Question

Marks obtained by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is:

• A. 48
• B. 44
• C. 40
• D. 52

Hint:

When solving for the total number of students using the median formula, always ensure the proper use of the class width and cumulative frequency before the median class.

Answer 1

The Correct Option is B

To solve this problem, we need to use the concept of the median in a grouped frequency distribution. Given that:

• Median = 14 
• Median Class Interval = 12-18
• Median Class Frequency = 12
• Number of students with marks less than 12 = 18

The formula for finding the median in a grouped data set is:

\(Median = L + \left( \frac{N/2 - F}{f} \right) \times h\)

• \(L\) is the lower limit of the median class (12 in this case).
• \(N\) is the total number of observations.
• \(F\) is the cumulative frequency of the class preceding the median class (students getting marks less than 12, which is 18).
• \(f\) is the frequency of the median class (12).
• \(h\) is the class width (18 - 12 = 6).

Substitute the known values into the formula:

\(14 = 12 + \left( \frac{N/2 - 18}{12} \right) \times 6\)

Solving for \(N\):

\(14 - 12 = \left( \frac{N/2 - 18}{12} \right) \times 6\)

\(2 = \left( \frac{N/2 - 18}{12} \right) \times 6\)

\(2/6 = \frac{N/2 - 18}{12}\)

\(1/3 = \frac{N/2 - 18}{12}\)

Convert this to fraction and solve:

\(12 \times 1/3 = N/2 - 18\)

\(4 = N/2 - 18\)

\(N/2 = 4 + 18\)

\(N/2 = 22\)

\(N = 44\)

Therefore, the total number of students is 44. This matches with the given correct answer option.

Answer 2

Step 1: The median of a grouped data is given by the formula: \[ \text{Median} = \ell + \left( \frac{\frac{N}{2} - F}{f} \right) \times h \] where: 
- \( \ell \) is the lower boundary of the median class, - \( N \) is the total number of observations, 
- \( F \) is the cumulative frequency before the median class, 
- \( f \) is the frequency of the median class, 
- \( h \) is the class width. From the problem, we are given: 
- Median class interval: 12-18, - Median class frequency \( f = 12 \), - \( \ell = 12 \), 
- Median = 14, 
- Number of students with marks less than 12 is 18. 
Step 2: Using the formula: \[ 14 = 12 + \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6 \] Simplifying the equation: \[ 14 - 12 = \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6 \] \[ 2 = \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6 \] \[ 2 = \frac{\frac{N}{2} - 18}{2} \] \[ 4 = \frac{N}{2} - 18 \] \[ \frac{N}{2} = 22 \quad \Rightarrow \quad N = 44 \]