For the month of February- a class teacher of Class IX has the following absentee record for 45 students- Find the mean number of days- a student was absent-Nbeginarray - c-c-c-c-c-c-c- Wline t e x t Number of days of absent - 0-4 - 4-8 - 8-12 - 12-16 - 16-20 - 20-241- Wlinetext Number of students - 18 - 3 - 6 - 2 - 0 - 1 - Wline e n d array

[ C.I. amp; f_i amp; x_i (mid value) amp; d = x_i A amp; f_i × d_i; 0 4 amp; ...

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Standard X

Mathematics

Assumed Mean Method of Finding Mean

For the month...

Question

For the month of February, a class teacher of Class IX has the following absentee record for 45 students. Find the mean number of days, a student was absent.
$\begin{matrix} Number of days of absent & 0 - 4 & 4 - 8 & 8 - 12 & 12 - 16 & 16 - 20 & 20 - 24 Number of students & 18 & 3 & 6 & 2 & 0 & 1 \end{matrix}$

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Solution

$\begin{matrix} C . I . & f_{i} & x_{i} (m i d - v a l u e) & d = x_{i} - A & f_{i} \times d_{i} 0 - 4 & 18 & 2 & - 12 & - 216 4 - 8 & 3 & 6 & - 8 & - 24 8 - 12 & 6 & 10 & - 4 & - 24 12 - 16 & 2 & A = 14 & 0 & 00 16 - 20 & 0 & 18 & 4 & 00 20 - 24 & 1 & 22 & 8 & 08 \sum f_{i} = 30 & \sum f_{i} d_{i} = - 256 \end{matrix}$
$M e a n = A + \frac{\sum f_{i} d_{i}}{\sum f_{i}} = 14 + (\frac{- 256}{30}) = 14 - 8.53 = 5.47$

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Question 8
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.

$\begin{matrix} Number of days & 0 - 6 & 6 - 10 & 10 - 14 & 14 - 20 & 20 - 28 & 28 - 38 & 38 - 40 Number of students & 11 & 10 & 7 & 4 & 4 & 3 & 1 \end{matrix}$

Q. 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.
(i)

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

(ii)

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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For the month of February, a class teacher of Class IX has the following absentee record for 45 students. Find the mean number of days, a student was absent. Number of days of absent0−44−88−1212−1616−2020−24Number of students1836201

C.I.fixi(mid−value)d=xi−Afi×di0−4182−12−2164−836−8−248−12610−4−2412−162A=1400016−2001840020−24122808∑fi=30∑fidi=−256 Mean=A+∑fidi∑fi=14+(−25630)=14−8.53=5.47