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Page No 165:

Question 1:

Calculate arithmetic averages of the following information:
(a) Marks obtained by 10 students:
30, 62, 47, 25, 52, 39, 56, 66, 12, 31
(b) Income of 7 families (in ₹): Also show Σ X-X¯ = 0
550, 490, 670, 890, 435, 590, 575
(c) Height of 8 students ( in cm)  :
140, 145, 147, 152, 148, 144, 150, 158
 

Answer:

(a) 

X 30 62 47 25 52 39 56 66 12 31

ΣX = 420
X=ΣXN=42010=42
Thus, mean marks is 42

(b)
X
550
490
670
890
435
590
575
ΣX = 4200

X=ΣXN=42007=600
Thus, mean income is equal to Rs 600.
 
X X - X
550
490
670
890
435
590
575
-50
-110
70
290
-165
-10
-25
  Σ (X - X) = 0

(c)
S.No. X
1
2
3
4
5
6
7
8
140
145
147
152
148
144
150
158
  ΣX = 1184

X=ΣXN=11848=148
Thus, mean height is equal to 148 cm.

Page No 165:

Question 2:

Batsman A, B , C and D  played three matches . Calculate the average runs scored by each batsman.

Name of batsman Match I Match II Match III
I II I II I II
  Inning Inning Inning Inning Inning Inning
A 60 20 26 10 100 40
B 40 50 60 36 70 80
C 100 10 8 18 100 140
D 20 40 46 84 42 52

 

Answer:

The data given for the marks scored by the batsmen in different innings can be summarised as follows.

Inning Score by batsman A
(XA)
Score by batsman B
(
XB)
Score by batsman C
(
XC)
Score by batsman D
(
XD)
1
2
3
4
5
6
60
20
26
10
100
40
40
50
60
36
70
80
100
10
8
18
100
140
20
40
46
84
42
52
  ΣXA = 256 ΣXB = 336 ΣXC = 376 ΣXD = 284

XA=ΣXAN=2566=42.67XB=ΣXBN=3366=56XC=ΣXCN=3766=62.67XD=ΣXDN=2846=47.33

Page No 165:

Question 3:

Calculate mean of the following series:

Size : 4 5 6 7 8 9 10
Frequency : 6 12 15 28 20 14 5
 

Answer:

X f fX
4
5
6
7
8
9
10
6
12
15
28
20
14
5
24
60
90
196
160
126
50
  Σf = 100 ΣfX = 706

X=ΣfXΣf=706100=7.06

Thus, mean of the series is 7.06.

Page No 165:

Question 4:

Find out mean of sales and expenses of following 10 firms:

Firms : 1 2 3 4 5 6 7 8 9 10
Sales (₹ in '000) : 50 50 55 60 65 65 65 60 60 50
Expenses (₹ in '000) : 11 13 14 16 16 15 15 14 13 13
 

Answer:

Firm Sales
(X)
(Rs in '000)
Expenses
(Y)
(Rs in '000)
1
2
3
4
5
6
7
8
9
10
50
50
55
60
65
65
65
60
60
50
11
13
14
16
16
15
15
14
13
13
  ΣX = 580 ΣY = 140

X=ΣXN=58010=58
So, mean sales is Rs 58,000

Y=ΣYN=14010=14
So, mean expenses is Rs 14,000

Page No 165:

Question 5:

Calculate mean of the following frequency distribution:

Values : 60 62 64 67 70 73 77 81 85 89
Frequency : 54 82 103 176 212 180 115 78 50 21
 

Answer:

X Frequency
(f)
fX
60
62
64
67
70
73
77
81
85
89
54
82
103
176
212
180
115
78
50
21
3240
5084
6592
11792
14840
13140
8855
6318
4250
1869
  Σf = 1071 ΣfX = 75980

X=ΣfXΣf=759801071=70.94
Thus, the mean for the given data is Rs 70.94.

