Question

13.
Find the median age of the patients from the following distribution:
5-14
15-24
25-34
35-44
45-54
55-64
Age(in years)
No. of patients
6
11
21
23
14
5​

Answer 1

Answer:

Here class intervals are not inclusive form. So, we first convert them in inclusive from by subtracting h/2 from the lower limit and adding h/2 to the upper limit of each class, where h is the difference between the lower limit of a class and the upper limit of the preceding class. The given frequency distribution in inclusive form is as follows.

Age ( in years) 4.5-14.5 14.5-24.5 24.5-34.5 34.5-44.5 44.5-54.5 54.5-64.5

No. of cases: 6 11 21 23 14 5

We observe that the class 34.5−44.5 has the maximum frequency. So, it is the modal class such that,

l=34.5,h=10,f=23,f

1

​

=21,andf

2

​

=14

∴Mode=l+

2f−f

1

​

−f

2

​

f−f

1

​

​

×h

⇒Mode=34.5+

46−21−14

23−21

​

×10

⇒Mode=34.5+

11

1

​

×10

⇒Mode=36.31

Step-by-step explanation: