Measures Of Central Tendency Arithmetic MeaN

Question 22:

The man height of 25 male workers in a factory is 61 inches and the mean height of 35 female workers in the same factory is 58 inches. Find the combined mean height of 60 workers in the factory

Question 37:

Locate the missing frequency, if arithmetic mean of the following series is 44.8.
X 20 30 40 50 60 70
f 5 ? 15 10 8 5

Question 38:

The average marks of 30 students in a class were 52. The top six students had an average of 31. What were the average marks of the other students?

Question 26:

The mean of 100 observation was found to be 40. Later on, it was discovered that tow items were wrongly taken as 30 and 27 instead of 3 and 72. Find correct mean.

Question 27:

The average height of a group of 40 students was calculated as 155 cm. It was later discovered that the height of one student was read as 157 instead of 137 cm. Calculate the correct average height.

Question 1:

Find average for following individual data.

ANSWER:

Calculating the average using direct method:

$Average (X) = \frac{Σ X}{N} or, X = \frac{2 + 3 + 5 + 6 + 8 + 10 + 11 + 13 + 17 + 20}{10} or, X = \frac{95}{10} \Rightarrow X = 9.5$

Hence, the average of the above individual data is 9.5

Page No 8.60:

Question 2:

Daily income of 10 families is given as follows:

S. No.	1	2	3	4	5	6	7	8	9	10
Daily Income (in ₹)	100	120	80	85	95	130	200	250	225	275

Calculate average daily income

ANSWER:

Serial No.	Daily Income (in Rs) (X)
1 2 3 4 5 6 7 8 9 10	100 120 80 85 95 130 200 250 225 275
N=10	$\sum_{} X$ = 1560

Calculating the average daily income using direct method:

Average (X) = \frac{\sum_{} X}{N} or, X = \frac{100 + 120 + 80 + 85 + 95 + 130 + 200 + 250 + 225 + 275}{10} or, X = \frac{1560}{10} \Rightarrow X = Rs 156

Hence, the average daily income is Rs 156

Page No 8.60:

Question 3:

The following table gives the marks obtained by 10 students of a class:

S. No.	1	2	3	4	5	6	7	8	9	10
Marks	43	60	37	48	65	48	57	78	31	59

Find out mean marks, using Direct Method as well as Short-Cut Method

ANSWER:

S.N.	Marks X	deviation from assumed mean (X − A)
1 2 3 4 5 6 7 8 9 10	43 60 37 48 $65 = A$ 48 57 78 31 59	−22 −5 −28 −17 0 −17 −8 13 −34 −6
N=10	$\sum_{} X$ =526	Σd=−124

(i) Calculating mean marks using direct method:

X = \frac{\sum_{} X}{N} or, X = \frac{43 + 60 + 37 + 48 + 65 + 48 + 57 + 78 + 31 + 59}{10} or, X = \frac{526}{10} \Rightarrow X = 52.6 m a r k s

(ii) Calculating mean marks using short cut method:

X = A + \frac{Σ d}{N}

Here,
A represents assumed mean
d represents the deviation of the values from the assumed mean i.e. X − A

Here, we take 65 as the assumed mean. So, we take deviations of each item in the series from 65.

X = 65 + \frac{(- 124)}{10} or, X = 65 - \frac{124}{10} \Rightarrow X = 52.6 marks

Thus, mean marks is 52.6

Page No 8.60:

Question 4:

The following figures are the heights in cms of 7 children chose at random:
64, 59, 67, 69, 65, 70, 68
Calculate the simple arithmetic mean of the heights by (i) Direct method, (ii) Short-c ut Method, and (iii) Step Deviation Method.

