Tr Jain & Vk Ohri (2017) for Class 11 Humanities Economics Chapter 9 - Measures Of Central Tendency Arithematic Mean

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Page No 148:

Question 1:

Following is the monthly income of eight families in a locality
Monthly Income (₹) : 70, 10, 500, 75, 13, 250, 8, 42
Find out arithmetic mean using Direct Method and Short-cut Method.

Answer:

Families	Monthly income (X)	Deviation from the Assumed average (d = X − A) (A = 75)
1 2 3 4 5 6 7 8	70 10 500 75 (A) 13 250 8 42	−5 −65 425 0 −62 175 −67 −33
(N) = 8	(ΣX)=968	Σd = −232 + 600 = 368

where,
N= Number of items

Σ X

= Total of monthly income of all the families

Σ d

= net sum of deviations

Calculation of Mean by Direct Method

X = \frac{Σ X}{N} = \frac{968}{8} = 121

Calculation of mean by short cut method

X = A + \frac{Σ d}{N} or, X = 75 + \frac{368}{8} or, X = 75 + 46 \Rightarrow X = 121

So, mean monthly income of a family is Rs 121.

Page No 148:

Question 2:

Students of Class XI secured following marks in their Statistics paper.
Calculate arithmetic mean using Direct Method and Short-cut Method.
20, 44, 65, 28, 45, 67, 30, 50, 68, 39, 53, 70, 40, 60, 75

Answer:

Sr. No.	Marks (X)	Deviation from the Assumed mean d = X − A where A = 50
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.	20 44 65 28 45 67 30 50 (A) 68 39 53 70 40 60 75	-30 -06 15 -22 -05 17 -20 0 18 -11 3 20 -10 10 25
N = 15	ΣX = 754	Σd = -104 + 108 = 4

where,
N= Number of items

Σ X

= Total of monthly income of all the families

Σ d

= net sum of deviations

Calculation of Mean by direct method

X = \frac{Σ X}{N} or, \bar{X} = \frac{754}{15} or, \bar{X} = 50.27 \Rightarrow \bar{X} = 50.27

Calculation of mean by short cut method

X = A + \frac{Σ d}{N} or, X = 50 + \frac{4}{15} or, X = 50 + 0.27 = 50.27 \Rightarrow X = 50.27

So, mean marks in statistics is 50.27.

Page No 148:

Question 3:

Find out mean height of girl-students in your school. Compare it with the mean height of boy-students.

Answer:

Sr. No.	Height of girls (in cm) (X)	Height of Boys (in cm) (Y)
1. 2. 3. 4. 5. 6. 7. 8. 9. 10.	152 153.2 151.4 152 156 154 152.7 158 152 153.5	170 172.5 168.2 175 172 170.3 176.5 178 180 172
N = 10	ΣX = 1534.8	ΣY = 1734.5

Mean height of girls (X) = \frac{Σ X}{N} = \frac{1534.8}{10} \Rightarrow \bar{X} = 153.48 cm Mean height of boyx (Y) = \frac{Σ Y}{N} = \frac{1734.5}{10} \Rightarrow \bar{Y} = 173.45 cm

Boys' mean height is more than girls's mean height.

Page No 152:

Question 1:

Calculate arithmetic mean of the following series.
Take 15 as assumed average and use Short-cut Method.

Size	20	19	18	17	16	15	14	13	12	11
Frequency	1	2	4	8	11	10	7	4	2	1

Answer:

Size (X)	Frequency (f)	Deviation d = X − A (A = 15)	Multiple of Deviation and Frequency (fd)
20 19 18 17 16 15 14 13 12 11	1 2 4 8 11 10 7 4 2 1	5 4 3 2 1 0 −1 −2 −3 −4	5 8 12 16 11 0 −7 −8 −6 −4
	Σf = 50		Σfd = 52 − 25 = 27

Mean size (X) = A + \frac{Σ f d}{Σ f} or, \bar{X} = 15 + \frac{27}{50} or, \bar{X} = 15 + 0.54 \Rightarrow \bar{X} = 15.54

So the average size is 15.54.

Page No 152:

Question 2:

Students of your class obtained following marks in Statistics in the weekly test:

Marks	4	5	6	7	8
Number of Students	12	10	18	20	24

Calculate arithmetic mean using Direct and Short-cut Methods.

