In a college, the number of absentees…

Question 1 Report

In a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table

Number of absentees	0-4	5-9	10-14	15-19	20-24
Number of Days	1	5	10	9	5

Calculate the : (a) Mean

(b) Standard deviation , correct to two decimal places.

Answer

Answer Details

Take the class mid-values \(x\) and build a working table (\(N=1+5+10+9+5=30\)):

Class	Mid-value x	f	fx	fx²
0 - 4	2	1	2	4
5 - 9	7	5	35	245
10 - 14	12	10	120	1440
15 - 19	17	9	153	2601
20 - 24	22	5	110	2420
Total		30	420	6710

(a) Mean.

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{420}{30}=14.\]

(b) Standard deviation.

\[s=\sqrt{\frac{\sum fx^{2}}{N}-\bar{x}^{2}}=\sqrt{\frac{6710}{30}-14^{2}}=\sqrt{223.67-196}=\sqrt{27.67}\approx 5.26.\]

The mean number of absentees is 14 and the standard deviation is 5.26 (to 2 d.p.).

In a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table Number of absentees 0-4 5-9 10-14 1...