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

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Solution

Class interval Frequency (f_i) Mid-point (x_i) f_ix_i
0-6 11 3 33
6-10 10 8 80
10-14 7 12 84
14-20 4 17 68
20-28 4 24 96
28-38 3 33 99
38-40 1 39 39
Sum f_i = 40 Sum f_ix_i = 499
The formula to find out the mean is
$M e a n = x̄ = \frac{\sum f i x i}{\sum f i}$
$= \frac{499}{40} = 12.48 d a y s$
Hence, the mean number of days a student was absent $= 12.48$

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Number of days:	0 − 6	6 − 12	12 − 18	18 − 24	24 − 30	30 − 36	36 − 42
Number of students:	10	11	7	4	4	3	1

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

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

No. of days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
No. of students	11	10	7	4	4	3	1

for this data what would be ' h ' inorder to find the mean through step deviation method

Number of days

0 6

6 10

10 14

14 20

20 28

28 38

38 40

Number of students

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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