Answer:

Marks (X)	Frequency (f)	fX	Deviation d = X − A (A = 6)	Multiplication of deviation and frequency (fd)
4 5 6(A) 7 8	12 10 18 20 24	48 50 108 140 192	−2 −1 0 1 2	−24 −10 0 20 48
	Σf = 84	ΣfX = 538		Σfd = −34 + 68 = 34

Calculation of mean by Direct method

X = \frac{Σ f X}{Σ f} or, \bar{X} = \frac{538}{84} \Rightarrow \bar{X} = 6.40

Calculation of mean by short cut method

X = A + \frac{Σ f d}{Σ f} or, \bar{X} = 6 + \frac{34}{84} or, \bar{X} = 6 + 0.40 \Rightarrow \bar{X} = 6.40

So, average statistics marks is 6.40.

Page No 152:

Question 3:

The following table gives height of certain persons:

Height (inches)	57	59	61	63	65	67	69	71	73
Number of Persons	1	3	7	10	11	15	10	7	2

Calculate arithmetic mean.

Answer:

Heights (inches) (X)	No. of Persons (f)	Multiplication of the value of X and frequency (fX)
57 59 61 63 65 67 69 71 73	1 3 7 10 11 15 10 7 2	57 177 427 630 715 1005 690 497 146
	Σf = 66	Σfx = 4344

Arithmetic mean will be:

X = \frac{Σ f x}{Σ f} or, \bar{X} = \frac{4344}{66} \Rightarrow \bar{X} = 65.82

So, the average height is 65.82 inches.

Page No 152:

Question 4:

Following is the number of heads of 8 coins when tossed 205 times:

Number of Heads	0	1	2	3	4	5	6	7	8
Frequency	2	19	46	62	47	20	5	2	2

Calculate arithmetic mean (mean number of heads per toss), using Direct Method and Short-cut Method.

Answer:

No. of Heads (X)	Frequency (f)	fX	Deviation d = X − A (A = 4)	Multiplication of deviation and frequency (fd)
0 1 2 3 4 (A) 5 6 7 8	2 19 46 62 47 20 5 2 2	0 19 92 186 188 100 30 14 16	−4 −3 −2 −1 0 1 2 3 4	−8 −57 −92 −62 0 20 10 6 8
	Σf = 205	Σfx = 645		Σfd = −219 + 44 = −175

Calculation of mean by direct method.

X = \frac{Σ f X}{Σ f} or, \bar{X} = \frac{645}{205} or, \bar{X} = 3.146 \Rightarrow \bar{X} = 3.15 a p p r o x

Calculation mean by short cut method

X = A + \frac{Σ f d}{Σ f} or, \bar{X} = 4 + \frac{- 175}{205} or, \bar{X} = 4 - 0.853 or, \bar{X} = 3.147 \Rightarrow \bar{X} = 3.15 a p p r o x

So, mean number of heads per toss is 3.15.

Page No 155:

Question 1:

Find out the mean of the following distribution:

Items	0−5	5−10	10−15	15−20	20−25	25−30	30−35	35−40
Frequency	2	5	7	13	21	16	8	3

Answer:

Items	Mid value $(m = \frac{l_{1} + l_{2}}{2})$	Frequency (f)	Multiplication of mid value and frequency (fm)
0 − 5	$\frac{0 + 5}{2} = 2.5$	2	5
5 − 10	$\frac{5 + 10}{2} = 7.5$	5	37.5
10 − 15	$\frac{10 + 15}{2} = 12.5$	7	87.5
15 − 20	$\frac{15 + 20}{2} = 17.5$	13	227.5
20 − 25	$\frac{20 + 25}{2} = 22.5$	21	472.5
25 − 30	$\frac{25 + 30}{2} = 27.5$	16	440
30 − 35	$\frac{30 + 35}{3} = 32.5$	8	260
35 − 40	$\frac{35 + 40}{2} = 37.5$	3	112.5
		Σf = 75	Σfm = 1642.5

X = \frac{Σ f m}{Σ f} or, \bar{X} = \frac{1642.5}{75} \Rightarrow \bar{X} = 21.9

Hence, the mean of the above distribution is 21.9

Page No 155:

Question 2:

Find arithmetic mean from the following distribution:

