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Let m be the mid-point and l be the upper class limit of a class in a continuous frequency distribution. Find the lower class limit of the class.

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m + l

2 m − l

m – l

2m – 2l

answer is B.

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Detailed Solution

It is given that, m be the mid-point and l be the upper class limit of a class.
We have to find the lower limit of the class.
Let the lower limit be x.
Mid-point m is given by,

m = upper limit + lower limit 2

Substitute the known values within the formula,

m = l + x 2 x = 2 m − l

The lower limit is

2 m − l

.
Hence, option 2 is correct.

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