saidabad	5	975	$$ \frac{975}{5}=195$$
Khairathabad	4	912	$$\frac{912}{4}=228$$
Secunderabad	3	600	$$\frac{600}{3}=200$$
Bahadurpura	9	7533	$$ \frac{7533}{6}=837$$

                             $$\sum f_{i}$$=33     $$\sum f_{i}x_{i}$$=14652

Mean=$$\frac{\sum f_{i}x_{i}}{\sum f_{i}}$$

$$\bar{x}=\frac{14652}{33}$$=Rs/-444(mean savings per school)

6.The heights of boys and girls of IX class of a school are given below.

Height (cm)	135	140	147	152	155	160
Boys	2	5	12	10	7	1
Girls	1	2	10	5	6	5

Compare the heights of the boys and girls [Hint : Find median heights of boys and girls]

Sol.

Height(cm)	Boys (f)	c.f.	Girls (f)	c.f.
135	2	2	1	1
140	5	7	2	3
147	12	19 $$\leftarrow$$	10	13
152	10	29	5	18 $$\leftarrow$$
155	7	36	6	24
160	1	37	5	29
		N=37	N=29

 

Boys median class = $$\frac{37+1}{2}=\frac{38}{2}$$ =19th observation

$$\therefore$$ median height of boys =147cm

Girls median class = $$\frac{29+1}{2}=\frac{30}{2}$$ = 15th observation

$$\therefore$$ median height of girls =152cm

7.Centuries scored and number of cricketers in the world are given below.

No.of centuries	5	10	15	20	25
No.of cricketers	56	23	39	13	8

Find the mean, median and mode of the given data.

Sol.

No.of centuries $$\sum x_{i}=14652$$	No.of cricketers $$f_{i}$$	c.f.	$$f_{i} x_{i}$$
5 10 15 20 25	56$$ \leftarrow$$ mode 23 $$\leftarrow$$ median 39 13 8	56 79 118 131 139	280 230 585 260 200

                                                   $$N=\sum f_{i}=139$$                                $$\sum f_{i}x_{i}=1555$$

Mean $$\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{1555}{139}$$=11.187

Median=$$\left ( \frac{N}{2}+1 \right )^{th} term =\frac{139+1}{2}=\frac{140}{2}=70^{th}$$ term=10

mode =5

8.On the occasion of New year’s day a sweet stall prepared sweet packets. Number of sweet packets and cost of each packet are given as follows.

Cost of packet (inRs/-)	Rs/25	Rs/50	Rs/75	Rs/100	Rs/125	Rs/150
No.of packets	20	36	32	29	22	11

Find the mean, median and mode of the data.

Sol.

Cost of packet (inRs/)$$x_{i}$$	No.of packets $$f_{i}$$	c.f.	$$f_{i} x_{i}$$
25 50 75 100 125 150	20mode 36 $$\leftarrow$$ 32 29 22 11	20 56 88$$\leftarrow$$ median 117 139 150	500 1800 2400 2900 2750 1650

                                           $$N=\sum f_{i}=150$$                                       $$\sum f_{i}x_{i}=12000$$

Mean =$$\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{12000}{150}$$ =80

Median=average of  $$\left ( \frac{N}{2}+1 and \frac{N}{2} \right )$$terms = average of 75 and 76 observations =75

Mode=50

9.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 Rahim’s weight.

Sol.Weight of Ranga=46kg

Weight of Reshma=weight of Rahim=xkg say

Average = $$\frac{sum of the weights}{number}$$=40kg

$$\therefore 40$$ = $$\frac{46+x+x}{3}$$

3X40=46+2x

2x=120-46=74

$$\therefore x=\frac{74}{2}$$=37

$$\therefore$$ Rahim's weight =37 kg.

10.The donations given to an orphanage home by the students of different classes of a secondary school are given below.

class	Donation by each student in(Rs)	No.of students donated
VI	5	15
VII	7	15
VIII	10	20
IX	15	16
X	20	14

Find the mean, median and mode of the data.

Sol.

Donation (inRs/)$$x_{i}$$	No.of students  $$f_{i}$$	c.f.	$$f_{i} x_{i}$$
5 7 10 15 20	15 15 20 $$\leftarrow$$ 16 14	15 30 50 $$\leftarrow$$ median 66 80	75 105 200 240 280

                                    $$\sum f^{i}=N=80                                                         \sum f^{i}x_{i}=900$$

$$mean \bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{900}{80}=11.25$$

$$median=Average of \left ( \frac{N}{2} \right )and \left ( \frac{N}{2}+1 \right )terms $$

=$$average of \frac{80}{2},\left ( \frac{80}{2}+1 \right )$$

=$$average of 40 and 41 terms =Rs/10$$

Mode=Rs/10

11.There are four unknown numbers. The mean of the first two numbers is 4 and the mean of the first three is 9. The mean of all four number is 15, if one of the four number is 2 find the other numbers.

Sol.We know that mean = $$\frac{sum}{number}$$

Given that

Mean of 4 numbers = 15

$$\Rightarrow$$ Sum of the 4 numbers =4 X 15 = 60

Mean of the first 3 numbers =9

$$\Rightarrow$$ sum of the first 3 numbers =3 X 9 = 27

Mean of the first 2 number = 4

$$\Rightarrow$$ sum of the first 2 numbers =2 X 4 = 8

Fourth number = sum of 4 numbers - sum of 3 numbers = 60-27= 33

Third number =sum of 3 numbers - sum of 2 numbers = 27-8=19

Second number = sum of 2 numbers - given number =8-2=6

$$\therefore$$ The other three numbers are 6,19,33

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