ANSWER:

S.N.	Height (X)	Deviation from assumed mean d=(X − A)	$d' = \frac{d_{}}{i} = \frac{d_{}}{2}$
1 2 3 4 5 6 7	64 59 67 $69 = A$ 65 70 68	−5 −10 −2 0 −4 1 −1	−2.5 −5 −1 0 −2 0.5 −0.5
N=7	$\sum_{} X$ =462	Σd=−21	Σd'= −10.5

(i) Calculating mean height using direct method:

X = \frac{\sum_{} X}{N} or, X = \frac{64 + 59 + 67 + 69 + 65 + 70 + 68}{7} or, X = \frac{462}{7} \Rightarrow X = 66 c m s

(ii) Calculating mean height using short cut method:

X = A + \frac{\sum_{}^{} d}{N}

Here,

A represents assumed mean.
d represents the deviation of the values from the assumed mean.

Here, we take 69 as the assumed mean. So, we take deviations of each item in the series from 69.

X = 69 - \frac{21}{7} or, X = 69 - 3 \Rightarrow X = 66 c m s

(iii) Calculating mean height using step-deviation method:

X = A + \frac{Σ d'}{N} \times i

X = 69 + \frac{(- 10.5)}{7} \times 2 or, X = 69 - \frac{21}{7} \Rightarrow X = 66 cms

Hence, the arithmetic mean of the heights is 66 cms.

Page No 8.61:

Question 5:

Find average for following discrete series.

X	3	5	6	7	8
f	2	4	3	8	10

ANSWER:

X	f	fX
3 5 6 7 8	2 4 3 8 10	6 20 18 56 80
	N=Σf= 27	ΣfX= 180

Calculating average for the above discrete series:

X = \frac{Σ f X}{Σ f}

X = \frac{180}{27} \Rightarrow X = 6.66

Thus, the average of the above series is 6.66

Page No 8.61:

Question 6:

Compute the arithmetic mean from the following frequency table:

Hight (in cms.)	58	60	62	64	66	68
Number of Plants	12	14	20	13	8	5

ANSWER:

Height (X)	Plants (f)	fX
58 60 62 64 66 68	12 14 20 13 8 5	696 840 1240 832 528 340
	N=Σf= 72	ΣfX=4476

Calculating the arithmetic mean of the above series by direct method:

X = \frac{Σ f X}{Σ f} or, X = \frac{4476}{72} \Rightarrow X = 62.16 cms

Hence, the mean height of the plant is 62.16 cms

Page No 8.61:

Question 7:

Compute mean marks from the data given below by: (i) Direct method, (ii) Short-cut Method, and (iii) Step Deviation Method.

Marks	5	15	25	35	45	55	65
Students	4	6	10	20	10	6	4

ANSWER:

Marks (X)	Students (f)	fX	d=X−A	fd	$d' = \frac{d}{i} = \frac{d}{10}$	*fd'*
5 15 25 $35 = A$ 45 55 65	4 6 10 20 10 6 4	20 90 250 700 450 330 260	$-$ 30 −20 −10 0 10 20 30	−120 −120 −100 0 100 120 120	−3 −2 −1 0 1 2 3	−12 −12 −10 0 10 12 12
	N=Σf= 60	ΣfX= 2100		$Σ f d$ =0		$Σ f d'$ =0

i) Calculating mean marks using direct method:

X = \frac{Σ f X}{Σ f} or, X = \frac{2100}{60} \Rightarrow X = 35 marks

(ii) Calculating mean marks using short cut method:

X = A + \frac{Σ f d}{Σ f}

Here,

A represents assumed mean.
d represents the deviation of the values from the assumed mean.

Here, we take 35 as the assumed mean. So, we take deviations of each item in the series from 35.

X = 35 + \frac{0}{60} \Rightarrow X = 35 marks

(iii) Calculating mean marks using step-deviation method:

X = A + \frac{Σ f d'}{Σ f} \times i

X = 35 + \frac{0}{60} \times 10 \Rightarrow X = 35 marks

Hence, the mean marks of the above data is 35

Page No 8.61:

Question 8:

The distribution fo age at marriage of 50 males is given below:

Age in Years	20	21	22	23	24	25
No. Males	1	2	4	5	15	23

Calculate arithmetic mean using direct method.

ANSWER:

Age (X)	No. of Males (f)	fX
20 21 22 23 24 25	1 2 4 5 15 23	20 42 88 115 360 575
	Σf= 50	ΣfX= 1200

Calculating mean age at marriage:

X = \frac{Σ f X}{Σ f} or, X = \frac{1200}{50} \Rightarrow X = 24 y ears

Hence, mean age at marriage is 24 years

Page No 8.61:

Question 9:

Compute the mean marks obtained by the students from the following data:

Marks	0−10	10−20	20−30	30−40	40−50
No. of Students	4	6	10	20	10

ANSWER:

Class Interval	Mid Values $m = \frac{l_{1} + l_{2}}{2}$	Students (f)	fm
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50	5 15 25 35 45	4 6 10 20 10	20 90 250 700 450
		Σf= 50	Σfm= 1510

Calculating mean marks using direct method:

X = \frac{Σ f m}{Σ f} Here, m = \frac{lower limit (l_{1}) + upper limit (l_{2})}{2} X = \frac{1510}{50} \Rightarrow X = 30.2 marks

Hence, mean marks obtained by the students are 30.2

Page No 8.61:

Question 10:

Compute the mean marks obtained by the students from the following data:

Marks	0−4	4−8	8−12	12−16	16−20	20−24
No. of Students	7	9	16	8	6	4

ANSWER:

Class Interval (Marks)	Mid-Values (m)	Students (f)	fm
0 − 4 4 − 8 8 − 12 12 − 16 16 − 20 20 − 24	2 6 10 14 18 22	7 9 16 8 6 4	14 54 160 112 108 88
		$Σ f$ =50	$Σ f m$ =586

Calculating the mean marks using direct method:

X = \frac{Σ f m}{Σ f} or, X = \frac{536}{50} \Rightarrow X = 10.72 marks

Hence, mean marks obtained by the students are 10.72

Page No 8.61:

Question 11:

Calculate the arithmetic average from the following data:

Daily wages (in ₹)	2−4	4−6	6−8	8−10	10−12	12−14	14−16	16−18
No. of Workers	11	14	20	32	25	7	5	2

ANSWER:

Class Interval (Wages)	Mid Values (m)	Workers (f)	fm
2 − 4 4 − 6 6 − 8 8 − 10 10 − 12 12 − 14 14 − 16 16 − 18	3 5 7 9 11 13 15 17	11 14 20 32 25 7 5 2	33 70 140 288 275 91 75 34
		$Σ f$ =116	$Σ f m$ =1006

Calculating the average wage using direct method:

X = \frac{Σ f m}{Σ f} or, X = \frac{1006}{116} \Rightarrow X = Rs 8.67

Hence, average daily wage is Rs 8.67

Page No 8.61:

Question 12:

Calculate mean from following data:

Marks	5−15	15−25	25−35	35−45	45−55	55−65
No. of Students	8	12	6	14	7	3

ANSWER:

Class Interval (Marks)	Mid-Values (m)	Students (f)	fm
5 − 15 15 − 25 25 − 35 35 − 45 45 − 55 55 − 65	10 20 30 40 50 60	8 12 6 14 7 3	80 240 180 560 350 180
		Σf= 50	Σfm= 1590

Calculating mean marks by using direct method:

X = \frac{Σ f m}{Σ f} or, X = \frac{1590}{50} \Rightarrow X = 31.8 marks

Hence, mean of the above series is 31.8 marks

Page No 8.62:

Question 13:

Find mean for the following data by using: (i) Direct Method: (ii) Short-cut Method; (iii) Step Deviation Method.

