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Question
In a frequency distribution, the mid-value of a class is 15 and the class intervals is 4. The lower limit of the class is
Options
10
12
13
14
Solution
Let l and m respectively be the lower and upper limits of the class. Then the mid-value of the class is `(l+m)/2` and the class-size is(m-l) .
Therefore, we have two equations
`(l+m)/2 = 15`
⇒ l + m = 30
m - l = 4
Subtracting the second equation from the first equation, we have
`(l+m)-(m-l)=30-4`
`⇒l+m-m+l=26`
`⇒2l=26`
`⇒l=13`
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