Marks	0−4	4−8	8−12	12−16
Number of Students	4	8	2	1

Answer:

Marks	Mid value (m) $(m = \frac{l_{1} + l_{2}}{2})$	No. of student (f)	Multiplication of mid-value and frequency (fm)
0 − 4	$\frac{0 + 4}{2} = 2$	4	8
4 − 8	$\frac{4 + 8}{2} = 6$	8	48
8 − 12	$\frac{8 + 12}{2} = 10$	2	20
12 − 16	$\frac{12 + 16}{2} = 14$	1	14
		Σf = 15	Σfm = 90

$X = \frac{Σ f m}{Σ f} or, \bar{X} = \frac{90}{15} \Rightarrow \bar{X} = 6$

Hence, the mean marks of the above distribution is 6.

Page No 160:

Question 1:

A polling agency interviewed 200 persons. The age distribution of those persons was recorded as under:

Age (Years)	80−89	70−79	60−69	50−59	40−49	30−39	20−29	10−19
Frequency	2	2	6	20	56	40	42	32

Calculate mean age of the persons by Direct Method.

Answer:

Age	Mid Value (m) $(m = \frac{l_{1} + l_{2}}{2})$	Frequency (f)	fm
10 − 19	$\frac{10 + 19}{2} = 14.5$	32	464
20 − 29	$\frac{20 + 29}{2} = 24.5$	42	1029
30 − 39	$\frac{30 + 39}{2} = 34.5$	40	1380
40 − 49	$\frac{40 + 49}{2} = 44.5$	56	2492
50 − 59	$\frac{50 + 59}{2} = 54.5$	20	1090
60 − 69	$\frac{60 + 69}{2} = 64.5$	6	387
70 − 79	$\frac{70 + 79}{2} = 74.5$	2	149
80 − 89	$\frac{80 + 89}{2} = 84.5$	2	169
		Σf = 200	Σfm = 7160

Calculating the mean by Direct method:

X = \frac{Σ f m}{Σ f} or, \bar{X} = \frac{7160}{200} \Rightarrow \bar{X} = 35.8

Hence,mean age of the persons is 35.8 years.

Page No 160:

Question 2:

Calculate mean of the following series by Direct Method:

Mid-value	10	12	14	16	18	20
Frequency	5	8	12	20	10	5

Answer:

Mid Value (m)	Frequency (f)	fm
10 12 14 16 18 20	5 8 12 20 10 5	50 96 168 320 180 100
	Σf = 60	Σfm = 914

Calculating the mean by Direct method:

X = \frac{Σ f m}{Σ f} or, \bar{X} = \frac{914}{60} \Rightarrow \bar{X} = 15.23

Hence, the mean of the above series is 15.23.

Page No 160:

Question 3:

In an examination in Hindi, the students of a class secured following marks. Calculate mean marks by Direct Method.

Marks	Number of Students
More than 0 More than 10 More than 20 More than 30 More than 40 More than 50	50 46 40 20 10 5

Answer:

Converting cumulative frequency Distribution into a simple frequency distribution

Marks	No. of students
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60	50 − 46 = 4 46 − 40 = 6 40 − 20 = 20 20 − 10 = 10 10 − 3 = 7 3 − 0 = 3

Marks	Mid Value (m)	No. of students (f)	Multiple of mid value and frequency (fm)
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60	5 15 25 35 45 55	4 6 20 10 7 3	20 90 500 350 315 165
		Σf = 50	Σfm = 1440

Calculating the mean by Direct method:

X = \frac{Σ f m}{Σ f} or, \bar{X} = \frac{1440}{50} \Rightarrow \bar{X} = 28.8

Hence, the mean marks in Hindi is 28.8

Page No 161:

Question 1:

The following table shows prices per 100g of tea of different brands. Using quantities as weight, find out weighted arithmetic mean of the prices.

Prices per 100g in ₹	2.50	3	3.50	4	4.25	5
Quantity	10	8	8	4	4	2

Answer:

Price (X)	Quantity (W)	WX
2.50 3 3.50 4 4.25 5	10 8 8 4 4 2	25 24 28 16 17 10
	∑W = 36	∑WX=120

Calculating the weighted arithmetic mean of the prices:

X_{W} = \frac{Σ W X}{Σ W} or, {\bar{X}}_{w} = \frac{120}{36} \Rightarrow {\bar{X}}_{w} = 3.33

Hence, the weighted arithmetic mean of the prices is 3.33.