Page No 165:

Question 6:

Calculate average of the following series:

X : 4 6 10 14 18 22
f : 7 9 16 8 6 4
 

Answer:

X f fX
4
6
10
14
18
22
7
9
16
8
6
4
28
54
160
112
108
88
  Σ= 50 ΣfX = 550

X=ΣfXΣf=55050=11
Thus, the mean is 11.



Page No 166:

Question 7:

Calculate arithmetic mean of the following data:

Profit ( in ₹) : 0-10 10-20 20-30 30-40 40-50 50-60
No of shops : 12 18 27 20 17 16
 

Answer:

Profit
(in Rs)

 
Mid Value
(m)
No. of Shops
Frequency
(f)
fm
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
5
15
25
35
45
55
12
18
27
20
17
16
60
270
675
700
765
880
    Σf = 110 Σfm = 3350

X=ΣfmΣf=3350110=30.45
Thus, the mean profit is Rs 30.45.

Page No 166:

Question 8:

Calculate the arithmetic mean from the following data:

Marks Less than 10 Less than 20 Less than 30 Less than 40  Less than 50
No. of Students 5 15 55 75 100

Answer:

Marks Mid Value
(m)
No. of student
(f)
fm
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
5
15
25
35
45
5
10
40
20
25
25
150
1000
700
1125
    Σf = 100 Σfm = 3000

For the calculation of mean first the given less than series is converted in continuous class intervals as above.

X=ΣfmΣf=3000100=30
Thus, mean is 30 marks.
 
Marks Mid Value
(m)
No. of student
(f)
X - X fX - X
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
5
15
25
35
45
5
10
40
20
25
-25
-15
-5
5
15
-125
-150
-200
100
375
    Σf = 100   ΣfX - X =0

Page No 166:

Question 9:

A candidate obtains 46 marks in English , 67 in Mathematics , 53 in Hindi , 72 in History and 58 in Economics. It is agreed to give triple weights to marks in English and Double weights to marks in Mathematics  as compared to other subjects . Calculate Weighted Mean. Also, compare it with simple Arithmetic Mean.

Answer:

Subject Mark (%)
(X)
Weight
(W)
WX
English
Mathematics
Hindi
History
Economics
46
67
53
72
58
3
2
1
1
1
138
134
53
72
58
  ΣX = 296 ΣW = 8 ΣWX = 455

The given information can be summarised as above.

Simple Arithmetic Mean X=ΣXN=2965=59.2Weighted Arithmetic MeanXw=ΣWXΣW=4558=56.9
Thus, the weighted arithmetic mean is greater than the simple arithmetic mean.

Page No 166:

Question 10:

There are two branches of an establishment employing 100 and 80 persons respectively. If the Arithmetic Means of the monthly salaries by the two branches are ₹ 275 and ₹ 225 respectively, find out the arithmetic mean of the salaries of the employees of the establishment as a whole.

Answer:

The given information can be summarised as follows.

N1=100, X1=275N2=80, X2=225X1,2=N1 X1+N2 X2N1+N2    =100 275+80 225100+80    =27500+18000180=45500180    =252.8
Thus, the combined mean salary is Rs 252.8

Page No 166:

Question 11:

The mean marks obtained in an examination by a group of 100 students were found to be 49.46. The mean marks obtained in the same examination by another group of 200 students were 52.32. Find out the mean of marks obtained by both  the groups of students taken together.

Answer:

The given information can be summarised as follows.
N1=100, X1=49.46N2=200, X2=52.32X1,2=N1 X1+N2 X2N1+N2    =100 49.46+200 52.32100+200    =4946+10464300=15410300=51.37Combined Mean X1,2=51.37 Marks

Thus, the mean marks of both the groups taken together is 51.37.

Page No 166:

Question 12:

The mean marks of 100 students were found to be 40 . Later on it was discovered that a score of 53 was misread as 83. Find the corrected mean corresponding to the corrected score .

