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Page No 8.51:

Question 1:

Find average for following individual data.

2 3 5 6 8 10 11 13 17 20

Answer:

Calculating the average using direct method:

Average (X)=ΣXNor, X  =2+3+5+6+8+10+11+13+17+2010 or, X =9510X=9.5

Hence, the average of the above individual data is 9.5

Page No 8.51:

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 X= 1560

Calculating the average daily income using direct method:

Average (X)=XNor, X=100+120+80+85+95+130+200+250+225+27510or, X=156010X= Rs 156

Hence, the average daily income is Rs 156

Page No 8.51:

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 X=526 Σd=−124

(i) Calculating mean marks using direct method:

X=XNor, X =43+60+37+48+65+48+57+78+31+5910or, X  =52610X=52.6 marks

(ii) Calculating mean marks using short cut method:

X=A+ΣdN
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+(-124)10or, X = 65-12410X = 52.6 marks

Thus, mean marks is 52.6



Page No 8.52:

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'=di=d2
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 X=462 Σd=−21 Σd'= −10.5

 (i) Calculating mean height using direct method:

X=XNor, X=64+59+67+69+65+70+687or, X =4627 X =66 cms

(ii) Calculating mean height using short cut method:

X=A+dN

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=69-217or, X =69-3 X =66 cms

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

X=A+Σd'N×i

X=69+-10.57×2or, X =69-217 X =66 cms

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

Page No 8.52:

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
  Nf= 27 ΣfX= 180

Calculating average for the above discrete series:

X=ΣfXΣf

X=18027X=6.66

Thus, the average of the above series is 6.66

Page No 8.52:

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
  Nf= 72 ΣfX=4476

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

X=ΣfXΣfor, X =447672 X = 62.16 cms

Hence, the mean height of the plant is 62.16 cms

Page No 8.52:

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'=di=d10 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   Σfd=0   Σfd'=0
 

i) Calculating mean marks using direct method:

X=ΣfXΣfor, X=210060 X= 35 marks

(ii) Calculating mean marks using short cut method:


X=A+ΣfdΣf

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+060X= 35 marks

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

X=A+Σfd'Σf×i

X=35+060×10X=35 marks

Hence, the mean marks of the above data is 35

Page No 8.52:

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=ΣfXΣfor, X = 120050 X =24 years

Hence, mean age at marriage is 24 years

Page No 8.52:

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=l1+l22
 
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=ΣfmΣfHere,m=lower limit (l1)+upper limit (l2)2X=151050X=30.2 marks

Hence, mean marks obtained by the students are 30.2

Page No 8.52:

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
    Σf=50 Σfm=586

Calculating the mean marks using direct method:

X=ΣfmΣfor, X =53650X =10.72 marks

Hence, mean marks obtained by the students are 10.72

Page No 8.52:

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
    Σf=116 Σfm=1006

Calculating the average wage using direct method:

X=ΣfmΣfor, X=1006116 X= Rs 8.67 

Hence, average daily wage is Rs 8.67



Page No 8.53:

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=ΣfmΣfor, X=159050X=31.8 marks

Hence, mean of the above series is 31.8 marks

Page No 8.53:

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'=di=d100 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
    Σf=100 Σfm=41200   Σfd=6200   Σfd'=62
 

(i) Calculating mean using direct method:

X=ΣfmΣf here,m=l1+l22X=41200100X =412

(ii) Calculating mean using short cut method:

X=A+ΣfdΣf

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+6200100or, X=350+62 X=412

(iii) Calculating mean height step-deviation method:

X=A+Σfd'Σf×i

X=A+Σfd'Σf×ior, X=350+62100×100X=412

Hence, the mean of the above series is 412.

Page No 8.53:

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=mA d'=di=d10 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
    Σf=90     Σfd'=−19

Calculating mean marks using step-deviation method:

X=A+Σfd'Σf×ior, X =45+-1990×10or, X =45-199 X=42.89 Marks

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

Page No 8.53:

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=ΣfmΣfor, X=3440100X =34.4 marks

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

Page No 8.53:

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
    Σf=150 Σfm=5890

Calculating the mean marks by using direct method:

X=ΣfmΣfor, X = 5890150X  = 39.27 marks

Hence, the mean marks of the above series is 39.27



Page No 8.54:

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=ΣfmΣfor, X=414080X=51.75 marks

Page No 8.54:

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
    Σf=30 Σfm=1135

Computing the average age of workers:

X=ΣfmΣfor, X = 113530X=37.83 years

Hence, average age of workers is 37.83 years

Page No 8.54:

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
      Σf=35 Σfm=835

Computing average age:

X=ΣfmΣfor, X =83535 X =23.85 years

Hence, average age is 23.85 years.