Page No 161:

Question 2:

Calculate weighted mean of the following data:

Items (X)	5	10	15	20	25	30
Weight (W)	8	4	5	10	7	6

Answer:

Item (X)	Weight (W)	WX
5 10 25 20 25 30	8 4 5 10 7 6	40 40 125 200 175 180
	ΣW = 40	ΣWX = 760

Calculating the weighted mean:

X_{W} = \frac{Σ W X}{Σ W} or, {\bar{X}}_{w} = \frac{760}{40} \Rightarrow {\bar{X}}_{w} = 19

Hence, the weighted mean of the above data is 19

Page No 164:

Question 1:

The arithmetic mean of the following series is 18, find out the missing frequency.

Class Interval	11−13	13−15	15−17	17−19	19−21	21−23	23−25
Frequency	3	6	9	13	?	5	4

Answer:

Class interval	Mid value (m)	Frequency (f)	Multiple of mid value and frequency (fm)
11 − 13 13 − 15 15 − 17 17 − 19 19 − 21 21 − 23 23 − 25	12 14 16 18 20 22 24	3 6 9 13 f 5 4	36 84 144 234 20f 110 96
		Σf = 40 + f	Σfm = 704 + 20f

X = \frac{Σ f m}{Σ f} or, 18 = \frac{704 + 20 f}{40 + f} or, 18 (40 + f) = 704 + 20 f or, 720 + 18f = 704 + 20f or, 720 - 704 = 20f - 18f or, 16 = 2f or, 2f = 16 or, f = \frac{16}{2} \Rightarrow f = 8

Hence, the missing frequency is 8.

Page No 164:

Question 2:

The following series gives the weekly wages. The mean wages are ₹ 50.75. Find out the missing value.

Weekly Wages	40−43	43−46	46−49	49−52	52−55
Frequency	3	6	9	13	?

Answer:

Weekly wages	Mid value (m)	Frequency (f)	Multiple of mid value and frequency (fm)
40 − 43 43 − 46 46 − 49 49 − 52 52 − 55	41.5 44.5 47.5 50.5 53.5	3 6 9 13 f	124.5 267 427.5 656.5 53.5f
		Σf = 31 + f	Σfm = 1475.5 + 53.5f

X = \frac{Σ f m}{Σ f} or, 50.75 = \frac{147.5 + 53.5 f}{31 + f} or, 50.75(31 + f) = 1475.5 + 53.5 f or, 1573.25 + 50.75 f = 1475.5 + 53.5 f or, 1573.25 − 1475.5 = 53.5 f - 50.75 f or, 97.75 = 2 .75 f or, 2.75 f = 97.75 or, f = \frac{97.75}{2.75} \Rightarrow f = 35.54 or 36 a p p r o x .

Hence, the missing frequency or value is 36.

Page No 164:

Question 3:

Mark the missing value of the following data. Mean marks are 11.66.

Weekly Wages	5	6	7	8	12	?	15	17	18
Frequency	3	10	5	10	12	15	11	13	1

Answer:

Marks (x)	Frequency (f)	fx
5 6 7 8 12 x 15 17 18	3 10 5 10 12 15 11 13 1	15 60 35 80 144 15x 165 221 18
	Σf = 80	Σfx = 738 + 15x

X = \frac{Σ f x}{Σ f} or, 11.66 = \frac{738 + 15 x}{80} or, 932.8 = 738 + 15 x or, 932.8 - 738 = 15 x or, 194.8 = 15 x or, x = \frac{194.8}{15} \Rightarrow x = 12.99 or 13 (approximately)

So, the missing weekly wage is Rs 13.

Page No 176:

Question 1:

Eight workers earn the following income:
30, 36, 34, 40, 42, 46, 54, 62
Find out arithmetic mean.

Answer:

$Arithmetic Mean = X = \frac{Σ X}{N} or, X = \frac{30 + 36 + 34 + 40 + 42 + 46 + 54 + 62}{8} or, X = \frac{344}{8} = 43 Thus, the mean income of the workers is Rs 43 .$

Page No 176:

Question 2:

Pocket allowance of 5 students respectively are:
125, 75, 150, 175, 200
Find out arithmetic mean.