Answer:

Given:
Incorrect Mean = 40
Number of students = 100
Correct value = 53
Incorrect value = 83

From the given information incorrect summation is calculated as follows:XI=ΣXIN40=ΣXI100ΣXI=4000Now, Correct ΣX = Incorrect summation X (ΣXI)  -Incorrect value +Correct valueCorrect ΣXC=4000-83+53=3970XC=ΣXCN=3970100=39.70
Thus, the corrected mean marks is 39.7.

Page No 166:

Question 13:

The mean weight of 25 boys in group A of a class is 61 Kg and the mean weight of 35 boys in group B of the same class is 58 Kg. Find the mean weight of 60 boys.

Answer:

The given information can be summarised as follows:

N1=25, X1=61N2=35, X2=58Combined Mean X1,2=N1 X1+N2X2N1+N2                                =2561+355825+35                                =1525+203060=355560                                =59.25 kg
Thus, the combined mean weight is 59.25 kg

Page No 166:

Question 14:

Calculate mean of the following data:

Marks Below : 10 20 30 40 50 60 70
No. of Students : 5 9 17 29 45 60 70
 

Answer:

Marks Mid Value
(m)
f fm
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
60 − 70
5
15
25
35
45
55
65
5
9 − 5 = 4
17 − 9 = 8
29 − 17 = 12
45 − 29 = 16
60 − 45 = 15
70 − 60 = 10
25
60
200
420
720
825
650
    Σf = 70 Σfm = 2900

For the calculation of mean, first the given less than series is converted in continuous series as above.

X=ΣfmΣf=290070=41.42
Thus, the mean marks is 41.42.

Page No 166:

Question 15:

Calculate Combined Mean:

Section Mean Marks No. of Students
A 75 50
B 60 60
C 55 50
 

Answer:

NA=50, XA=75NB=60, XB=60NC=50, XC=55XA,B,C=NA XA+NBXB+NC XCNA+NB+NC=5075+6060+505550+60+50       =3750+3600+2750160=10100160=63.125
Thus, the combined mean marks is 63.125.

Page No 166:

Question 16:

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

Answer:

The given information can be summarised as follows:

Combined average marks = 52
N1=6, X1=31X2=?, N2=24X1,2=N1X1+N2X2N1+N252 =631+24X26+2452×30=186+24X21560-186=24X224X2=1374X2=137424=57.25
So, average marks of the remaining 24 students is 57.25.



Page No 167:

Question 17:

The mean salary paid to 1000 workers of a factory was found to be ₹ 180.4 . Later on it was discovered that the wages of two workers were wrongly taken as 297 and 165 instead of 197 and 185. Find the correct mean.

Answer:

Given:
Number of workers = 1000
Incorrect values = 297, 165
Correct values = 197, 185
Incorrect mean = 180.4

Incorrect Mean, XI=ΣXIN180.4=ΣXI1000ΣXI=180400

Correct ΣXCXI + correct values - incorrect values
                     = ΣXI + 197 + 185 − 297 − 165
                     = 180400 + 197 + 185 − 297 − 165
                     = 180320
Correct X=ΣXCN=1803201000=180.32
Thus, the correct mean wage is Rs 180.32.

Page No 167:

Question 18:

Calculate arithmetic measure from the following data:

Temp.( 0C) No. of days (f)
−40 to −30 10
−30 to −20 28
−20 to −10 30
−10 to 0 42
0 to 10 65
10 to 20 180
20 to 30 10
  Σf = 365
 

Answer:

Temprature Mid Value
(m)
No. of Days
(f)
fm
(−40) − (−30)
(−30) − (−20)
(−20) − (−10)
(−10) − 0
0 − 10
10 − 20
20 − 30
−35
−25
−15
−5
5
15
25
10
28
30
42
65
180
10
−350
−700
−450
−210
325
2700
250
    Σf = 365 Σfm = 1565

X=ΣfmΣf=1565365=4.29°C

Thus, the mean temperature us 4.29oC.

