Class Project Marks and a Missing Score
A Grade 7 class of 25 students completed a project scored out of 10. The teacher recorded the marks in a frequency table: score 4 - 2 students; score 5 - 3; score 6 - 5; score 7 - 6; score 8 - 5; score 9 - 3; score 10 - 1. One student, Mira, was absent and her mark is not yet included, so the table covers only 24 students. The teacher says the class average must be at least 7.0 to qualify for a school award. A colleague argues that because most marks are 6 or 7, the class cannot reach 7.0 even with Mira's mark.
1. Compute the mean of the 24 recorded marks, then find the minimum mark Mira needs for the class average of all 25 students to reach at least 7.0.
Read the worked answer
Sum of 24 marks: 4×2 + 5×3 + 6×5 + 7×6 + 8×5 + 9×3 + 10×1 = 8 + 15 + 30 + 42 + 40 + 27 + 10 = 172. Mean of 24 = 172 ÷ 24 ≈ 7.17.
For 25 students to average at least 7.0, the total must be at least 25 × 7 = 175. Mira needs 175 − 172 = 3. Since marks are out of 10, a mark of 3 is possible, so any mark ≥ 3 works. The colleague is wrong.
2. Find the median and mode of the 24 recorded marks. Does the median support the colleague's claim?
Read the worked answer
Ordered, the 24 marks occupy positions 1–24. The median is the average of the 12th and 13th values. Cumulative counts: 4 → 2; 5 → 5; 6 → 10; 7 → 16. So positions 11 and 12 are both 7, and position 13 is 7. The 12th and 13th values are both 7, so median = 7.
Mode: score 7 appears 6 times, the most, so mode = 7.
The median of 7 equals the 7.0 target; it does not show the class is below target. The colleague confused 'most marks are 6 or 7' with 'average below 7'; the high marks (8, 9, 10) pull the mean above 7.
3. If Mira's mark were 2 instead of 3, what would the class average be, and how would you describe the effect of one low mark on the mean versus the median?
Read the worked answer
New total = 172 + 2 = 174, mean = 174 ÷ 25 = 6.96, below 7.0.
With 25 marks the median is the 13th value. Cumulative with Mira's 2: 2 → 1; 4 → 3; 5 → 6; 6 → 11; 7 → 17. The 13th value is 7, so the median stays 7.
The mean drops by about 0.21 while the median is unchanged. One extreme low mark shifts the mean noticeably but barely affects the median, which is why the median is called resistant to outliers.