Question

Let \(l\) be the lower class limit of a class-interval in a frequency distribution and \(m\) be the mid point of the class. Then the upper class limit of the class is :

• A. \(m + \frac{l+m}{2}\)
• B. \(l + \frac{m+l}{2}\)
• C. \(2m - l\)
• D. \(m - 2l\)

Hint:

The mid-point (\(m\)) of a class is exactly in the middle of the lower limit (\(l\)) and upper limit (\(u\)). So, \(m = \frac{l+u}{2}\). To find the upper limit (\(u\)) if you know \(l\) and \(m\): 1. Multiply mid-point by 2: \(2m = l+u\) 2. Subtract the lower limit: \(u = 2m - l\)

Answer 1

The Correct Option is C

Concept: In a frequency distribution, for any class interval:
Let \(l\) be the lower class limit.
Let \(u\) be the upper class limit.
The mid-point (\(m\)) of the class interval is the average of the lower and upper class limits. Mid-point formula: \(m = \frac{l + u}{2}\) Step 1: Use the mid-point formula We are given the mid-point formula: \[ m = \frac{l + u}{2} \] We are given \(l\) (lower class limit) and \(m\) (mid-point), and we need to find \(u\) (upper class limit). Step 2: Rearrange the formula to solve for \(u\) Multiply both sides of the equation by 2: \[ 2m = l + u \] Now, isolate \(u\) by subtracting \(l\) from both sides: \[ 2m - l = u \] So, the upper class limit \(u\) is equal to \(2m - l\). Step 3: Compare with the options The derived expression for the upper class limit is \(2m - l\). This matches option (3). Example: Let a class interval be 10-20. Lower class limit, \(l = 10\). Upper class limit, \(u = 20\). Mid-point, \(m = \frac{10+20}{2} = \frac{30}{2} = 15\). Now let's use the formula \(u = 2m - l\) to check: \(u = 2(15) - 10 = 30 - 10 = 20\). This is correct.