1. The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.
Expenditure (in rupees) (x) | Frequency (fi) | Expenditure (in rupees) (xi) | Frequency (fi) |
100 – 150 | 24 | 300 – 350 | 30 |
150 – 200 | 40 | 350 – 400 | 22 |
200 – 250 | 33 | 400 – 450 | 16 |
250 – 300 | 28 | 450 – 500 | 7 |
Find the average expenditure (in rupees) per household
Sol: Let the assumed mean (A) = 275
Class interval | Mid value (xi) | di = xi – 275 | ui = (xi-275)/50 | Frequency fi | fiui |
100 – 150 | 125 | -150 | -3 | 24 | -12 |
150 – 200 | 175 | -100 | -2 | 40 | -80 |
200 – 250 | 225 | -50 | -1 | 33 | -33 |
250 – 300 | 275 | 0 | 0 | 28 | 0 |
300 – 350 | 325 | 50 | 1 | 30 | 30 |
350 – 400 | 375 | 100 | 2 | 22 | 44 |
400 – 450 | 425 | 150 | 3 | 16 | 48 |
450 – 500 | 475 | 200 | 4 | 7 | 28 |
N = 200 | Sum = -35 |
We have
A = 275, h = 50
Mean = A + h * sum/N = 275 + 50 * -35/200 = 275 – 8.75 = 266.25
2. A survey was conducted by a group of students as a part of their environmental awareness program, in which they collected the following data regarding the number of plants in 200 houses in a locality. Find the mean number of plants per house.
Number of plants: | 0-2 2-4 4-6 6-B 8-10 10-12 12-14 |
Number of houses: | 1 2 1 5 6 2 3 |
Which method did you use for finding the mean, and why?
Sol:
Let us find class marks (xi) = (upper class limit + lower class limit)/2
Now we may compute xi and fixi as following.
Number of plants | Number of house (fi) | xi | Fixi |
0 – 2 | 1 | 1 | 1 |
2 – 4 | 2 | 3 | 6 |
4 – 6 | 1 | 5 | 5 |
6 – 8 | 5 | 7 | 35 |
8 – 10 | 6 | 9 | 54 |
10 – 12 | 2 | 11 | 22 |
12 – 14 | 3 | 13 | 39 |
Total | N = 20 | Sum = 162 |
From the table we may observe that
N = 20
Sum = 162
Mean = Sum/N
162/20 = 8.1
So mean number of plants per house is 8.1
We have used for the direct method values Xi and fi are very small
3. Consider the following distribution of daily wages of workers of a factory
Daily wages (in Rs) | 100-120 120-140 140-160 160-180 180-200 |
Number of workers: | 12 16 8 6 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Sol: Let the assume mean (A) = 150
Class interval | Mid value xi | di = xi – 150 | ui = (xi-150)/20 | Frequency fi | Fiui |
100 – 120 | 110 | -40 | -2 | 12 | -24 |
120 – 140 | 130 | -20 | -1 | 14 | -14 |
140 – 160 | 150 | 0 | 0 | 8 | 0 |
160 – 180 | 170 | 20 | 1 | 6 | 6 |
180 – 200 | 190 | 40 | 2 | 10 | 20 |
N = 50 | Sum = -12 |
We have
N = 50, h = 20
Mean = A + h x sum/N
= 150 + 2 x (-15)/5
= 150 – 4.8
= 145.2
4.Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Number of heart
Beats Per minute: | 65 – 68 68 – 71 71 – 74 74 – 77 77 – 80 80 – 83 83 – 86 |
Number of women: | 2 4 3 8 7 4 2 |
Sol: we may find marks of each interval (xi) by using the relation
(xi) = (upper class limit + lower class limit)/2
Class size of this data = 3
Now talking 75.5 as assumed mean (a)
We may calculate di, ui, fiui as following
Number of heart beats per minute | Number of women (xi) | xi | di = xi – 75.5 | ui = (xi – 755)/h | fiui |
65-68 | 2 | 66.5 | -9 | -3 | -6 |
68-71 | 9 | 69.5 | -6 | -2 | -8 |
71-74 | 3 | 72.5 | -3 | -1 | -3 |
74-77 | 8 | 75.5 | 0 | 0 | 0 |
77-80 | 7 | 78.5 | 3 | 1 | 7 |
80-83 | 4 | 81.5 | 6 | 2 | 8 |
83-86 | 2 | 84.5 | 9 | 3 | 6 |
N = 30 | Sum = 4 |
Now we may observe from table that N = 30, sum = 4
Mean = 75.5 + (4/3)x3
= 75.5 + 0.4
= 75.9
So mean heart beats per minute for those women are 75.9 beats per minute
5.Find the mean of each of the following frequency distributions: (5 – 14)
Class interval: | 0-6 6-12 12-18 18-24 24-30 |
Frequency: | 6 8 10 9 7 |
Sol:
Let us assume mean be 15
Class interval | Mid – value | di = xi – 15 | ui = (xi – 15)/6 | fi | fiui |
0 – 6 | 3 | -12 | -2 | 6 | -12 |
6 – 12 | 9 | -6 | -1 | 8 | -8 |
12 – 18 | 15 | 0 | 0 | 10 | 0 |
18 – 24 | 21 | 6 | 1 | 9 | 9 |
24 – 30 | 27 | 18 | 2 | 7 | 14 |
N = 40 | Sum = 3 |
A = 15, h = 6
Mean = A + h(sum/A)
= 15 + 6(3/40)
= 15 + 0.45
= 15.45
6.