Page No 167:

Question 19:

A candidate obtains the following percentage of marks: Sanskrit 75, Mathematics 84, Economics 56, English 78, Politics 57, History 54, Geography 47,. It is agreed to give double weights to marks in English , Mathematics and Sanskrit . What is the weighted and simple arithmetic mean?

Answer:

Subject Marks
(X)
Weight
(W)
WX
Sanskrit
Mathematics
Economics
English
Politics
History
Geography
75
84
56
78
57
54
47
2
2
1
2
1
1
1
150
168
56
156
57
54
47
  ΣX = 451 ΣW = 10 ΣWX = 688

X=ΣXN=4517=64.43 MarksXW=ΣWXΣW=68810=68.8 Marks
Thus, the weighted mean and simple arithmetic mean are 68.8 and 64.43 respectively.

Page No 167:

Question 20:

Calculate weighted mean by weighting each price by the quantity consumed:

Food items Quantity Consumed Price in Rupees (per Kg)
Flour 500 kg 1.25
Ghee 200 kg 20.00
Sugar 30 kg 4.50
Potato 15 kg 0.50
Oil 40 kg 5.50
 

Answer:

Food Item Price
(RS)
(X)
Quantity Consumed
(W)
WX
Flour
Ghee
Sugar
Potato
Oil
1.25
20
4.50
0.50
5.50
500 kg
200 kg
30 kg
15 kg
40 kg
625
4000
135
7.5
220
    ΣW = 785 ΣWX = 4987.5

XW=ΣWXΣW=4987.5785=6.35
Thus, the weighted mean is Rs 6.35.

Page No 167:

Question 21:

Comment on the performance of the students of three universities given below using weighted mean:

No. of Students are in hundreds
Courses of study Mumbai Kolkata Chennai

pass
No. of students %
pass
No. of Students
pass
No. of Students
M.A 71 3 82 2 81 2
M.COM. 83 4 76 3 76 3.5
B.A 73 5 73 6 74 4.5
B.COM. 74 2 76 7 58 2
B.Sc 65 3 65 3 70 7
M.Sc 66 3 60 7 73 2

 

Answer:

Courses Mumbai Kolkata Chennai
X1 W1 X1W1 X2 W2 X2W2 X3 W3 X3W3
M.A
M.Com
B.A
B.Com
B.Sc
M.Sc
71
83
73
74
65
66
3
4
5
2
3
3
213
332
365
148
195
198
82
76
73
76
65
60
2
3
6
7
3
7
164
228
438
532
195
420
81
76
74
58
70
73
2
3.5
4.5
2
7
2
162
266
333
116
490
146
    ΣW1 = 20 ΣX1W1 = 1451   ΣW2 = 28 ΣX2W2 = 1977   ΣW3 = 21 ΣX3W3 = 1513

XW1=ΣX1W1ΣW1=145120=72.55XW2=ΣX2W2ΣW2=197728=70.6XW3=ΣX3W3ΣW3=151321=72.04
Average score of Mumbai is more than Kolkata and Chennai. So performance of Mumbai is better.

Page No 167:

Question 22:

A distribution consists of three components with total frequencies of 200, 250 and 300 having means of 25, 10 and 15 respectively . Find out the mean of combined distribution.

Answer:

N1=200, X1=25N2=250, X2=10N3=300, X3=15X1,2,3=X1 N1+X2 N2+X3 N3N1+N2+N3     =25×200+10×250+15×300200+250+300    =5000+25000+45000750    =12000750=16
Thus, mean of combined distribution is 16.



Page No 168:

Question 23:

Find the average age of 500 people in a town.

Age (in Years) 0-10 10-20 20-30 30-60 60-90
No. of people 50 90 200 120 40
 

Answer:

Age
(in years)

 
Mid Value
(m)
No. of People
Frequency
(f)
fm
0 − 10
10 − 20
20 − 30
30 − 60
60 − 90
5
15
25
45
75
50
90
200
120
40
250
1350
5000
5400
3000
    Σf = 500 Σfm = 15000

X=ΣfmΣf=15000500=30
Thus, the mean is 30 years.

