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

Finding Median and Mode from the Given Data (Corrected):

We are given a cumulative frequency table of marks obtained by 80 students. Let's solve step by step.

Step 1: Convert 'Number of Students' to 'Frequency (f)'
Frequency for each class is calculated as:
f = Previous cumulative frequency − Current cumulative frequency

Class Interval (Marks) | Cumulative Frequency (CF) | Frequency (f)
0-10                   | 80                        | 3    (80-77)
10-20                  | 77                        | 5    (77-72)
20-30                  | 72                        | 7    (72-65)
30-40                  | 65                        | 10   (65-55)
40-50                  | 55                        | 12   (55-43)
50-60                  | 43                        | 15   (43-28)
60-70                  | 28                        | 12   (28-16)
70-80                  | 16                        | 6    (16-10)
80-90                  | 10                        | 2    (10-8)
90-100                 | 8                         | 8    (8-0)

Step 2: Find Median
• Total students (N) = 80
• Median class = The class containing N/2 = 40th student
• From cumulative frequency, 50-60 class contains the 40th student (CF before = 43−15 = 28, cumulative frequency just before median = 43)
• Median formula:

Median = L + [(N/2 − CF_prev) / f_m] × h

Where:
L = lower boundary of median class = 50
CF_prev = cumulative frequency before median class = 43−15 = 28
f_m = frequency of median class = 15
h = class width = 10

Median = 50 + [(40 − 28)/15] × 10
       = 50 + (12/15) × 10
       = 50 + 8
       = 58

**Correction check:**
Actually, median lies in class 40-50 (cumulative frequency before = 30, frequency = 25?), let's verify cumulative frequencies step-by-step: - 0-10 → 80-77 = 3 - 10-20 → 77-72 = 5 - 20-30 → 72-65 = 7 - 30-40 → 65-55 = 10 - 40-50 → 55-43 = 12 - 50-60 → 43-28 = 15 - 60-70 → 28-16 = 12 - 70-80 → 16-10 = 6 - 80-90 → 10-8 = 2 - 90-100 → 8-0 = 8 N/2 = 80/2 = 40 → Lies in 40-50 class
CF_prev = cumulative frequency before median class = 30+3+5+7+10 = 25? Wait let's sum properly: - 0-10: f = 3 → cumulative frequency = 3 - 10-20: f = 5 → cumulative frequency = 3+5=8 - 20-30: f = 7 → cumulative frequency = 8+7=15 - 30-40: f = 10 → cumulative frequency = 15+10=25 - 40-50: f = 12 → cumulative frequency = 25+12=37 - 50-60: f = 15 → cumulative frequency = 37+15=52 N/2=40 → lies in 50-60 class (since CF just before = 37, CF at end of class = 52)
Median class = 50-60, L = 50, CF_prev = 37, f_m = 15, h = 10

Median = L + [(N/2 − CF_prev) / f_m] × h
       = 50 + [(40 − 37)/15] × 10
       = 50 + (3/15) × 10
       = 50 + 2
       = 52

Step 3: Find Mode
• Modal class = Class with highest frequency = 50-60 (f = 15)
• Mode formula:

Mode = L + [(f1 − f0) / (2f1 − f0 − f2)] × h

Where:
L = lower boundary of modal class = 50
f1 = frequency of modal class = 15
f0 = frequency of previous class = 12
f2 = frequency of next class = 12
h = class width = 10

Mode = 50 + [(15 − 12) / (2×15 − 12 − 12)] × 10
     = 50 + [3 / (30 − 24)] × 10
     = 50 + (3/6) × 10
     = 50 + 5
     = 55

Step 4: Summary
• Median = 52
• Mode = 55

Conclusion:
The median represents the middle score of the students, while the mode represents the most frequently obtained score.