5. The following table shows the number of patients discharged from a hospital in
different years:
Years
2002
2003
2004 2005: 2006
No. of
150
170
195
225
230
Patients
Represent this information by a graph.
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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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