X	100−200	200−300	300−400	400−500	500−600
f	10	18	12	20	40

ANSWER:

Class Interval	Mid values (m)	(f)	fm	d=m−A	fd	$d' = \frac{d}{i} = \frac{d}{100}$	fd'
100−200 200 −300 300 −400 400 −500 500 −600	150 250 $350 = A$ 450 550	10 18 12 20 40	1500 4500 4200 9000 22000	−200 −100 0 100 200	−2000 −1800 0 2000 8000	−2 −1 0 1 2	−20 −18 0 20 80
		$Σ f$ =100	$Σ f m$ =41200		$Σ f d$ =6200		$Σ f d'$ =62

(i) Calculating mean using direct method:

X = \frac{Σ f m}{Σ f} h e r e, m = \frac{l_{1} + l_{2}}{2} X = \frac{41200}{100} \Rightarrow X = 412

(ii) Calculating mean using short cut method:

X = A + \frac{Σ f d}{Σ f}

Here,

A represents assumed mean.
d represents the deviation of the values from the assumed mean.

Here, we take 350 as the assumed mean. So, we take deviations of each item in the series from 350.

X = 350 + \frac{6200}{100} o r, X = 350 + 62 \Rightarrow X = 412

(iii) Calculating mean height step-deviation method:

X = A + \frac{Σ f d'}{Σ f} \times i

X = A + \frac{Σ f d'}{Σ f} \times i o r, X = 350 + \frac{62}{100} \times 100 \Rightarrow X = 412

Hence, the mean of the above series is 412.

Page No 8.62:

Question 14:

The following table shows the marks obtained by 90 students in a certain examination. Calculate the average marks per students by step deviation method.

Marks	10−20	20−30	30−40	40−50	50−60	60−70	70−80	80−90
No. of students	7	13	20	25	10	8	6	1

ANSWER:

Class Interval (Marks)	Mid Values (m)	Students (f)	d=m−A	$d' = \frac{d}{i} = \frac{d}{10}$	fd'
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90	15 25 35 $45 = A$ 55 65 75 85	7 13 20 25 10 8 6 1	−30 −20 −10 0 10 20 30 40	−3 −2 −1 0 1 2 3 4	−21 −26 −20 0 10 16 18 4
		$Σ f$ =90			$Σ f d'$ =−19

Calculating mean marks using step-deviation method:

X = A + \frac{Σ f d'}{Σ f} \times i o r, X = 45 + \frac{(- 19)}{90} \times 10 o r, X = 45 - \frac{19}{9} \Rightarrow X = 42.89 Marks

Hence, the mean of the above series is 42.89 marks.

Page No 8.62:

Question 15:

Find the mean form the following data:

Find the mean form the following data	No. of Students
Less than 10	5
Less than 20	20
Less than 30	45
Less than 40	70
Less than 50	80
Less than 60	88
Less than 70	98
Less than 80	100

ANSWER:

Converting less than cumulative frequency distribution into a simple frequency distribution

Marks	No. of students (c.f)
Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80	5 20 45 70 80 88 98 100

Class Interval (Marks)	Mid-Values (m)	Students (f)	fm
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	5 15 25 35 45 55 65 75	0-5=5 20-5=15 45-20=25 70-45=25 80-70=10 88-80=8 98-88=10 100-98=2	25 225 625 875 450 440 650 150
		Σf=100	Σfm= 3440

Calculating mean from the above data by using direct method:

X = \frac{Σ f m}{Σ f} o r, X = \frac{3440}{100} \Rightarrow X = 34.4 marks

Hence, the mean of the above data is 34.4 marks.