Class interval: | 50-70 70-90 90-110 110-130 130-150 150-170 |
Frequency: | 18 12 13 27 8 22 |
Sol: Let us assumed mean be 100
Class interval | Mid-value xi | di = xi – 100 | ui = (xi – 100)/20 | fi | fiui |
50 – 70 | 60 | -40 | -2 | 18 | -36 |
70 – 90 | 80 | -20 | -1 | 12 | -12 |
90 – 110 | 100 | 0 | 0 | 13 | 0 |
110 – 130 | 120 | 20 | 1 | 27 | 27 |
130 – 150 | 140 | 40 | 2 | 8 | 16 |
150 – 170 | 160 | 60 | 3 | 22 | 66 |
61 |
A = 100, h = 20
Mean = 100 + 20 (61/100)
= 100 + 12.2
= 112.2
7.
Class interval: | 0-8 8-16 16-24 24-32 32-40 |
Frequency: | 6 7 10 8 9 |
Sol: Let the assumed mean (A) = 20
Class interval | Mid- value | di= xi – 20 | ui = (xi – 20)/8 | fi | fiui |
0-8 | 4 | -16 | -2 | 6 | -12 |
8-16 | 12 | -8 | -1 | 7 | -7 |
16-24 | 20 | 0 | 0 | 10 | 0 |
24-32 | 28 | 8 | 1 | 8 | 8 |
32-40 | 36 | 16 | 2 | 9 | 18 |
N = 40 | Sum = 7 |
We have A = 20, h = 8
Mean= A + h (sum/N)
= 20 + 8 (7/40)
= 20 + 1.4
= 21.4
8.
Class interval: | 0 – 6 6 – 12 12 – 18 18 – 24 24 – 30 |
Frequency: | 7 5 10 12 6 |
Sol: Let the assumed mean be (A) = 15
Class interval | Mid – value | di = xi -15 | ui = (xi -15)/6 | Frequency fi | fiui |
0 – 6 | 3 | -12 | -2 | -1 | -14 |
6 – 12 | 9 | -6 | -1 | 5 | -5 |
12 – 18 | 15 | 0 | 0 | 10 | 0 |
18 – 24 | 21 | 6 | 1 | 12 | 12 |
24 – 30 | 27 | 12 | 2 | 6 | 12 |
N = 40 | Sum = 5 |
We have
A = 15, h = 6
Mean = A + h(sum/N)
= 15 + 6 (5/40)
= 15 + 0.75
= 15.75
9.
Class interval: | 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 |
Frequency: | 9 12 15 10 14 |
Sol: Let the assumed mean (A) = 25
Class interval | Mid – value | di = xi -25 | ui = (xi -25)/10 | Frequency fi | fiui |
0 – 10 | 5 | -20 | -2 | 9 | -18 |
10 – 20 | 15 | -10 | -1 | 10 | -12 |
20 – 30 | 25 | 0 | 0 | 15 | 0 |
30 – 40 | 35 | 10 | 1 | 10 | 10 |
40 – 50 | 45 | 20 | 2 | 14 | 28 |
N = 60 | Sum = 8 |
We have A = 25, h = 10
Mean = A + h(sum/N)
= 25 + 19 (8/60)
= 25 + (4/3)
= 26.333
10.
