The arithmetic mean (A.M) or simply the mean or average of n observations x1, x2, …, xn is defined to be the number x such that the sum of the deviations of the observations from x is 0. That is, the arithmetic mean x of n observations x1, x2, …, xn is given by the equation
(x1 − x) +(x2 − x) + ... +(xn − x) = 0 or (x1 +x2 + ... +xn )- n x = 0
Mean =[(x1 +x2 + ... +xn )]/n
1: Calculate the mean of the data 9, 11, 13, 15, 17, 19.
2. Compute the A.M. of the following data:
x
|
10
|
11
|
13
|
15
|
16
|
19
|
f
|
4
|
5
|
8
|
6
|
4
|
3
|
3. Calculate the A.M. for the following data:
Marks
|
80
|
85
|
90
|
95
|
100
|
No. of students
|
5
|
6
|
6
|
2
|
1
|
4. If A .M for the following data is 28 find x
Class Interval
|
0-10
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10-20
|
20-30
|
30-40
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40-50
|
50-60
| |
Marks
|
12
|
18
|
x
|
20
|
17
|
6
|
N=100
|
5. Find the median of 23, 25, 29, 30, 39.
6. Find the median of 3, 4, 10, 12, 27, 32, 41, 49, 50, 55, 60, 63, 71, 75, 80.
7. Find the median of 29, 23, 25, 29, 30, 25, 28.
8. Calculate the median of the following table:
Variable ( x)
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5
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10
|
15
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20
|
25
|
30
|
Frequency( f )
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3
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6
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10
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8
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2
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3
|
9. Find the mode of 7, 4, 5, 1, 7, 3, 4, 6,7.
10. Find the mode for 12, 15, 11, 12, 19, 15, 24, 27, 20, 12, 19, 15.
11. Find the mode from the following frequency table:
Wage
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45
|
50
|
55
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60
|
65
|
70
|
75
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No. of Employees
|
12
|
11
|
14
|
13
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12
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10
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9
|
12. Compute the A.M. of the following data
Compute the A.M. of the following data: x
|
10
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11
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13
|
15
|
16
|
19
|
f
|
4
|
5
|
8
|
6
|
4
|
3
|
x
|
f
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f × x
|
10
|
4
|
40
|
11
|
5
|
55
|
13
|
8
|
104
|
15
|
6
|
90
|
16
|
4
|
64
|
19
|
3
|
57
|
Total
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N = 30
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Σ fx = 410
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Mean = Σ fx/n=410/30=13.67
13. Calculate the A.M. for the following data:
Marks
|
80
|
85
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90
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95
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100
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No. of students
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5
|
6
|
6
|
2
|
1
|
Take A = 90 , c = 5 and d= [x-A/c]
X
|
f
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d
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f × d
|
80
|
5
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−2
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−10
|
85
|
6
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−1
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−6
|
90
|
6
|
0
|
0
|
95
|
2
|
1
|
2
|
100
|
1
|
2
|
2
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N = 20
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Σ fd = −12
| ||
A.M.= A+ c(Σ fd / N)= 90 + 5 ×[-12/20] =90 − 3 = 87.
14. Calculate A .M for the following data
Calculate A .M for the following data: Class Interval
|
0-10
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10-20
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20-30
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30-40
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40-50
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50-60
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Marks
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12
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18
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27
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20
|
17
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6
|
solution
Class
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Mid-value x
|
Frequency f
|
f × x
|
0-10
|
5
|
12
|
60
|
10-20
|
15
|
18
|
270
|
20-30
|
25
|
27
|
675
|
30-40
|
35
|
20
|
700
|
40-50
|
45
|
17
|
765
|
50-60
|
55
|
6
|
330
|
N = 100
|
Σ fx = 2800
| ||
mean= [Σ fx]/N= 2800/100=28
Short-cut Method:
Let A = 30, c = 5 and d=[x-A]/c
Class
|
Mid-value x
|
d
|
Frequency f
|
f × d
| |
0-10
|
5
|
−5
|
12
|
−60
| |
10-20
|
15
|
−3
|
18
|
−54
| |
20-30
|
25
|
−1
|
27
|
−27
| |
30-40
|
35
|
1
|
20
|
20
| |
40-50
|
45
|
3
|
17
|
51
| |
50-60
|
55
|
5
|
6
|
30
| |
N = 100
|
Σ fd = 101−141= −40
| ||||
A.M. =x= A+ c ×[ Σ fd /n]=28
15. Find the median of 23, 25, 29, 30, 39.
Solution: The given values are already in the ascending order. No. of observations N = 5. This is an odd number. So the median = [(n+1)2]th term=[5+1]/2 =3rd term=29
= 3 rd term =29.
∴ Median = 29.
16. Find the median of 26, 25, 29, 23, 25, 29, 30, 25, 28, 30.
Solution: Arranging the observations in the ascending order, we get
23, 25, 25, 25, 26, 28, 29, 29, 30, 30.
N = No. of observations = 10, an even integer
Median = average of (N/2)th and (n/2 +1)th term
=average of 5th and 6th terms= average of 26 and 28 = (26 + 28)/2=27
16. Calculate the median of the following table:
Variable ( x)
|
5
|
10
|
15
|
20
|
25
|
30
|
Frequency( f )
|
3
|
6
|
10
|
8
|
2
|
3
|
solution:
x
|
f
|
Cumulative frequency
|
5
|
3
|
3
|
10
|
6
|
9
|
15
|
10
|
19
|
20
|
8
|
27
|
25
|
2
|
29
|
30
|
3
|
32
|
Total frequency = N = Σf = 32 and so N/2 = 16.
The median is the [N/2]th value = 16th value. But the 16th value occurs in the class whose cumulative frequency is 19. The corresponding value of the variate is 15.
Hence, the median = 15
17. Find the mode of 7, 4, 5, 1, 7, 3, 4, 6,7.
Solution: Arranging the data in the ascending order, we get
1, 3, 4, 4, 5, 6, 7, 7, 7.
In the above data 7 occurs maximum number of times. Hence mode = 7.
18: Find the mode of 19, 20, 21, 24, 27, 30.
Solution: The data are already in the ascending order. Each value occurs exactly one time in the series. Hence there is no mode for this data.
19: Find the mode for 12, 15, 11, 12, 19, 15, 24, 27, 20, 12, 19, 15.
Solution: Arranging the data in the ascending order, we get
11, 12, 12, 12, 15, 15, 15, 19, 19, 20, 24, 27.
Here 12 occurs 3 times and 15 also occurs 3 times.
∴ both 12 and 15 are the modes for this data. We observe that there are two modes for the given data.
∴ both 12 and 15 are the modes for this data. We observe that there are two modes for the given data.
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