Question

If x be the mid-point and 1 be the upper-class limit of a class in a continuous frequency

distribution. What is the lower limit of the class?​

Answer 1

Answer:

(upper limit + lower limit)/2 = mid point

given; mid point = x, upper limit = 1

to find lower class limit => LL

using the above formula, MP = 1/2[UL + LL]

=> (1 + LL)/2 = x

=> 1 + LL = 2x

=> LL = (2x - 1)