Class interval: | 0-8 8-16 16-24 24-32 32-40 |
Frequency: | 5 9 10 8 8 |
Sol: Let the assumed mean (A) = 20
Class interval | Mid value xi | di= xi – 20 | ui = (xi -20)/8 | Frequency fi | fiui |
0-8 | 4 | -16 | -2 | 5 | -10 |
8-16 | 12 | -8 | -1 | 9 | -9 |
16-24 | 20 | 0 | 0 | 10 | 0 |
24-32 | 28 | 8 | 1 | 8 | 8 |
32-40 | 36 | 16 | 2 | 8 | 16 |
N = 40 | Sum = 5 |
We have,
A = 20, h = 8
Mean = A + h (sum/N)
= 20 + 8 (5/ 40)
= 20 + 1
= 21
11.
Class interval: | 0-8 8-16 16-24 24-32 32-40 |
Frequency: | 5 6 4 3 2 |
Sol: Let the assumed mean (A) = 20
Class interval | Mid value xi | di= xi – 20 | ui = (xi -20)/8 | Frequency fi | fiui |
0-8 | 4 | -16 | -2 | -2 | -10 |
8-16 | 12 | -8 | -1 | -1 | -6 |
16-24 | 20 | 0 | 0 | 0 | 0 |
24-32 | 28 | 8 | 1 | 1 | 3 |
32-40 | 36 | 16 | 2 | 2 | 4 |
N = 20 | Sum = -9 |
We have,
A = 20, h = 8
Mean = A + h (sum/N)
= 20 + 8 (-9/ 20)
= 20 – (72/20)
= 20 – 3.6
= 16.4
12.
Class interval: | 10-30 30-50 50-70 70-90 90-110 110-130 |
Frequency: | 5 8 12 20 3 2 |
Sol: Let the assumed mean (A) = 60
Class interval | Mid value xi | di= xi – 60 | ui = (xi -60)/20 | Frequency fi | fiui |
10 – 30 | 20 | -40 | -2 | 5 | -10 |
30 – 50 | 40 | -20 | -1 | 8 | -8 |
50 – 70 | 60 | 0 | 0 | 12 | 0 |
70 – 90 | 80 | 20 | 1 | 20 | 20 |
90 – 110 | 100 | 40 | 2 | 3 | 6 |
110 – 130 | 120 | 60 | 3 | 2 | 6 |
N = 50 | Sum = 14 |
We have
A = 60, h = 20
Mean = A + h (sum/N)
= 60 + 20 (14/ 5)
= 60 + 5.6
= 65.6
13.
Class interval: | 25-35 35-45 45-55 55-65 65-75 |
Frequency: | 6 10 8 10 4 |
Sol: Let the assumed mean (A) = 50
Class interval | Mid value xi | di= xi – 50 | ui = (xi – 50)/ 10 | Frequency fi | fiui |
25 – 35 | 30 | -20 | -2 | 6 | -12 |
35 – 45 | 40 | -10 | -1 | 10 | -10 |
45 – 55 | 50 | 0 | 0 | 8 | 0 |
55 – 65 | 60 | 10 | 1 | 12 | 12 |
65 – 75 | 70 | 20 | 2 | 4 | 8 |
N = 40 | Sum = -2 |
We have
A = 50, h = 10
Mean = A + h (sum/N)
= 50 + 10 (-2/ 40)
= 50 – 0.5
= 49.5
14.
Class interval: | 25-29 30-34 35-39 40-44 45-49 50-54 55-59 |
Frequency: | 14 22 16 6 5 3 4 |
Sol: Let the assumed mean (A) = 42
Class interval | Mid value xi | di= xi – 42 | ui = (xi – 42)/ 5 | Frequency fi | fiui |
25 – 29 | 27 | -15 | -3 | 14 | -42 |
30 – 34 | 32 | -10 | -2 | 22 | -44 |
35 – 39 | 37 | -5 | -1 | 16 | -16 |
40 – 44 | 42 | 0 | 0 | 6 | 0 |
45 – 49 | 47 | 5 | 1 | 5 | 5 |
50 – 54 | 52 | 10 | 2 | 3 | 6 |
55 – 59 | 57 | 15 | 3 | 4 | 12 |
N = 70 | Sum = -79 |
We have
A = 42, h = 5
Mean = A + h (sum/N)
= 42 + 5 (-79/70)
= 42 – 79/14
= 36.357
15.For the following distribution, calculate mean using all suitable methods:
Size of item: | 1 – 4 4 – 9 9 – 16 16 – 20 |
Frequency: | 6 12 26 20 |
Sol: By direct method
Class interval | Mid value xi | Frequency fi | fixi |
1 – 4 | 2.5 | 6 | 15 |
4 – 9 | 6.5 | 12 | 18 |
9 – 16 | 12.5 | 26 | 325 |
16 – 27 | 21.5 | 20 | 430 |
N = 64 | Sum = 848 |
Mean = (sum/N) + A
= 848/ 64
= 13.25
By assuming mean method
Let the assumed mean (A) = 65
Class interval | Mid value xi | ui = (xi – A) =xi– 65 | Frequency fi | fiui |
1 – 4 | 2.5 | -4 | 6 | -25 |
4 – 9 | 6.5 | 0 | 12 | 0 |
9 – 16 | 12.5 | 6 | 26 | 196 |
16 – 27 | 21.5 | 15 | 20 | 300 |
N = 64 | Sum = 432 |
Mean = A + sum/N
= 6.5 + 6.75
= 13.25
16.The weekly observation on cost of living index in a certain city for the year 2004 – 2005 are given below. Compute the weekly cost of living index.
