Question

The following table shows the ages of tha year:  

Age (in years)	5 - 15	15 - 25	25 - 35	35 - 45	45 - 55	55 - 65
Number of patients	6	11	21	23	14	5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Answer 1

To find the class mark (\(x_i\)) for each interval, the following relation is used.  

Class mark  \((x_i)\) = \(\frac {\text{Upper \,limit + Lower \,limit}}{2}\)

Taking 30 as assumed mean (a), \(d_i\), and \(f_id_i\) can be calculated as follows.   

Age (in years)	Number of patients \(\bf{f_i}\)	Class mark \(\bf{x_i}\)	\(\bf{d_i = x_i -30}\)	\(\bf{f_id_i}\)
5 - 15	6	10	-20	-120
15 - 25	11	20	-10	-110
25 - 35	21	30	0	0
35 - 45	23	40	10	230
45 - 55	14	50	20	280
Total	80			430

From the table, We obtain

\(\sum f_i = 80\)
\(\sum f_id_i = 430\)

Mean, \(\overset{-}{x} = a + (\frac{\sum f_id_i}{\sum f_i})\)

x = \(30 + (\frac{430}{80})\)

x = 30 + 5.375
x = 35.375
x = 35.38

Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years.

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It can be observed that the maximum class frequency is 23 belonging to class interval 35 - 45.
Modal class = 35 − 45 
Lower limit (\(l\)) of modal class = 35
Frequency (\(f_1\)) of modal class = 23
Class size (\(h\)) = 10  
Frequency (\(f_0\)) of class preceding the modal class = 21 
Frequency (\(f_2\)) of class succeeding the modal class = 14    

Mode = \(l\) + \((\frac{f_1 - f_0 }{2f_1 - f_0 - f_2)} \times h\)

Mode = 3\(5+ (\frac{23 - 21 }{ 2(23) - 21 - 14})\)

Mode =\(35 + [\frac{2}{46 - 35}] \times 10\)

Mode = \(35 + \frac{20}{ 11}\)
Mode = 35 + 1.81
Mode = 36.8

Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.