Question

The marks obtained by 80 students of class X in a mock test of Mathematics are given below in the table. Find median and the mode of the data :

Hint:

When given "more than" or "less than" data, always cross-check that the sum of your calculated frequencies equals the total number of students. Here, $3+5+7+10+12+15+12+6+2+8 = 80$. Correct!

Answer 1

Step 1: Understanding the Concept:
The given data is in "more than" cumulative frequency format. First, convert it into a standard frequency distribution table with class intervals.
Step 2: Key Formula or Approach:
Class Interval Frequency = (CF of current class) - (CF of next class).
Median $= l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Step 3: Detailed Explanation:
Convert to standard frequency table:

Finding Median:
$N/2 = 40$. The cumulative frequency just greater than 40 is 52, corresponding to class interval 50-60.
$l = 50$, $cf = 37$ (of preceding class), $f = 15$, $h = 10$.
Median $= 50 + \left( \frac{40 - 37}{15} \right) \times 10 = 50 + \left( \frac{3}{15} \right) \times 10 = 50 + 2 = 52$.
(Correction: Checking calculation again, $50 + 2 = 52$).
Finding Mode:
Maximum frequency is 15 in class 50-60.
$l = 50$, $f_1 = 15$, $f_0 = 12$, $f_2 = 12$, $h = 10$.
Mode $= 50 + \left( \frac{15 - 12}{2(15) - 12 - 12} \right) \times 10 = 50 + \left( \frac{3}{30 - 24} \right) \times 10 = 50 + \left( \frac{3}{6} \right) \times 10 = 50 + 5 = 55$.
Step 4: Final Answer:
The Median is 52 and the Mode is 55.