Cost of living index | Number of students | Cost of living index | Number of students |
1400 – 1500 | 5 | 1700 – 1800 | 9 |
1500 – 1600 | 10 | 1800 – 1900 | 6 |
1600 – 1700 | 20 | 1900 – 2000 | 2 |
Sol: Let the assumed mean (A) = 1650
Class interval | Mid value xi | di= xi – A = xi – 1650 | ui = (xi – 1650) 100 | Frequency fi | fiui |
1400 – 1500 | 1450 | -200 | -2 | 5 | -10 |
1500 – 1600 | 1550 | -100 | -1 | 10 | -10 |
1600 – 1700 | 1650 | 0 | 0 | 20 | 0 |
1700 – 1800 | 1750 | 100 | 1 | 9 | 9 |
1800 – 1900 | 1850 | 200 | 2 | 6 | 12 |
1900 – 2000 | 1950 | 300 | 3 | 2 | 6 |
N = 52 | Sum = 7 |
We have
A = 16, h = 100
Mean = A + h (sum/N)
= 1650 + 100 (7/52)
= 1650 + (175/13)
= 21625/13
= 1663.46
17.The following table shows the marks scored by 140 students in an examination of a certain paper:
Marks: | 0-10 10-20 20-30 30-40 40-50 |
Number of students: | 20 24 40 36 20 |
Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.
Sol: (i) Direct method:
Class interval | Mid value xi | Frequency fi | fixi |
0 – 10 | 5 | 20 | 100 |
10 – 20 | 15 | 24 | 360 |
20 – 30 | 25 | 40 | 1000 |
30 – 40 | 35 | 36 | 1260 |
40 – 50 | 45 | 20 | 900 |
N = 140 | Sum = 3620 |
Mean = sum/ N
= 3620/ 140
= 25.857
(ii)Assumed mean method:
Let the assumed mean = 25
Mean = A + (sum/ N)
Class interval | Mid value xi | ui = (xi – A) | Frequency fi | fiui |
0 – 10 | 5 | -20 | 20 | -400 |
10 – 20 | 15 | -10 | 24 | -240 |
20 – 30 | 25 | 0 | 40 | 0 |
30 – 40 | 35 | 10 | 36 | 360 |
40 – 50 | 45 | 20 | 20 | 400 |
N = 140 | Sum = 120 |
Mean = A + (sum/ N)
= 25 + (120/ 140)
= 25 + 0.857
= 25.857
(iii)Step deviation method:
Let the assumed mean (A) = 25
Class interval | Mid value xi | di= xi – A
= xi – 25 | ui = (xi – 25)
10 | Frequency fi | fiui |
0 – 10 | 5 | -20 | -2 | 20 | -40 |
10 – 20 | 15 | -10 | -1 | 24 | -24 |
20 – 30 | 25 | 0 | 0 | 40 | 0 |
30 – 40 | 35 | 10 | 1 | 36 | 36 |
40 – 50 | 45 | 20 | 2 | 20 | 40 |
N = 140 | Sum = 12 |
Mean = A + h(sum/ N)
= 25 + 10(12/ 140)
= 25 + 0.857
= 25.857
18.The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the miss frequency f1 and f2.
