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If m is the mid-point and l is the upper limit of a class in a continous frequency distribution, then lower class limit of the class isA. 2m-uB. 2m+uC. m-uD. m+u |
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Answer» Correct Answer - A Let l be the lower limit. Then, `m=(l+u)/(2)Rightarrow l=2m-u` |
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