Page No 8.62:

Question 16:

Calculate arithemtic eman from the foolowing data:

Marks	No. Students
More than 0	150
More than 10	140
More than 20	100
More than 30	80
More than 40	80
More than 50	70
More than 60	30
More than 70	14

ANSWER:

Converting more than cumulative frequency distribution into a simple frequency distribution

Marks	No. of students (c.f)
More than 0 More than 10 More than 20 More than 30 More than 40 More than 50 More than 60 More than 70	150 140 100 80 80 70 30 14

Class Interval	Mid-Values (m)	No. of students (f)	fm
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	5 15 25 35 45 55 65 75	150-140=10 140-100=40 100-80=20 80-80=0 80-70=10 70-30=40 30-14=16 14-0=14	50 600 500 0 450 2200 1040 1050
		$Σ f$ =150	$Σ f m$ =5890

Calculating the mean marks by using direct method:

X = \frac{Σ f m}{Σ f} o r, X = \frac{5890}{150} \Rightarrow X = 39.27 m a r k s

Hence, the mean marks of the above series is 39.27

Page No 8.63:

Question 17:

Find the mean from the following data:

Marks	No. of Students
Above 0	80
Above 10	77
Above 20	72
Above 30	65
Above 40	55
Above 50	43
Above 60	28
Above 70	16
Above 80	10
Above 90	8
Above 100	0

ANSWER:

Marks	Students
Above 0 Above 10 Above 20 Above 30 Above 40 Above 50 Above 60 Above 70 Above 80 Above 90 Above 100	80 77 72 65 55 43 28 16 10 8 0

Class Interval (Marks)	Mid-Values (m)	Students (f)	fm
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 90 − 100 100 − 110	5 15 25 35 45 55 65 75 85 95 105	80−77=3 77−72=5 72−-65=7 65−55=10 55−43=12 43−28=15 28−16=12 16−10=6 10−8=2 8−0=8 0−0=0	15 75 175 350 540 825 780 450 170 760 0
		Σf= 80	Σfm= 4140

Calculating mean marks by using direct method:

X = \frac{Σ f m}{Σ f} o r, X = \frac{4140}{80} \Rightarrow X = 51.75 m arks

Page No 8.63:

Question 18:

The following table shows the age of workers in a factory. Find out the average age of workers.

Age (in Years)	20−29	30−39	40−49	50−59	60−69
Workers	10	8	6	4	2

ANSWER:

Calculation of mean in an inclusive series:

Class Interval (Age)	Mid-Values (m)	Workers (f)	fm
20 − 29 30 − 39 40 − 49 50 − 59 60 − 69	24.5 34.5 44.5 54.5 64.5	10 8 6 4 2	245 276 267 218 129
		$Σ f$ =30	$Σ f m$ =1135

Computing the average age of workers:

X = \frac{Σ f m}{Σ f} o r, X = \frac{1135}{30} \Rightarrow X = 37.83 y ears

Hence, average age of workers is 37.83 years

Page No 8.63:

Question 19:

Find the average age from the following data:

Age (in Years)	Less than 10	10−20	20−30	30−40	More than 40
No. of Persons	5	8	12	6	4

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 (40 - 50), respectively

Age	Class Interval	Mid-Values (m)	Persons (f)	fm
Less than 10 10 − 20 20 − 30 30 − 40 more than 40	0 − 10 10 − 20 20 − 30 30 − 40 40 − 50	5 15 25 35 45	5 8 12 6 4	25 120 300 210 180
			$Σ f$ =35	$Σ f m$ =835

Computing average age:

X = \frac{Σ f m}{Σ f} o r, X = \frac{835}{35} \Rightarrow X = 23.85 years

Hence, average age is 23.85 years.

Page No 8.63:

Question 20:

Calculate the mean from the following data:

Mid-value	10	20	30	40	50
Frequency	8	12	15	9	6

ANSWER:

Mid-Values(m)	(f)	fm
10 20 30 40 50	8 12 15 9 6	80 240 450 360 300
	$Σ f$ =50	$Σ f m$ =1430

Calculating mean using direct method:

X = \frac{Σ f m}{Σ f} o r, X = \frac{1430}{50} \Rightarrow X = 28.6

Hence, mean of the above series is 28.6

Page No 8.63:

Question 21:

Calculate mean form the following data:

X	0−10	10−20	20−30	30−60	60−90
Frequency	5	9	20	12	4

ANSWER:

Class Interval	Mid-Values (m)	(f)	fm
0 − 10 10 − 20 20 − 30 30 − 60 60 − 90	5 15 25 45 75	5 9 20 12 4	25 135 500 540 300
		$Σ f$ =50	$Σ f m$ =1500

From the following data, calculate the weighted mean.

