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Presentation of data — Statistics STD 11 Commerce — Question

Gujarat BoardEnglish MediumSTD 11 CommerceStatisticsPresentation of data5 Marks
Question
Obtain an inclusive continuous frequency distribution from the following data :
Lower Boundary Point or more $44.5$ $49.5$ $54.5$ $59.5$ $64.5$ $69.5$ $74.5$ $79.5$ -
Cumulative Frequency  $500$ $470$ $390$ $290$ $240$ $90$ $10$ $0$ -
✓

Answer

Given data is ‘more than’ type cumulative frequency distribution.
$\therefore $ Class length $=$ Difference between the adjoining lower boundary points $= 54.5 – 49.5 = 5$
Now, upper boundary point of the class $=$ lower boundary point $+$ class length
For initial class lower boundary point $= 44.5$
$\therefore $ initial class is $44.5 – 49.5$.
For inclusive continuous frequency distribution lower limit of initial class $= 44.5 + 0.5 = 45$ and
upper limit of initial class $= 49.5 – 0.5 = 49.$
Thus, for given data the initial class in inclusive form, we get $45-49.$
In this manner, we will obtain the class for each lower boundary point.
We will find the frequency of each class from the given ‘more than’ cumulative frequency as follows :
Frequency of a class $= \ ($‘more than’ cumulative frequency of a class$) \ – \ ($‘more than cumulative frequency of immediate following class$)$
Frequency of class for $44.5 = ($‘more than’ cumulative frequency of a class for $44.5) \ – \ ($‘more than cumulative frequency of immediate following class for $49.5)$
$= (500 – 470) = 30$
In this manner we will obtain the frequency for the rest of classes.
For the given data, we get the inclusive continuous frequency distribution as follows :
Lower Boundrary Point or more More then cumulative frequency $CF$ class Frequency $f$
$44.5$ $500$ $45-49$ $500-470=30$
$49.5$ $470$ $50-54$ $470-390=80$
$54.5$ $390$ $55-59$ $390-290=100$
$59.5$ $290$ $60-64$ $290-240=50$
$64.5$ $240$ $65-69$ $240-90=150$
$69.5$ $90$ $70-74$ $90-10=80$
$74.5$ $10$ $75-79$ $10-0=10$
$79.5$ $0$ $80-84$ $=0$
- - Total $n=500$

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