Page No 168:

Question 24:

Calculate mean of the following data:

Values more than 0 10 20 30 40
Frequency 100 95 85 45 25
 

Answer:

Class Intervals
 
c.f. Mid Value
(m)
Frequency
(f)
fm
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
100
95
85
45
25
5
15
25
35
45
5
10
40
20
25
25
150
1000
700
1125
      Σf = 100 Σfm = 3000

X=ΣfmΣf=3000100=30
Thus, the mean is 30.

Page No 168:

Question 25:

Find the missing value , if mean is 23.

X 5 15 ? 35 45
Frequency 4 8 10 6 2
 

Answer:

X
 
Frequency
(f)
fX
5
15
? (x)
35
45
4
8
10
6
2
20
120
10x
210
90
  Σf = 30 ΣfX = 440+10x

X=23X=ΣfXΣf23=440+10x30690=440+10x250=10xx=25010=25
Thus, the missing value of X is 25.

Page No 168:

Question 26:

Find the missing frequency, if mean is 62.

C.I 0-40 40-80 80-120 120-160
Frequency 6 9 ? 2
 

Answer:

C.I.
 
Mid-Point
(m)
Frequency
(f)
fm
0 − 40
40 − 80
80 − 120
120 − 160
20
60
100
140
6
9
? (f)
2
120
540
100f
280
    Σf = 17+f Σfm = 940+100f

X=62X=ΣfmΣf62=940+100f17+f1054+62f=940+100f38f=114f=11438=3
Thus, the missing freuency is 3.

Page No 168:

Question 27:

Calculate arithmetic mean:

Marks (More than) 0 10 20 30 40 50
No. of Students 100 88 70 43 23 6

Answer:

Marks Mid Value
(m)
f fm
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
5
15
25
35
45
55
100 − 88 = 12
88 − 70 = 18
70 − 43 = 27
43 − 23 = 20
23 − 6 = 17
6
60
270
675
700
765
330
    Σf = 100 Σfm = 2800

For the calculation of mean, first the given more than series is converted in continuous series as above.

X=ΣfmΣf=2800100=28
Thus, the mean marks is 28.

Page No 168:

Question 28:

Calculate mean of the following data:

Values below 10 20 20 40 50
Frequency 20 44 84 120 140
 

Answer:

Marks Mid Value
(m)
f fm
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
5
15
25
35
45
20
44 − 20 = 24
84 − 44 = 40
120 − 84 = 36
140 − 120 = 20
100
360
1000
1260
900
    Σf = 140 Σfm = 3620

For the calculation of mean, first the given less than series is converted in continuous series as above.

X=ΣfmΣf=3620140=25.86
Thus, the mean is 25.86.

Page No 168:

Question 29:

Calculate missing frequency, if mean of the distribution is 28.

C.I 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 12 ? 27 20 17 6
 

Answer:

C.I.
 
Mid-Point
(m)
Frequency
(f)
fm
0 − 10
10 − 20
20 − 30
30 − 40
40 − 50
50 − 60
5
15
25
35
45
55
12
? (f)
27
20
17
6
60
15f
675
700
765
330
    Σf = 82+f Σfm = 2530+15f

X=28X=ΣfmΣf28=2530+15f82+f2296+28f=2530+15f13f=234f=23413=18
Thus, the missing freuency is 18.

Page No 168:

Question 30:

Calculate mean from the information given below.

C.I 5-6 8-10 11-13 14-16 17-19
Frequency 10 8 7 3 2
 

Answer:

C.I.
 
Mid-Point
(m)
Frequency
(f)
fm
4.5 − 7.5
7.5 − 10.5
10.5 − 13.5
13.5 − 16.0
16.5 − 19.50
6
9
12
15
18
10
8
7
3
2
60
72
84
45
36
    Σf = 30 Σfm = 297

X=ΣfmΣfX=29730=9.9
Thus, the mean is 9.9.



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