Class: | 0-20 20-40 40-60 60-80 80-100 100-120 |
Frequency: | 5 f1 10 f2 7 8 |
Sol:
Class interval | Mid value xi | Frequency fi | fixi |
0 – 20 | 10 | 5 | 50 |
20 – 40 | 30 | f1 | 30f1 |
40 – 60 | 50 | 10 | 500 |
60 – 80 | 70 | f2 | 70f2 |
80 – 100 | 90 | 7 | 630 |
100 – 120 | 110 | 8 | 880 |
N = 50 | Sum = 30f1+70f2+ 2060 |
Given,
sum of frequency = 50
5 + f1 + 10 + f2 + 7 + 8 = 50
f1 + f2 = 20
3f1 + 3f2 = 60 —- (1) [multiply both side by 3]
And mean = 62.8
Sum/ N = 62.8
(30f1+70f2+ 2060)/ 50 = 62.8
30f1+70f2 = 3140 – 2060
30f1+70f2 = 1080
3f1 + 7f2 = 108 —- (2) [divide it by 10]
subtract equation (1) from equation (2)
3f1 + 7f2 – 3f1 – 3f2 = 108 – 60
4f2 = 48
f2 = 12
Put value of f2 in equation (1)
3f1 + 3(12) = 60
f1 = 24/3 = 8
f1 = 8, f2 = 12
19.The following distribution shows the daily pocket allowance given to the children of a multistory building. The average pocket allowance is Rs 18.00. Find out the missing frequency.
Class interval: | 11-13 13-15 15-17 17-19 19-21 21-23 23-25 |
Frequency: | 7 6 9 13 – 5 4 |
Sol: Given mean = 18,
Let the missing frequency be v
Class interval | Mid value xi | Frequency fi | fixi |
11 – 13 | 12 | 7 | 84 |
13 – 15 | 14 | 6 | 88 |
15 – 17 | 16 | 9 | 144 |
17 – 19 | 18 | 13 | 234 |
19 – 21 | 20 | x | 20x |
21 – 23 | 22 | 5 | 110 |
23 – 25 | 14 | 4 | 56 |
N =44 + x | Sum = 752 + 20x |
Mean = sum/ N
792 + 18x = 752 + 20x
2x = 40
x = 20
20.If the mean of the following distribution is 27. Find the value of p.
Class: | 0-10 10-20 20-30 30-40 40-50 |
Frequency: | 8 p 12 13 10 |
Sol:
Class interval | Mid value xi | Frequency fi | fixi |
0 – 10 | 5 | 8 | 40 |
10 – 20 | 15 | P | 152 |
20 – 30 | 25 | 12 | 300 |
30 – 40 | 35 | 13 | 455 |
40 – 50 | 45 | 16 | 450 |
N = 43 + P | Sum = 1245 + 15p |
Given mean =27
Mean = sum/ N
1245 + 15p = 1161 + 27p
12p = 84
P = 7
21.In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contain varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes: | 50-52 53-55 56-58 59-61 62-64 |
Number of boxes: | 15 110 135 115 25 |
Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?
Sol:
Number of mangoes | Number of boxes (fi) |
50 – 52 | 15 |
53 – 55 | 110 |
56 – 58 | 135 |
59 – 61 | 115 |
62 – 64 | 25 |
We may observe that class internals are not continuous
There is a gap between two class intervals. So we have to add ½ from lower class limit of each interval and class mark (xi) may be obtained by using the relation
Class size (h) of this data = 3
Now taking 57 as assumed mean (a) we may calculated di ,ui, fiui as follows
Class interval | Frequency fi | Mid value xi
| di= xi – A
= xi – 25 | ui = (xi – 25)
10 | fiui |
49.5 – 52.5 | 15 | 51 | -6 | -2 | -30 |
52.5 – 55.5 | 110 | 54 | -3 | -1 | -110 |
55.5 – 58.5 | 135 | 57 | 0 | 0 | 0 |
58.5 – 61.5 | 115 | 60 | 3 | 1 | 115 |
61.5 – 64.5 | 25 | 63 | 6 | 2 | 50 |
Total | N = 400 | Sum = 25 |
Now we have N = 400
Sum = 25
Mean = A + h (sum/ N)
= 57 + 3 (45/400)
= 57 + 3/ 16
= 57+ 0.1875
= 57.19
Clearly mean number of mangoes kept in packing box is 57.19
22.The table below shows the daily expenditure on food of 25 households in a locality
Daily expenditure (in Rs): | 100-150 150-200 200-250 250-300 300-350 |
Number of households: | 4 5 12 2 2 |
Find the mean daily expenditure on food by a suitable method.