Page No 8.54:

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
  Σf=50 Σfm=1430

Calculating mean using direct method:

X=ΣfmΣfor, X=143050X = 28.6

Hence, mean of the above series is 28.6

Page No 8.54:

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
    Σf=50 Σfm=1500

Calculating the mean using direct method:

X=ΣfmΣfor, X =150050 X =30

Hence, mean of the above series is 30

Page No 8.54:

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

Answer:


Combined Mean ( Xm,f) = Nm Xm+ Nf XfNm+Nf
Here,

Xm=61 inches,  Xf=58 inchesNm=25,  Nf=35

Combined Mean ( X¯m,f) =25×61 + 35×5825+35or,  X¯m,f =355560  Xm,f = 59.25 inches

Hence, the combined mean height of the factory is 59.25 inches.



Page No 8.55:

Question 23:

The mean age of a combined group of men and women is 30.5 years. If the mean age of the group of men is 35 and that of the group of women is 25, find out the percentage of men and women in the group.

Answer:

Let the percentage of male in the combined mean be x
∵ percentage of women = (100 − x) 

Note: Total of the percentage= 100

Given:

Xm,w=30.5Xm=35Xw=25

We know,
Combined Mean  Xm,w =Nm× Xm +Nw×Xw Nm+Nwor, 30.5=35× x+25× (100-x)100or, 3050=35 x-25 x + 2500or, 10 x =550or, x=55010 x=55

∴ Males = 55%
∴ Women= (100 − x) = (100 − 55) = 45%

Therefore, there are 55% men and 45% women in the group.

Page No 8.55:

Question 24:

The mean weight of 150 students in a class is 60 kg. The mean of boys in the class is 70 kg and that of girls is 55 kg. Find the number of boys and girls in the class.

Answer:

Given:

Nb + Ng= 150     (Equation-1)

Xb, g= 60 kgXb=70 kgXg=55 kg
                                
Ng
= 150 − Nb      (From eqn-1)

We know,
Combined Mean (Xb, g) = Nb×Xb+Ng×XgNb+Ngor, 60=70Nb+55150-Nb1509000=70Nb-55Nb +825015Nb=750Nb=50Ng=150-Nb=150-50=100

Therefore,

No. of boys = Nb= 50
No. of Girls = Ng= 100

Page No 8.55:

Question 25:

The average weight of a group of 20 boys was calculated to be 89.4. It was later discovered that one weight was misread as 78 kg instead of the correct on of 87 kg. Calculate the correct average weight.

Answer:


Given:

Mean, X=89.4
Number of observations, N=20
Incorrect observation= 78
Correct observation = 87

Wrong X=Wrong ΣXNor, 89.4=Wrong ΣX20Wrong ΣX=1788

Thus, wrong summation of X, i.e., ΣXwrong is 1788.
Now, Correct mean is calculated using the following formula.

Correct X = ΣXwrong+corrected value-incorrect valueNSubstituting the given values in the formula.Correct X =1788-78+8720=179720=89.95

Hence, the correct average weight is 89.95 kgs

Page No 8.55:

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.

Answer:

Given:

Mean, X=40
Number of observations, N=100
Incorrect observations= 30 and 27
Correct observations = 3 and 72
Wrong X=Wrong ΣXNor, 89.4=Wrong ΣX20Wrong ΣX=1
( X)wrong=(ΣX)wrongNor, 40 =(ΣX)wrong100(ΣX)wrong=40×100=4000

Thus, wrong summation of X, i.e., (ΣX)wrong=4000
Now, correct mean is calculated using the following formula.

Correct X=ΣX(wrong)+correct values-incorrect valuesNSubstituting the given values in the formula.Correct X=4000-30-27+3+72100Correct X=4018100=40.18

Hence, the correct mean is 40.18

Page No 8.55:

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.

