Question
The following table shows the ages of the patients admitted in a hospital during a 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.
Solution
To find the class marks (x_{i}), the following relation is used.
`x_i = ("Upper class limit + Lower class limit")/2`
Taking 30 as assumed mean (a), d_{i} and f_{i}d_{i}are calculated as follows.
Age (in years) |
Number of patients f_{i} |
Class mark x_{i} |
d_{i}_{ }= x_{i} − 30 |
f_{i}d_{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 |
55 − 65 |
5 |
60 |
30 |
150 |
Total |
80 |
430 |
From the table, we obtain
`sumf_i = 80`
`sumf_iu_i = 430`
`"Mean " barx = a+ ((sumf_iu_i)/(sumf_i))xh`
= 30+(430/80)
= 30+5.375
= 35.375
= 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.
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 + ((f_1-f_0)/(2f_1-f_0-f_2)xh)`
= `35+((23-21)/(2(23)-21-14))xx10`
=`35+[2/(46-35)]xx10`
= `35+20/11`
= 36.8
Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years