Question

Let m be the mid-point and 1 be
the lower class limit of a class in a
continuous frequency distribution. The
upper class limit of the class is:
(a) 2m +1 (b) 2m-1 (c) m-1 (d) m-2​

Answer 1

Answer:

Ans = (b) 2m-1

Step-by-step explanation:

Upper limit = Mid value + (Mid value-Lower limit)

Answer 2

Upper class limit of the class is 2m - 1 if m is mid point , 1 is lower class limit

Given:

• Continuous Frequency Distribution
• mid-point is m
• Lower class limit  1

To Find:

• Upper class limit

Solution:

• mid Value = ( Lower class limit + Upper Class Limit)/2

Step 1:

Substitute mid value =  m  and Lower class limit = 1 in the formula

m = (1  + Upper Class Limit)/2

Step 2:

Solve for Upper Class Limit

2m = (1  + Upper Class Limit)     ( Multiply both sides by 2)

2m - 1 = Upper class Limit  ( Subtract 1 from both sides)

Hence upper class limit of the class is 2m - 1

Correct option is  b) 2m - 1