Answer:

Given:

Mean, X= 155
Number of observations =40
Incorrect observation= 157
Correct Observation = 137
  
Xwrong=ΣXwrongNor, 155=ΣXwrong40or, ΣXwrong= 155×40 ΣXwrong =6200

Thus, wrong summation of X, i.e., ΣX(wrong) is 6200
Now, correct mean is calculated using the following method.

Correct X =ΣXwrong+correct value-incorrect valueNSubstituting the given values in the formula.Correct X= 6200+137-15740or, Correct X=618040 Correct X = 154.5 

Hence, the correct average height is 154.5 cm

Page No 8.55:

Question 28:

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
  ΣW=12 ΣWX=745


XW=ΣWXΣW

Xw=74512Xw=62.08

Hence, the weighted mean is 62.08 marks

Page No 8.55:

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
  ΣW=59   ΣWX= 2325

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

XW=ΣWXΣWor, XW=232559 XW=39.40

Hence, the weighted mean of the above series is 39.40

Page No 8.55:

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, RamXR=ΣWX1ΣW=63310=63.3Weighted Mean Marks, ShyamXS=ΣWX2ΣW=62410=62.4Weighted Mean Marks, ManishXM=ΣWX3ΣW=64810=64.8

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



Page No 8.56:

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
  Σf=60 ΣfX=1960 + 10x

Given:

Mean, X= 41

X=ΣXΣf41=1960+10x60or, 41×60=1960+10xor, 2460-1960=10xor, 500=10xor, x=50010 x =50 marks

Hence, the missing item is 50 marks

Page No 8.56:

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
    Σf=80 + f Σfm=2100 + 35f

Let the missing frequency be f

Given:
Mean, X=28

X=ΣfmΣfor, 28=2100+35f80+for, 2240+28f=2100+35for, 7f=140f=1407=20

Hence, the missing frequency is 20.

Page No 8.56:

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
    Σf=47+f Σfm=1135 + 45 f

Let missing frequency be f

Given:

Mean, X =25.4

X=ΣfmΣfor, 25.4=1135+45f47+for, 1193.8 + 25.4f=1135 +45for, 58.8 =19.6 ff=3

Hence, missing frequency is 3

Page No 8.56:

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
    Σf=240 Σfm=11,900

Calculating average marks using direct method:

X=ΣfmΣfor, X=11900240 X=49.58 marks

Hence, the average marks are 49.58

Page No 8.56:

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
    Σf=200 Σfm=7160

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

X=ΣfmΣfor, X =7160200 X = 35.8

Hence, the mean of the above series is 35.8

Page No 8.56:

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=ΣXN

ΣX of 20 students = X20×N= 20 × 16 = 320
ΣX of 19 students = ΣX20- age of the student who left the class=320 − 18 = 302

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

X of 19 students=30219=15.89 years

Hence, the correct average age is 15.89 years

Page No 8.56:

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
  Σf=43 + f ΣfX=2030 + 30f

Let missing frequency be f

Given:
Mean, X=44.8

We know,

X=ΣfXΣfor, 44.8=2030+30f43+for, 1926.4+44.8f=2030+30for, 14.8f=103.6f=103.614.8=7

Hence, the missing frequency is 7



Page No 8.57:

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
Mean, X30=52ΣX30=X30× Nor, ΣX30 =30×52 ΣX30 =1560

Average of top 6 students = 31

ΣX of top 6 students, ΣX6= 31 × 6 = 186

ΣX of remaining 24 (30-6) students, ΣX24 = ΣX30-ΣX6= 1560 − 186 = 1374

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

X of 24 students=ΣX24Nor, X24 =137424  X24= 57.25 marks

Page No 8.57:

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
  Σf=70 ΣfX=612

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

X=ΣfXΣfor, X =61270 X = 8.74

Hence, the mean of the above series is 8.74

Page No 8.57:

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
    ΣW=10 ΣWX=615


Xw=ΣWXΣWor, Xw=61510Xw=61.5 Marks

Hence, the weighted arithmetic mean is 61.5 marks

Page No 8.57:

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
  Σf=30 ΣfX30=2024

Given:
Σf = 50

X50=68

 ΣfX50=X50×Σf=68×50=3400
ΣfX of remaining 20 students

ΣfX20= ΣfX50-ΣfX30=3400 − 2024 = 1376

X of 20 students=ΣfX20Σf=137620=68.8 inches
Hence, mean height of 20 students is 68.8 inches



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