Question

Statement (1): Let m be the mid-value and x be the upper limit of a class in a continuous frequency distribution, then lower limit of this class is 2m − x.

Statement (2): For a given class: `"lower limit + upper limit"/2` = mid-value of the class.

Options

• Both the statements are true.

• Both the statements are false.

• Statement 1 is true, and statement 2 is false.

• Statement 1 is false, and statement 2 is true.

Answer 1

Both the statements are true.

Explanation:

Statement 2 correctly states the standard formula for calculating the mid-value of a class interval. By algebraically rearranging this exact formula multiplying the mid-value (m) by 2 and subtracting the upper limit (x) we can solve for the lower limit as 2m − x, confirming that Statement 1 is also mathematically true.