The Project Survey: A Missing Frequency and a Data Entry Error
Thirty students in a Form 5 class recorded the number of hours each spent on a mathematics project during one week. The teacher spilled coffee on one entry of the summary table:
| Hours | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Number of students | 3 | 8 | x | 7 | 4 | 2 |
The teacher is certain the class has exactly 30 students and that the hours were recorded in whole numbers only, but cannot recover x from memory.
A neighbouring class of 20 students did the same project; their mean time was 2.6 hours.
Later, while checking the original sheets, the teacher discovers that one student wrote '5 hours' when the true figure was 15 hours (the student had worked over two weekends). The reconstructed value of x is confirmed as correct. The teacher must now write a single sentence for the school newsletter describing how long the class typically spent on the project, and is unsure whether to quote the mean or the median.
1. Find x, then calculate the mean, median and mode number of hours for the class.
Read the worked answer
Finding x The frequencies must total 30: 3 + 8 + x + 7 + 4 + 2 = 30 24 + x = 30 → x = 6
Mean Total hours = (0×3) + (1×8) + (2×6) + (3×7) + (4×4) + (5×2) = 0 + 8 + 12 + 21 + 16 + 10 = 67 Mean = 67 ÷ 30 = 2.23 hours (2 d.p.)
Median With 30 values, the median is the average of the 15th and 16th values in order. Cumulative frequencies: 3, 11, 17 - so both the 15th and 16th values fall in the '2 hours' group. Median = 2 hours
Mode = 1 hour (frequency 8, the highest)
The three averages disagree: the data is skewed to the right by the small number of students who spent 4 or 5 hours, which pulls the mean above the median.
2. Calculate the mean time for the two classes combined, and explain why it is not simply the average of 2.23 and 2.6.
Read the worked answer
Total hours available Form 5 class: 67 hours (30 students) Neighbouring class: mean 2.6 × 20 students = 52 hours
Combined total = 67 + 52 = 119 hours over 30 + 20 = 50 students Combined mean = 119 ÷ 50 = 2.38 hours
Why not just average the two means? (2.23 + 2.6) ÷ 2 = 2.415, which is wrong. A mean is a total divided by a count, so combining groups means combining totals and counts. Simply averaging the two means treats the classes as if they were the same size; the larger class (30 students) should carry more weight.
A weighted check confirms it: (30 × 2.2333 + 20 × 2.6) ÷ 50 = (67 + 52) ÷ 50 = 2.38.
The 0.035-hour difference here looks small, but the same mistake on groups of very different sizes - for example 200 students against 5 - can produce a seriously misleading figure.
3. Correct the recording error and recompute the mean. State, with reasons, which average the teacher should quote in the newsletter.
Read the worked answer
Corrected mean Replace one value of 5 with 15: New total = 67 − 5 + 15 = 77 hours Mean = 77 ÷ 30 = 2.57 hours (2 d.p.)
Other averages after the correction The ordered data still runs 3 zeros, 8 ones, 6 twos, so the 15th and 16th values are still 2 → median unchanged at 2 hours. Mode still 1 hour.
The mean has risen by 0.33 hours because of a single student.
Which to quote?
- The mean (2.57) uses every value but is dragged upward by one unusual student; a reader would form a picture of a class working far longer than most actually did.
- The median (2 hours) is resistant to that one extreme value and describes the typical student accurately: half the class did 2 hours or less.
Recommendation: quote the median - 'the typical student spent about 2 hours' - and, if space allows, note the mean of 2.57 alongside it. The 15-hour figure is a genuine value, not an error to be deleted, so it must stay in the data set; it simply should not be allowed to define the headline. Reporting both is the honest option, because quoting only the median hides the fact that some students did far more.