Marks	62	77	65	62	57
Weights	2	1	2	3	4

ANSWER:

Marks (X)	Weight (W)	WX
62 77 65 62 57	2 1 2 3 4	124 77 130 186 228
	$Σ W$ =12	$Σ W X$ =745

X_{W} = \frac{Σ W X}{Σ W}

X_{w} = \frac{745}{12} \Rightarrow X_{w} = 62.08

Hence, the weighted mean is 62.08 marks

Page No 8.64:

Question 29:

Find out weighted mean by weighting each price by the quantity consumed:

Items	Quantity consumed (in kg)	Price in Rupees (per kg)
Apple	12	30
Mango	10	50
Tomato	10	40
Orange	15	35
Grapes	12	45

ANSWER:

Items	Quantity (W)	Price (X)	WX
Apple Mango Tomato Orange Grapes	12 10 10 15 12	30 50 40 35 45	360 500 400 525 540
	$Σ W$ =59		*ΣWX=* 2325

Now, weighted mean can be calculating using the following formula.

X_{W} = \frac{Σ W X}{Σ W} o r, X_{W} = \frac{2325}{59} \Rightarrow X_{W} = 39.40

Hence, the weighted mean of the above series is 39.40

Page No 8.65:

Question 30:

An examination was held to decide the most eligible student for scholarship. The weights of various subjects along with the marks obtained by the three candidate (out of 100) are given below:

Subjects	Weighs	Marks of Ram	Marks of Shyam	Marks of Manish
Maths	3	65	64	70
Statistics	4	63	60	65
Economics	1	70	80	52
English	2	58	56	63

ANSWER:

Subjects	Weights (W)	Ram (X1)	Shyam (X2)	Manish (X3)	WX1	WX2	WX3
Math Stats Eco English	3 4 1 2	65 63 70 58	64 60 80 56	70 65 52 63	195 252 70 116	192 240 80 112	210 260 52 126
	10				633	624	648

Weighted Mean Marks, Ram X_{R} = \frac{Σ W X_{1}}{Σ W} = \frac{633}{10} = 63.3 Weighted Mean Marks, Shyam X_{S} = \frac{Σ W X_{2}}{Σ W} = \frac{624}{10} = 62.4 Weighted Mean Marks, Manish X_{M} = \frac{Σ W X_{3}}{Σ W} = \frac{648}{10} = 64.8

Manish should get the scholarship as he has the maximum weighted mean marks.

Page No 8.65:

Question 31:

Arithmetic mean of the following series is 41:

Marks	20	30	40	?	60	70
No. of Students	8	12	20	10	6	4

Find the missing item

ANSWER:

Marks (X)	No. of Students (f)	fX
20 30 40 x 60 70	8 12 20 10 6 4	160 360 800 10x 360 280
	$Σ f =$ 60	$Σ f X$ =1960 + 10x

Given:

Mean,

X

= 41

X = \frac{Σ X}{Σ f} 41 = \frac{1960 + 10 x}{60} o r, 41 \times 60 = 1960 + 10 x o r, 2460 - 1960 = 10 x o r, 500 = 10 x o r, x = \frac{500}{10} \Rightarrow x = 50 m a r k s

Hence, the missing item is 50 marks

Page No 8.65:

Question 32:

Calculate the number of students against the class 30−40 of the following data, where $X$ = 28.

