10,9,12,11,8,15,9,7,8,6,12,10
- Solution:
Let the observation be X.
Sum of the observation = n .
There for n = 12.
Observation = 10+9+12+11+8+15+9+7+8+6+12+10
Ther for ∑xi = 117.
A.M. = X = ∑xi
n
= 117
12
X = 9.75 -------------------------Ans.
Q.2 A relay race was completed by 7 participants, & their race times are given below.(in seconds) What is the range of race time?
13.2, 14.5, 12.9,13.9,15.6,14.1,12.3
Sum of the observation = n .
There for n = 12.
Observation = 10+9+12+11+8+15+9+7+8+6+12+10
Ther for ∑xi = 117.
A.M. = X = ∑xi
n
= 117
12
X = 9.75 -------------------------Ans.
Q.2 A relay race was completed by 7 participants, & their race times are given below.(in seconds) What is the range of race time?
13.2, 14.5, 12.9,13.9,15.6,14.1,12.3
- Solution:
Let the observation be X.
Sum of the observation = n .
There for n = 7.
Observation = 13.2+14.5+12.9+13.9+15.6+14.1+12.3
Ther for ∑xi = 96.5.
A.M. = X = ∑xi
n
= 96.5
7
X = 13.7857 -------------------------Ans.
Q.3 What is the mean test score for the six scores listed below?
89,93,87,86,85,94
Sum of the observation = n .
There for n = 7.
Observation = 13.2+14.5+12.9+13.9+15.6+14.1+12.3
Ther for ∑xi = 96.5.
A.M. = X = ∑xi
n
= 96.5
7
X = 13.7857 -------------------------Ans.
Q.3 What is the mean test score for the six scores listed below?
89,93,87,86,85,94
- Solution:
Let the observation be X.
Sum of the observation = n .
There for n = 6.
Observation = 89+93+87+86+85+94
There for ∑xi = 534.
A.M. = X = ∑xi
n
= 534
6
X = 89. -------------------------Ans.
Q.4 The average annual wind speed for the 5 windiest cities in the U.S. is given below in miles per hour. What is the mean of these annual wind speeds?
15.4 , 14.0 , 13.5 , 13.1 , 12.9
Sum of the observation = n .
There for n = 6.
Observation = 89+93+87+86+85+94
There for ∑xi = 534.
A.M. = X = ∑xi
n
= 534
6
X = 89. -------------------------Ans.
Q.4 The average annual wind speed for the 5 windiest cities in the U.S. is given below in miles per hour. What is the mean of these annual wind speeds?
15.4 , 14.0 , 13.5 , 13.1 , 12.9
- Solution:
Let the observation be X.
Sum of the observation = n .
There for n = 7.
Observation = 15.4 +14.0 +13.5 +13.1 +12.9
Ther for ∑xi = 68.9
A.M. = X = ∑xi
n
= 68.9
5
X = 13.78 -------------------------Ans.
Q.5 Calculate arithmetic mean from the following data :
Marks : (less than) 80 70 60 50 40 30 20
No of students : 100 90 80 60 32 20 13
Sum of the observation = n .
There for n = 7.
Observation = 15.4 +14.0 +13.5 +13.1 +12.9
Ther for ∑xi = 68.9
A.M. = X = ∑xi
n
= 68.9
5
X = 13.78 -------------------------Ans.
Q.5 Calculate arithmetic mean from the following data :
Marks : (less than) 80 70 60 50 40 30 20
No of students : 100 90 80 60 32 20 13
- Solution:
Let the observation be X.
Sum of the marks = n .
There for n = f.
Let students be f.
Marks : (less than) No of students ∑xfi = x*f
X f
80 100 80*100 = 8000
70 90 70*90 = 6300
60 80 60*80 = 4800
50 60 50*60 = 3000
40 32 40*32 = 1280
30 20 30*20 = 600
20 13 20*13 = 260
10 5 10*5 = 50
n = 400 ∑xfi = 24290
A.M. = X = ∑xfi
n
= 24290
400
X = 60.725 ------------------------Ans.