Answer:

$Arithmetic Mean = X = \frac{Σ X}{N} or, X = \frac{125 + 75 + 150 + 175 + 200}{5} or, X = \frac{725}{5} = 145 Thus, the mean pocket allowance is Rs 145 .$

Page No 176:

Question 3:

Following is the height of 10 students:

Students	A	B	C	D	E	F	G	H	I	J
Height (cm)	155	153	168	160	162	166	164	180	157	165

Calculate arithmetic mean using DIrect and Short-cut Methods.

Answer:

Students	Height (X)	X − A (d)
A B C D E F G H I J	155 153 168 160 162 = A 166 164 180 157 165	−7 −9 6 −2 0 4 2 18 −5 3
	∑ X = 1630	∑d = 10

Direct Method:

A . M . = X = \frac{Σ X}{N} = \frac{1630}{10} = 163

Shortcut Method:

A . M . = X = A + \frac{Σ d}{N} Here, A represents assumed mean d represents the deviations of the values from the assumed mean

Here, we take 162 as the assumed mean. So, we take deviations of each item in the series from 162.

X = 162 + \frac{10}{10} = 162 + 1 = 163

Thus, the mean height is 163 cm.

Page No 176:

Question 4:

Weight of 15 persons is as follows:

Weight (kg)

Find out mean weight, using Direct Method as well as Short-cut Method.

Answer:

S. No.	Weight (X)	*d = X − A*
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15	20 28 34 39 42 50 = A 53 54 59 64 72 74 74 78 79	−30 −22 −16 −11 −8 0 3 4 9 14 22 24 24 28 29














	∑X = 820	∑d = 70

Direct Method:

A . M . = X = \frac{Σ X}{N} = \frac{820}{15} = 54.67

Shortcut Method:

A . M . = X = A + \frac{Σ d}{N} Here, A represents the assumed mean d represents the deviations of the values from the assumed mean .

Here, we take 50 as assumed mean. Thus, we take deviations of the values from 50.

X = 50 + \frac{70}{15} = 50 + 4.67 = 54.67

Thus, the mean weight is 54.67 kg.

Page No 177:

Question 5:

Calculate average of the following discrete series. Use Short-cut Method by taking 25 as assumed average.

Size	30	29	28	27	26	25	24	23	22	21
Frequency (f)	2	4	5	3	2	7	1	4	5	7

Answer:

Size (X)	Frequency (f)	Deviation d = X − A	fd
30 29 28 27 26 25=A 24 23 22 21	2 4 5 3 2 7 1 4 5 7	5 4 3 2 1 0 −1 −2 −3 −4	10 16 15 6 2 0 −1 −8 −15 −28









	∑f = 40		∑fd = −3

A.M. = X=ΣXN =163010=163
Shortcut Method:-
Here, we take 25 as the assumed mean, so we tale the deviation of each item in the series from 25.

X = A + \frac{Σ f d}{Σ f}

= 25 + (\frac{- 3}{40}) = 25 - 0.075 = 24.925

Thus, the mean of the series is 24.925

Page No 177:

Question 6:

Marks secured by 42 students in Economics are:

Marks	15	20	22	23	27	35	18
Number of Students	8	4	7	3	8	7	5

Find average marks.

Answer:

Marks (X)	No. of students (f)	fX
15 20 22 23 27 35 18	8 4 7 3 8 7 5	120 80 154 69 216 245 90






	∑f = 42	∑fX = 974

Here, we use the direct method to find the mean of the given series.

A . M . = X = \frac{Σ f X}{Σ f} or, X = \frac{974}{42} = 23.20

Hence, the mean marks of the students is 23.20

Page No 177:

Question 7:

Average age of the people of a country is shown in the following table:

Age (Years)	10−20	20−30	30−40	40−50	50−60
People ('000)	30	32	15	12	9

Find out mean age by Direct Method.

Answer:

Age (X)	Mid-Values (m)	People (f)	fm
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60	15 25 35 45 55	30 32 15 12 9	450 800 525 540 495




		∑f = 98	∑fm = 2810

Here, we use the direct method to compute the value of mean.

X = \frac{Σ f m}{Σ f} or, X = \frac{2810}{98} = 28.7

Hence, the average age of the people in the country is 28.7 years.

