Question

Let 'm' be the mid value and 'l' be the upper limit of a class in a frequency distribution. The lower limit of the class is : A. 2m +l. B. 2m-l. C. m-l. D. m-2l. option B is the correct option...... Pls explain how ....... ​

Answer 1

Step-by-step explanation:

m is the mid value therefore m is also the class mark

formula for class mark is

$class mark =( lower limit + upper limit) \div 2$

thefore it is given that class mark is m and upper limit is I

therefore

m=(lower limit+ I)/2

2m=lower limit + I

therefore

lower limit =2m-I

Hope u find it helpful

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Answer 2

Step-by-step explanation:

hope u understand....