Marks	0−10	10−20	20−30	30−40	40−50	50−60
Frequency	12	18	27	?	17	6

ANSWER:

Class Interval (Marks)	Mid-Values (m)	Frequency (f)	fm
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60	5 15 25 35 45 55	12 18 27 f 17 6	60 270 675 35f 765 330
		$Σ f$ =80 + f	$Σ f m$ =2100 + 35f

Let the missing frequency be f

Given:
Mean,

X

=28

X = \frac{Σ f m}{Σ f} o r, 28 = \frac{2100 + 35 f}{80 + f} o r, 2240 + 28 f = 2100 + 35 f o r, 7 f = 140 \Rightarrow f = \frac{140}{7} = 20

Hence, the missing frequency is 20.

Page No 8.65:

Question 33:

Find the missing frequency form the following data, if athematic mean is 25.4.

Class-interval	10−20	20−30	30−40	40−50	50−60
Frequency	20	15	10	?	2

ANSWER:

Class Interval	Mid-Values (m)	Frequency (f)	fm
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60	15 25 35 45 55	20 15 10 f 2	300 375 350 45f 110
		$Σ f$ =47+f	$Σ f m$ =1135 + 45 f

Let missing frequency be f

Given:

Mean, $X$ =25.4

X = \frac{Σ f m}{Σ f} o r, 25.4 = \frac{1135 + 45 f}{47 + f} o r, 1193.8 + 25.4 f = 1135 + 45 f o r, 58.8 = 19.6 f \Rightarrow f = 3

Hence, missing frequency is 3

Page No 8.65:

Question 34:

Find the mi9ssing frequency from the followinjg table:

Marks	No. of Students
Below 10	25
Below 20	40
Below 30	60
Below 40	75
Below 50	95
Below 60	125
Below 70	190
Below 80	240

ANSWER:

Marks	Students
Below 10 Below 20 Below 30 Below 40 Below 50 Below 60 Below 70 Below 80	25 40 60 75 95 125 190 240

Class Interval	Mid-Values (m)	No. of Students (f)	fm
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	5 15 25 35 45 55 65 75	25-0=25 40-25=15 60-40=20 75-60=15 95-75=20 125-95=30 190-125=65 240-190=50	125 225 500 525 900 1650 4225 3750
		$Σ f$ =240	$Σ f m$ =11,900

Calculating average marks using direct method:

X = \frac{Σ f m}{Σ f} o r, X = \frac{11900}{240} \Rightarrow X = 49.58 marks

Hence, the average marks are 49.58

Page No 8.65:

Question 35:

Calculate the mean from the following frequency distribution:

X	10−19	20−29	30−39	40−49	50−59	60−69	70−79	80−89
No. Persons	32	42	40	56	20	6	2	2

ANSWER:

Class interval (X)	Mid-Values (m)	No. of persons (f)	fm
10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69 70 − 79 80 − 89	14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5	32 42 40 56 20 6 2 2	464 1029 1380 2492 1090 387 149 169
		$Σ f$ =200	$Σ f m$ =7160

Calculating the mean from the above inclusive series by using direct method:

X = \frac{Σ f m}{Σ f} o r, X = \frac{7160}{200} \Rightarrow X = 35.8

Hence, the mean of the above series is 35.8

Page No 8.66:

Question 36:

The Average age of 20 students in a class is 16 yars.One student whose age is 18 years has left the class. Find out average age of rest of the students.

ANSWER:

Given:
Mean, $X$ = 16 years
Number of observations, N= 20

We know,
$X = \frac{Σ X}{N}$

ΣX of 20 students = $X_{20} \times N$ = 20 × 16 = 320
ΣX of 19 students = $Σ X_{20} - a g e o f t h e s t u d e n t w h o l e f t t h e c l a s s$ =320 − 18 = 302

Thus, the average age of rest of the students is:

$X of 19 students = \frac{302}{19} = 15.89 y e a r s$

Hence, the correct average age is 15.89 years

Page No 8.66:

Question 37:

Locate the missing frequency, if arithmetic mean of the following series is 44.8.