Page No 177:

Question 8:

Calculate the arithmetic mean of the following frequency distribution by Direct Method:

Class Interval	10−20	20−40	40−70	70−120	120−200
Frequency	4	10	26	8	2

Answer:

Class Interval	Mid-Value (m)	Frequency (f)	fm
10 − 20 20 − 40 40 − 70 70 − 120 120 − 200	15 30 55 95 160	4 10 26 8 2	60 300 1430 760 320




		∑f = 50	∑fm = 2870

Using the direct method, arithmetic mean is calculated using the following formula.

X = \frac{Σ f m}{Σ f} or, X = \frac{2870}{50} = 57.4

Hence, the arithmetic mean of the series is 57.4

Page No 177:

Question 9:

Calculate arithmetic mean from the following data by Short-cut Method:

Class Interval	20−25	25−30	30−35	35−40	40−45	45−50	50−55
Frequency	10	12	8	20	11	4	5

Answer:

Class Interval	Mid-Value (m)	Frequency (f)	Deviation m − A (d)	fd
20 − 25 25 − 30 30 − 35 35 − 40 40 − 45 45 − 50 50 − 55	22.5 27.5 32.5 37.5 = A 42.5 47.5 52.5	10 12 8 20 11 4 5	−15 −10 −5 0 5 10 15	−150 −120 −40 0 55 40 75






		∑f = 70		∑fd = −140

Here, the assumed mean is decided from the mid-points. Let us take 37.5 as the assumed mean. So, we calculate the deviations of the mid-values from 37.5

X = A + \frac{Σ f d}{Σ f} Here, A represents the assumed mean d represents the deviation of the mid - values from the assumed mean X = 37.5 + (\frac{- 140}{70}) or, X = 37.5 - 2 = 35.5

Hence, the mean of the series is 35.5

Page No 177:

Question 10:

Find out arithmetic mean from the following distribution by Short-cut Method:

Items	10−8	8−6	6−4	4−2	2−0
Frequency	10	8	6	4	2

Answer:

Items (X)	Mid-Values (m)	Frequency (f)	Deviation m − A (d)	fd
0 − 2 2 − 4 4 − 6 6 − 8 8 − 10	1 3 5 = A 7 9	2 4 6 8 10	−4 −2 0 2 4	−8 −8 0 16 40



		∑f = 30		∑fd = 40

Here, the assumed mean is decided from the mid-points. Let us take 5 as the assumed mean. So, we calculate the deviations of the mid-values from 5

X = A + \frac{Σ f d}{Σ f} H e r e, A represents the assumed mean d represents the deviations of the mid - values from the assumed mean = 5 + \frac{40}{30} = 5 + 1.33 = 6.33

Hence, the mean of the given distribution is 6.33

Page No 177:

Question 11:

Sachin made the following runs in different matches:

Runs	5−15	15−25	25−35	35−45	45−55
Frequency	10	12	17	19	22

Calculate the average mean of the runs by Step-deviation Method.

Answer:

Run (X)	Mid-Values (m)	Frequency (f)	Deviation m − A (d)	Step deviation $d' = \frac{m - A}{i}$	fd'
5 − 15 15 − 25 25 − 35 35 − 45 45 − 55	10 20 30 = A 40 50	10 12 17 19 22	−20 −10 0 10 20	−2 −1 0 1 2	−20 −12 0 19 44




		∑f = 80			∑fd' = 31

Using the step - deviation method, arithmetic mean is calcualted using the following formula . X = A + \frac{Σ f d'}{Σ f d} \times i where, d' = \frac{m - A}{i} Substituting the calculated values in the formula X = 30 + \frac{31}{80} \times 10 = 30 + 3.875 = 33.875

Hence, the average runs scored by Sachin is 33.875

Page No 177:

Question 12:

Calculate arithmetic mean of the following frequency distribution:

Class	less than 10	10−20	20−30	30−40	40−50	50−60	more than 60
Frequency	5	12	18	22	6	4	3

Answer:

It must be noted that the given distribution is an open ended distribution, that is, the first and the last class interval are not explicitly defined. As mean is based on all the items in the series, so for such distributions mean cannot be calculated accurately.
However, going by the symmetry of the distribution we assume the first and the last class interval to be (0- 10) and (60 - 70), respectively

Class Interval	Mid-Values (m)	Frequency (f)	fm
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70	5 15 25 35 45 55 65	5 12 18 22 6 4 3	25 180 450 770 270 220 195






		∑f = 70	∑fm = 2110

Using the direct method to find the mean, we get

X = \frac{Σ f m}{Σ f} = \frac{2110}{70} = 30.14

Hence, the mean of the given distribution is 30.14

Page No 177:

Question 13:

Mean marks obtained by a student in his five subjects are 15. In English he secures 8 marks, in Economics 12, in Mathematics 18, and in Commerce 9. Find out the marks he secured in Statistics.

