A) \[2m+l\]
B) \[2m-l\]
C) \[m-l\]
D) \[m-2l\]
Correct Answer: B
Solution :
Given, mid point of the class =m The upper class limit \[=l\] We know that, \[\frac{\text{upper}\,\text{limit+lower}\,\text{limit}}{\text{2}}\text{=mid}\,\text{point}\] \[\Rightarrow \]\[\text{m=}\frac{l\text{+lower}\,\text{limit}}{\text{2}}\] \[\Rightarrow \]\[\text{lower}\,\text{limit}=2m=l\] Hence, the required lower class limit is \[2m-l.\]
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