Question

6.   If x be the mid-point and 1 be the upper class limit of a class in a containous frequency distribution. What is the lower limit of the class?         
(a) x – 1
(b) 3x + 8
(c) 2x + 2
(d) 2x – 1​

Answer 1

Answer:

2m-l

Step-by-step explanation:

Let x and y be the lower and upper class limit of a continuous frequency distribution

Now, mid-point of a class = (x+y) /2=m (given)

X+y=2m=x+l=2m

[ therefore y=l=upper class limit (given)]

X=2m-l

Hence, the lower class limit of the class is 2m-l

Answer 2

Answer:

(b) Let x and y be the lower and upper class limit of a continuous frequency distribution. Hence, the lower class limit of the class is 2m – l.