X	20	30	40	50	60	70
f	5	?	15	10	8	5

ANSWER:

X	f	fX
20 30 40 50 60 70	5 f 15 10 8 5	100 30f 600 500 480 350
	$Σ f$ =43 + f	$Σ f X$ =2030 + 30f

Let missing frequency be f

Given:
Mean,

X = 44.8

We know,

X = \frac{Σ f X}{Σ f} o r, 44.8 = \frac{2030 + 30 f}{43 + f} o r, 1926.4 + 44.8 f = 2030 + 30 f o r, 14.8 f = 103.6 \Rightarrow f = \frac{103.6}{14.8} = 7

Hence, the missing frequency is 7

Page No 8.66:

Question 38:

The average marks of 30 students in a class were 52. The top six students had an average of 31. What were the average marks of the other students?

ANSWER:

Given:

N= 30
$M e a n, X_{30} = 52 Σ X_{30} = X_{30} \times N o r, Σ X_{30} = 30 \times 52 \Rightarrow Σ X_{30} = 1560$

Average of top 6 students = 31

ΣX of top 6 students, $Σ X_{6}$ = 31 × 6 = 186

ΣX of remaining 24 (30-6) students, $Σ X_{24}$ = $Σ X_{30} - Σ X_{6}$ = 1560 − 186 = 1374

Now, we will calculate the average marks of the other students by using the direct method:

$X of 24 students = \frac{Σ X_{24}}{N} o r, X_{24} = \frac{1374}{24} \Rightarrow X_{24} = 57.25 marks$

Page No 8.66:

Question 39:

Calculate the mean of the given data:

Size of item	2	4	6	8	10	12	14	16
Frequency	5	7	10	13	17	9	6	3

ANSWER:

X	f	fX
2 4 6 8 10 12 14 16	5 7 10 13 17 9 6 3	10 28 60 104 170 108 84 48
	$Σ f$ =70	$Σ f X$ =612

Calculating the mean of the above data by using the direct method:

X = \frac{Σ f X}{Σ f} or, X = \frac{612}{70} \Rightarrow X = 8.74

Hence, the mean of the above series is 8.74

Page No 8.66:

Question 40:

A Candidate obtained the following percentage following of marks in an examination: English 60; Business Studies 75; Maths 63; Accounts 59; Economics 55. Find the candidate's weighted arithmetic mean, if weights 1, 2, 1, 3, 3 respectively are allotted to the subjects

ANSWER:

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

Subject	Marks (X)	Weights (W)	WX
English Business studies Maths Accounts Economics	60 75 63 59 55	1 2 1 3 3	60 150 63 177 165
		$Σ W$ =10	$Σ W X$ =615

X_{w} = \frac{Σ W X}{Σ W} o r, X_{w} = \frac{615}{10} \Rightarrow X_{w} = 61.5 Marks

Hence, the weighted arithmetic mean is 61.5 marks

Page No 8.66:

Question 41:

The mean height of 50 student of a college is 68 inches. The height of 30 students is given below. Find the mean height of the remaining 20 students.

Height (in inches)	64	66	68	70	72
Frequency	4	12	4	8	2

ANSWER:

Height (X)	Frequency (f)	fX
64 66 68 70 72	4 12 4 8 2	256 792 272 560 144
	$Σ f$ =30	$Σ f X_{30}$ =2024

Given:
Σf = 50

X_{50} = 68

Σ f X_{50} = X_{50} \times Σ f = 68 \times 50 = 3400

ΣfX of remaining 20 students

Σ f X_{20}

Σ f X_{50} - Σ f X_{30}

=3400 − 2024 = 1376

X of 20 students = \frac{Σ f X_{20}}{Σ f} = \frac{1376}{20} = 68.8 inches

Hence, mean height of 20 students is 68.8 inches