Question

in a continous frequency
distribution table if the mid point of a class is 42 and class size is 10 then let us write the upper and lower limit of the class.


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Answer 1

Answer:

Let upper limit and lower limit be x and y

⇒

2

x+y

=42 [ Given ]

∴ x+y=84 ------ ( 1 )

Now, according to the question,

⇒ x−y=10 ------ ( 2 )

Adding equation ( 1 ) and ( 2 ) we get,

2x=94

∴ x=47

And 47−y=10

∴ y=37

∴ Required class interval =37−47

Step-by-step explanation:

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