Comparing Two Quiz Sections
Two Grade 10 sections, A and B, each have five students selected for an optional mathematics quiz marked out of 100. Section A scores: 40, 50, 60, 70, 80. Section B scores: 55, 58, 60, 62, 65. The optional mathematics teacher claims Section A is stronger because its top score is 80 and its marks are more spread out. The principal replies that for a team event consistency matters more than a single high score. The school defines a 'high performer' as a student scoring more than 60. One of the ten students is to be chosen at random for a scholarship interview. The school also plans to send a two-student team to a quiz where the total score decides the winner, and must choose one section to represent the school.
1. Calculate the mean and population standard deviation of each section. Does the teacher’s claim that Section A is stronger stand up?
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Section A mean = (40 + 50 + 60 + 70 + 80)/5 = 300/5 = 60. Section B mean = (55 + 58 + 60 + 62 + 65)/5 = 300/5 = 60. The means are equal, so Section A is not stronger on average.
For Section A, deviations from 60 are −20, −10, 0, 10, 20. Their squares are 400, 100, 0, 100, 400, summing to 1,000. Population variance = 1,000/5 = 200, so standard deviation = √200 ≈ 14.14.
For Section B, deviations from 60 are −5, −2, 0, 2, 5. Their squares are 25, 4, 0, 4, 25, summing to 58. Population variance = 58/5 = 11.6, so standard deviation = √11.6 ≈ 3.41.
Section B is far more consistent: its scores cluster tightly around 60. Section A has a higher top score but also much greater spread, including a score of 40. The teacher has confused variability and a high maximum with overall strength. If sample standard deviations were used, A would be about 15.81 and B about 3.81, giving the same conclusion.
2. Find the probability that a randomly chosen student is a high performer (score more than 60), and the probability that such a high performer is from Section A.
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A high performer scores more than 60, so 60 itself does not count.
In Section A, the scores above 60 are 70 and 80: 2 students. In Section B, the scores above 60 are 62 and 65: 2 students. Altogether 4 of the 10 students are high performers.
P(high performer) = 4/10 = 2/5 = 0.4.
Given that a student is a high performer, 2 of the 4 come from Section A, so P(Section A | high performer) = 2/4 = 1/2.
A common error is to count the score 60 as a high score. If the definition were 'at least 60', there would be 6 high performers and the first probability would be 6/10, but under the stated definition the correct probability is 4/10.
3. For a two-student team whose total score decides the winner, which section should the school choose? Use the means and standard deviations to justify your recommendation, and mention the risk.
Read the worked answer
Both sections have mean 60, so for two students the expected total is 60 + 60 = 120 from either section. The difference is in the spread. Section B has standard deviation about 3.41, so its two scores are likely to be close to 60 each, giving a total close to 120 with little variation. Section A has standard deviation about 14.14, so a pair could total around 150 (for example, 70 + 80) but could also total around 90 (for example, 40 + 50).
For a reliable team total, Section B is the better choice because its lower standard deviation means predictable performance near the mean. The risk with Section B is that it is unlikely to produce an exceptionally high total. Section A offers a higher ceiling but also a much higher chance of a low total. If the school values consistency, it should choose Section B; choosing Section A is a gamble on a high pair of scores.