Marks Table with a Smudged Frequency: Mean, Median and Probability
In a Class X mathematics test marked out of 50, 40 students were awarded marks grouped into class intervals. The frequency table recorded: 0–10, 4 students; 10–20, 6 students; 20–30, f students; 30–40, 10 students; 40–50, 8 students. The teacher knows the class has exactly 40 students on the register, but the value of f was smudged. A student claims that the median can be read off as 25, the midpoint of the 20–30 class, because that class contains the middle mark. Another student wants to use the grouped mean as if it were the exact mean of every student's mark. The teacher needs a defensible value of f, a grouped mean, a grouped median, and a probability statement about a student chosen at random from the class.
1. Find f and calculate the grouped mean. Comment on the precision of the mean.
Read the worked answer
The frequencies must total 40:
4 + 6 + f + 10 + 8 = 28 + f = 40, so f = 12.
Using class midpoints 5, 15, 25, 35 and 45:
- 0–10: 4 × 5 = 20
- 10–20: 6 × 15 = 90
- 20–30: 12 × 25 = 300
- 30–40: 10 × 35 = 350
- 40–50: 8 × 45 = 360
Total = 20 + 90 + 300 + 350 + 360 = 1120.
Grouped mean = 1120/40 = 28 marks.
This is an estimate, because each student's exact mark is replaced by the midpoint of the interval. It would equal the exact mean only if the marks in each class averaged exactly the midpoint, which is not guaranteed. The value 28 is defensible as a grouped mean but should not be presented as the exact mean of all 40 individual marks.
2. Calculate the grouped median and evaluate the student's claim that it is 25.
Read the worked answer
The median position is n/2 = 40/2 = 20. Cumulative frequencies:
- 0–10: 4
- 10–20: 10
- 20–30: 22
Since the 20th observation falls in 20–30, that is the median class. With lower boundary L = 20, cumulative frequency before the class cf = 10, frequency f = 12 and class width h = 10:
Median = L + [(n/2 − cf)/f] × h = 20 + [(20 − 10)/12] × 10 = 20 + (10/12) × 10 = 20 + 8.33 = 28.33 marks (to 2 d.p.).
The student's 25 is merely the midpoint of the median class. The median is the mark of the 20th student, and the class contains 12 students spread across the interval. After the first 10 students, 10 more are needed to reach the 20th, so the median lies 10/12 of the way through the class, not halfway. The formula gives about 28.3, not 25.
3. Find the probability that a randomly chosen student scored 30 or above, and compare it with the probability of scoring below 20.
Read the worked answer
There are 40 students in total. Those scoring 30 or above are in the 30–40 and 40–50 classes:
10 + 8 = 18 students.
So P(30 or above) = 18/40 = 9/20 = 0.45.
Those scoring below 20 are in the 0–10 and 10–20 classes:
4 + 6 = 10 students.
So P(below 20) = 10/40 = 1/4 = 0.25.
A randomly chosen student is therefore more likely to have scored 30 or above than below 20; the difference in probabilities is 0.45 − 0.25 = 0.20. The events are mutually exclusive, and together with the 20–30 group they account for all 40 students: 10 + 12 + 18 = 40.