Sol: we may calculate class mark (xi) for each interval by using the relation
Class size = 50
Now, talking 225 as assumed mean (xi) we may calculate di ,ui, fiui as follows
Daily expenditure | Frequency fi | Mid value xi | di= xi – 225 | ui = (xi – 225) 50 | fiui |
100 – 150 | 4 | 125 | -100 | -2 | -8 |
150 – 200 | 5 | 175 | -50 | -1 | -5 |
200 – 250 | 12 | 225 | 0 | 0 | 0 |
250 – 300 | 2 | 275 | 50 | 1 | 2 |
300 – 350 | 2 | 325 | 100 | 2 | 4 |
N = 25 | Sum = -7 |
Now we may observe that
N = 25
Sum = -7
225 + 50 (-7/ 25)
225 – 14 = 211
So, mean daily expenditure on food is Rs 211
23.To find out the concentration of SO2 in the air (in parts per million i.e ppm) the data was collected for localities for 30 localities in a certain city and is presented below:
Concentration of SO2 (in ppm) | Frequency |
0.00 – 0.04 | 4 |
0.04 – 0.08 | 9 |
0.08 – 0.12 | 9 |
0.12 – 0.16 | 2 |
0.16 – 0.20 | 4 |
0.20 – 0.24 | 2 |
Find the mean concentration of SO2 in the air
Sol: we may find class marks for each interval by using the relation
Class size of this data = 0.04
Now taking 0.04 assumed mean (xi) we may calculate di ,ui, fiui as follows
Concentration of SO2 | Frequency fi | Class interval xi
| di = xi – 0.14 | ui | fiui |
0.00 – 0.04 | 4 | 0.02 | -0.12 | -3 | -12 |
0.04 – 0.08 | 9 | 0.06 | -0.08 | -2 | -18 |
0.08 – 0.12 | 9 | 0.10 | -0.04 | -1 | -9 |
0.12 – 0.16 | 2 | 0.14 | 0 | 0 | 0 |
0.16 – 0.20 | 4 | 0.18 | 0.04 | 1 | 4 |
0.20 – 0.24 | 2 | 0.22 | 0.08 | 2 | 4 |
Total | N = 30 | Sum = -31 |
From the table we may observe that
N = 30
Sum = -31
= 0.14 +(0.04)(-31/30)
= 0.099 ppm
So mean concentration of SO2 in the air is 0.099 ppm
24.A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days: | 0-6 6-10 10-14 14-20 20-28 28-38 38-40 |
Number of students: | 11 10 7 4 4 3 1 |
Sol: We may find class mark of each interval by using the relation
Now, taking 16 as assumed mean (a) we may
Calculate di and fidi as follows
Number of days | Number of students fi | Xi | d = xi + 10 | fidi |
0 – 6 | 11 | 3 | -13 | -143 |
6 – 10 | 10 | 8 | -8 | -280 |
10 – 14 | 7 | 12 | -4 | -28 |
14 – 20 | 7 | 16 | 0 | 0 |
20 – 28 | 8 | 24 | 8 | 32 |
28 – 36 | 3 | 33 | 17 | 51 |
30 – 40 | 1 | 39 | 23 | 23 |
Total | N = 40 | Sum = -145 |
Now we may observe that
N = 40
Sum= -145
= 16 + (-145/ 40)
= 16 – 3.625
= 12.38
So mean number of days is 12.38 days, for which student was absent
25.The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Literacy rate (in %): | 45-55 55-65 65-75 75-85 85-95 |
Number of cites: | 3 10 11 8 3 |
Sol: We may find class marks by using the relation
Class size (h) for this data = 10
Now taking 70 as assumed mean (a) wrong
Calculate di ,ui, fiui as follows
Litracy rate (in %) | Number of cities (fi) | Mid value xi
| di= xi – 70 | ui = di
50 | fiui |
45 – 55 | 3 | 50 | -20 | -2 | -6 |
55 – 65 | 10 | 60 | -10 | -1 | -10 |
65 – 75 | 11 | 70 | 0 | 0 | 0 |
75 – 85 | 8 | 80 | 10 | 1 | 8 |
85 – 95 | 3 | 90 | 20 | 2 | 6 |
Total | N = 35 | Sum = -2 |
Now we may observe that
N = 35
Sum = -2
= 70 + (-2/35)
= 70 – 4/7
= 70 – 0.57
= 69.43
So, mean literacy rate is 69.43 %
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