Answer:

Given:
Mean marks of five subjects = $X$ = 15
Marks in English = 8
Marks in Economics = 12
Marks in Mathematics = 18
Marks in Commerce = 9

To find: Marks in Statistics
Let marks scored in statistics be S

Substituting the given values in the formula for mean
$X = \frac{Σ X}{N} 15 = \frac{S + 8 + 12 + 18 + 9}{5}$
15 × 5 = S + 47
75 − 47 = S
S = 28
Hence, marks scored in statistics is 28

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Question 14:

Mean value of the weekly income of 40 families is 265. But in the calculation, income of one family was read as 150 instead of 115. Find the "Corrected" mean.

Answer:

Given:
Mean , $X$ = 265
Number of observations, N=40
Incorrect observation = 150
Correct observation = 115

${(X)}_{W r o n g} = \frac{Σ X}{N} or, Σ X = (X)$ _Wrong ×N or, ΣX = 265 ×40 = 10,600

Thus, wrong summation X, i.e., ${(Σ X)}_{w r o n g}$ is 10,600.
Now, correct mean is calculated using the following formula.
$C o r r e c t X = \frac{Σ X_{(w r o n g)} + c o r r e c t v a l u e - i n c o r r e c t v a l u e}{N} Substituting the given values in the formula . C o r r e c t X = \frac{10600 + 115 - 150}{40} = \frac{10565}{40} = 264.125$
Hence, the correct mean is 264.125

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Question 15:

Average pocket allowance of 6 students is ₹45. Of these, pocket allowance of 5 students is 20, 30, 22, 24 and 32 respectively. What is the pocket allowance of the sixth student?

Answer:

Given:
Average, $X$ = 45
Number of students, N = 6

Let sixth student's allowance be Z

Substituting the given values in the formula for mean
$X = \frac{Σ X}{N} 45 = \frac{20 + 30 + 22 + 24 + 32 + Z}{6}$
or, 45 × 6 = 128 + Z
or, 270 − 128 = Z
or, Z = 142
Hence, sixth student's allowance is Rs. 142

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Question 16:

The following table shows wages of the workers. Calculate the average wage of the workers.

Wages (₹)	10−19	20−29	30−39	40−49	50−59
Number of Workers	8	9	12	11	6

Answer:

Wages (X)	Mid Value (m)	Frequency (f)	Deviation d = m − A	$d' = \frac{d}{i}$	fd'
10 − 19 20 − 29 30 − 39 40 − 49 50 − 59	14.5 24.5 34.5 =A 44.5 54.5	8 9 12 11 6	−20 −10 0 10 20	−2 −1 0 1 2	−16 −9 0 11 12




		∑f = 46			∑fd = −2

We use the step-deviation method to find the value of mean.

X = A + \frac{Σ f d'}{Σ f} \times i = 34.5 + (\frac{- 2}{46}) \times 10 = 34.07

Hence, the mean of the given distribution is 34.07

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Question 17:

Ten players of the Australian team made an average of 63 runs and ten players of the Indian team made an average of 77 runs. Calculate the average run made by both the teams.

Answer:

The information given in the question can be summarised as follows.

	Average Runs	Number of Players
Australian Team	$(X_{1})$ = 63	N₁= 10
Indian Team	$(X_{2})$ = 77	N₁= 10

The combined average is calculated using the following formula.

X_{12} = \frac{N_{1} X_{1} + N_{2} X_{2}}{N_{1} + N_{2}} Substituting the given values in the formula . X_{12} = \frac{10 \times 63 + 10 \times 77}{10 + 10} = \frac{1400}{20} = 70

Hence, the average scored by both the teams is 70.

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Question 18:

Average income of 50 families is ₹3,000. Of these average income of 12 families is ₹1,800. Find out the average income of the remaining families.

