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
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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:
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