Q.6 The monthly profits in $ of 100, shop are distributed as follows:
Profit per shop : 0 -100 0-200 0-300 0– 400 0-500 0-600
No of shop : 12 30 57 77 94 100
Sum of the marks = n .
There for n = f.
Let students be f.
Marks : (less than) No of students ∑xfi = x*f
X f
80 100 80*100 = 8000
70 90 70*90 = 6300
60 80 60*80 = 4800
50 60 50*60 = 3000
40 32 40*32 = 1280
30 20 30*20 = 600
20 13 20*13 = 260
10 5 10*5 = 50
n = 400 ∑xfi = 24290
A.M. = X = ∑xfi
n
= 24290
400
X = 60.725 ------------------------Ans.
Q.6 The monthly profits in $ of 100, shop are distributed as follows:
Profit per shop : 0 -100 0-200 0-300 0– 400 0-500 0-600
No of shop : 12 30 57 77 94 100
- Solution:
Let the profit per shop be X.
Let the no of shop be = f .
Sum of no of shop = n.
There for n = ∑f.
Sum of profit per shop & no of shop is a product = ∑xfi.
There for ∑xfi = x*f.
Profit per shop : No of shop : Mead value of ∑xfi =
X f X X*f
0-100 12 50 12*50 = 600
0-200 30 100 30*100 = 3000
0-300 57 150 57*150 = 8550
0-400 77 200 77* 200 = 15400
0-500 94 250 94*250 = 23500
0-600 100 300 100*300 = 30000
n = 370 ∑xfi = 81050
A.M. = X = ∑xfi
n
= 81050
370
X = 219.054 ------------------------Ans.
Q.7
Income Group : 150-300 300-500 500-800 800– 1200 1200-1800
No of firms : 40 32 26 28 42
Average no : 8 12 7.5 8.5 4
Find the average salary pain in the whole market.
Let the no of shop be = f .
Sum of no of shop = n.
There for n = ∑f.
Sum of profit per shop & no of shop is a product = ∑xfi.
There for ∑xfi = x*f.
Profit per shop : No of shop : Mead value of ∑xfi =
X f X X*f
0-100 12 50 12*50 = 600
0-200 30 100 30*100 = 3000
0-300 57 150 57*150 = 8550
0-400 77 200 77* 200 = 15400
0-500 94 250 94*250 = 23500
0-600 100 300 100*300 = 30000
n = 370 ∑xfi = 81050
A.M. = X = ∑xfi
n
= 81050
370
X = 219.054 ------------------------Ans.
Q.7
Income Group : 150-300 300-500 500-800 800– 1200 1200-1800
No of firms : 40 32 26 28 42
Average no : 8 12 7.5 8.5 4
Find the average salary pain in the whole market.
- Solution:
Let the no of firm be X.
Let the average no of workers be = f .
Let sum of the average no of workers be = n.
There for n = 40.
Sum of No of firms & average no of workers is a product = ∑xfi.
There for ∑xfi = x*f.s
No of firms : Average no of workers: ∑xfi =
X f X*f
40 8 40*8 = 320
32 12 32*12 = 384
26 7.5 26*7.5 = 195
28 8.5 28* 8.5 = 238
42 4 42*4 = 168
n = 40 ∑xfi =1305
A.M. = X = ∑xfi
n
= 1305
40
X = 32.62500 ------------------------Ans.
Q.8 Find the missing frequency from the following data:
Marks Frequency
5-10 12
10-15 16
15-20 5
20-25 14
25-30 10
30-35 8
Let the average no of workers be = f .
Let sum of the average no of workers be = n.
There for n = 40.
Sum of No of firms & average no of workers is a product = ∑xfi.
There for ∑xfi = x*f.s
No of firms : Average no of workers: ∑xfi =
X f X*f
40 8 40*8 = 320
32 12 32*12 = 384
26 7.5 26*7.5 = 195
28 8.5 28* 8.5 = 238
42 4 42*4 = 168
n = 40 ∑xfi =1305
A.M. = X = ∑xfi
n
= 1305
40
X = 32.62500 ------------------------Ans.
Q.8 Find the missing frequency from the following data:
Marks Frequency
5-10 12
10-15 16
15-20 5
20-25 14
25-30 10
30-35 8
- Solution:
Let the marks be X.
Let the frequency be = f .
Sum of frequency = n.
There for n = ∑f.
Sum of marks & frequency is a product = ∑xfi.
There for ∑xfi = x*f.
Marks: Frequency : Mead value of ∑xfi =
X f X : x*f.
5-10 12 7.5 12*705 = 90
10-15 16 12.5 16*1205 = 200
15-20 5 17.5 5*1705 = 87.5
20-25 14 22.5 14*22.5 = 315
25-30 10 27.5 10*27.5 = 275
30-35 8 32.5 8* 32.5 = 260
n = 65 ∑xfi = 1227.5
A.M. = X = ∑xfi
n
= 1227.5
65
X = 18.88462 ------------------------Ans.
Q.9 The following is the frequency distribution of the marks obtained by 250 students in an examination. Compute the mean, median & mode.
Marks No of
Obtained : Students:
0-10 15
10-20 20
20-30 24
30-40 24
40-50 12
50-60 31
60-70 71
70-80 52
Let the frequency be = f .
Sum of frequency = n.
There for n = ∑f.
Sum of marks & frequency is a product = ∑xfi.
There for ∑xfi = x*f.
Marks: Frequency : Mead value of ∑xfi =
X f X : x*f.
5-10 12 7.5 12*705 = 90
10-15 16 12.5 16*1205 = 200
15-20 5 17.5 5*1705 = 87.5
20-25 14 22.5 14*22.5 = 315
25-30 10 27.5 10*27.5 = 275
30-35 8 32.5 8* 32.5 = 260
n = 65 ∑xfi = 1227.5
A.M. = X = ∑xfi
n
= 1227.5
65
X = 18.88462 ------------------------Ans.
Q.9 The following is the frequency distribution of the marks obtained by 250 students in an examination. Compute the mean, median & mode.
Marks No of
Obtained : Students:
0-10 15
10-20 20
20-30 24
30-40 24
40-50 12
50-60 31
60-70 71
70-80 52
- Solution:
Let the marks be X.
Let the no of students be = f .
Sum of no of students = n.
There for n = ∑f.
Sum of marks & no of students is a product = ∑xfi.
There for ∑xfi = x*f.
Marks: No of students : Mead value of ∑xfi =
X f X : x*f.
0-10 15 5 15*5 = 75
10-20 20 15 20*15 = 300
20-30 24 25 24*25 = 600
30-45 24 35 24* 35 = 840
40-50 12 45 12*45 = 540
50-60 31 55 31*55 = 1705
60-70 71 65 71*65 = 4615
70-80 52 75 52*75 = 3900
n = 250 ∑xfi = 12575
A.M. = X = ∑xfi
n
= 12575
250
X = 50.3 ------------------------Ans.
Let the no of students be = f .
Sum of no of students = n.
There for n = ∑f.
Sum of marks & no of students is a product = ∑xfi.
There for ∑xfi = x*f.
Marks: No of students : Mead value of ∑xfi =
X f X : x*f.
0-10 15 5 15*5 = 75
10-20 20 15 20*15 = 300
20-30 24 25 24*25 = 600
30-45 24 35 24* 35 = 840
40-50 12 45 12*45 = 540
50-60 31 55 31*55 = 1705
60-70 71 65 71*65 = 4615
70-80 52 75 52*75 = 3900
n = 250 ∑xfi = 12575
A.M. = X = ∑xfi
n
= 12575
250
X = 50.3 ------------------------Ans.