Answer:

The information given in the question can be summarised as follows.
Average income of all the 50 families = $X_{12}$ = 3000

	Average income	Number of families
12 families	$(X_{1})$ = 1800	N₁= 12
*38 families*	$(X_{2})$ = ?	N₁= 38

Substituting the given values in the formula for combined mean

X_{12} = \frac{N_{1} X_{1} + N_{2} X_{2}}{N_{1} + N_{2}} 3000 = \frac{12 \times 1800 + 38 \times X_{2}}{50} or, 3000 × 50 = 21600 + 38 X

₂or, 150000 − 21600 = 38X₂or, X₂ = 3378.95

Hence, the average income of remaining 38 families is Rs 3,378.95

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Question 19:

In the following frequency distribution, ifthe arithmetic mean is 45.6, find out missing frequency.

Wages (₹)	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of Workers	5	6	7	X	4	3	9

Answer:

Wages (X)	Mid Value (m)	No. of worker (f)	fm
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	15 25 35 45 55 65 75	5 6 7 X 4 3 9	75 150 245 45 220 195 675






		∑f = 34 + X	∑fm = 1560 + 45X

Substituting the values in the formula for mean.

X = \frac{Σ f m}{Σ f} 45.6 = \frac{1560 + 45 X}{34 + X} or, 45.6 (34 + X) = 1560 + 45 X or, 1550.4 + 45.6 X = 1560 + 45 X or, 45.6 X − 45 X = 1560 - 1550.4 or, 0 .6X = 9.6 or, X = 16

X = \frac{9.6}{. 6}

X = 16
Hence, the missing frequency is 16.

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Question 20:

Calculate the weighted mean of the following data:

Items	96	102	104	124	148	164
Weight	5	6	3	7	12	9

Answer:

Items (X)	Weight (W)	WX
96 102 104 124 148 164	5 6 3 7 12 9	480 612 312 868 1776 1476





	∑W = 42	∑WX = 5524

X_{W} = \frac{Σ W X}{Σ X} = \frac{5524}{42} = 131.52

Hence, the weighted mean is 131.52

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Question 21:

A student obtained 60 marks in English, 75 in Hindi, 63 in Mathematics, 59 in Economics and 55 in Statistics. Calculate weighted mean of the marks if weights are respectively 2, 1, 5, 5 and 3.

Answer:

The information given in the question can be presented as follows.

Subject	Marks (X)	Weight (W)	WX
English Hindi Mathematics Economics Statistics	60 75 63 59 55	2 1 5 5 3	120 75 315 295 165




		∑W = 16	∑WX = 970

X_{W} = \frac{Σ W X}{Σ W} = \frac{970}{16} = 60.625

Hence, the weighted mean marks is 60.625

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Question 22:

A housewife uses 10 kg of wheat, 20 kg of Fuel, 5 kg of Sugar, and 2 kg of oil. Prices (per kg) of these items are ₹1.50, 50 paise, ₹ 2.80 and ₹10 respectively. Taking quantities used as weights find out the weighted arithmetic average of the prices.

Answer:

The information given in the question can be summarised as follows.

Price (X)	Quantity (W)	WX
1.5 0.5 2.8 10	10 20 5 2	15 10 14 20



	∑W = 37	∑WX = 59

X_{W} = \frac{Σ W X}{Σ W} = \frac{59}{37} = 1.59

Hence, the weighted average of the prices is Rs 1.59

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Question 23:

Calculate weighted mean of the following data by using Direct and Short-cut Methods:

Items	81	76	74	58	70	73
Weight	2	3	6	7	3	7

Answer:

Items (X)	Weight (W)	WX	Deviation d = X − A	Wd
81 76 74 = A 58 70 73	2 3 6 7 3 7	162 228 444 406 210 511	7 2 0 −16 −4 −1	14 6 0 −112 −12 −7





	∑W = 28	∑WX = 1961		∑Wd = −111

Direct Method

X_{w} = \frac{Σ W X}{Σ W} = \frac{1961}{28} = 70.04

Shortcut Method

X_{w} = A + \frac{Σ W d}{Σ W} = 74 + (\frac{- 111}{28}) = 74 - 3.964 = 70.04

Hence, the weighted mean is 70.04

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