This set has twelve original questions, ordered from easier to harder, covering all five lessons in statistics and distributions. Questions 1 to 4 are warm-ups on averages, 5 to 8 cover weighted and grouped data, and 9 to 12 mix skills and wording.
Attempt each question on paper, and write a one-line conclusion wherever the question asks for a reason. Only then open the answer.
Mark the ones you got wrong and use the routing list at the end. The statistics and distribution explorer can check a small data set after you have tried it by hand.
Questions
1. Find the mean, median, mode and range of: 3, 5, 5, 6, 8, 9, 5. Give the mean to 2 decimal places.
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Sum = 3 + 5 + 5 + 6 + 8 + 9 + 5 = 41, and there are 7 values, so the mean is 41 ÷ 7 = 5.857…, which is 5.86.
In order: 3, 5, 5, 5, 6, 8, 9. The middle (4th) value is 5, so the median is 5. The value 5 appears three times, so the mode is 5. The range is 9 − 3 = 6.
2. Five values are 1, 2, 2, 3, 12. Find the mean and the median, and say which describes a typical value better.
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Mean: 1 + 2 + 2 + 3 + 12 = 20, and 20 ÷ 5 = 4. Median: the middle value is 2.
The median describes a typical value better. The 12 is far from the others and pulls the mean up to 4, which is larger than four of the five values.
3. A shop sells T-shirts in sizes S (4), M (15), L (9) and XL (2), with the number sold in brackets. Which average should the owner use to decide which size to order most of, and what is the answer?
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The question asks for the most popular choice, so use the mode. The highest frequency is 15, so the modal size is M. A mean or median has no meaning for sizes written as letters.
4. Five house prices (RM thousand) are 310, 320, 330, 340 and 1700. The seller says the typical price is about RM600 thousand. Is this a fair claim? Show the numbers.
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Mean: 310 + 320 + 330 + 340 + 1700 = 3000, and 3000 ÷ 5 = 600. Median = 330.
The seller quoted the mean. Four of the five houses cost RM340 thousand or less, so RM600 thousand is not typical. The median, RM330 thousand, is fairer because the one expensive house pulls the mean up.
5. A final grade is 60% written test and 40% project. A student scores 55 on the test and 80 on the project. Find the final mark.
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0.6 × 55 = 33 and 0.4 × 80 = 32. Adding gives 33 + 32 = 65.
Check: 65 lies between 55 and 80 and is nearer 55, because the test has the larger weight.
6. A class has 12 boys with a mean height of 165 cm and 18 girls with a mean height of 160 cm. Find the mean height of all 30 students.
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Boys: 12 × 165 = 1980. Girls: 18 × 160 = 2880. Total = 4860 for 30 students.
4860 ÷ 30 = 162 cm. Not 162.5, which is what averaging the two means would give, because there are more girls than boys.
7. The mean of six numbers is 8. A seventh number is added and the mean becomes 9. Find the seventh number.
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Total of six numbers = 6 × 8 = 48. Total of seven numbers = 7 × 9 = 63.
The seventh number is 63 − 48 = 15.
Check: (48 + 15) ÷ 7 = 63 ÷ 7 = 9.
8. The weights of 20 parcels are grouped as follows.
| Weight w (kg) | Frequency |
|---|---|
| 0 < w ≤ 2 | 6 |
| 2 < w ≤ 4 | 10 |
| 4 < w ≤ 6 | 3 |
| 6 < w ≤ 10 | 1 |
(a) Estimate the mean weight. (b) State the modal class. (c) In which class is the median?
(d) Can you say the heaviest parcel weighs exactly 10 kg?
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(a) Midpoints are 1, 3, 5 and 8. Products: 6 × 1 = 6, 10 × 3 = 30, 3 × 5 = 15, 1 × 8 = 8. Sum = 59. Estimated mean = 59 ÷ 20 = 2.95 kg.
(b) The highest frequency is 10, so the modal class is 2 < w ≤ 4.
(c) The median lies between the 10th and 11th values. Running totals are 6, 16, 19, 20, so both are in 2 < w ≤ 4.
(d) No. The heaviest parcel is only known to be more than 6 kg and at most 10 kg.
9. Using the table in question 8, find the smallest and largest values the mean could possibly take.
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Smallest: use the lower class limits. (0 × 6 + 2 × 10 + 4 × 3 + 6 × 1) ÷ 20 = (0 + 20 + 12 + 6) ÷ 20 = 38 ÷ 20 = 1.9 kg.
Largest: use the upper limits. (2 × 6 + 4 × 10 + 6 × 3 + 10 × 1) ÷ 20 = (12 + 40 + 18 + 10) ÷ 20 = 80 ÷ 20 = 4 kg.
The estimate 2.95 lies between 1.9 and 4, as it must.
10. Data set P is 10, 12, 14, 16, 18. Data set Q is 2, 8, 14, 20, 26. Compare the two sets using an average and the range.
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P: total 70, mean 14, median 14, range 18 − 10 = 8. Q: total 70, mean 14, median 14, range 26 − 2 = 24.
Comparison: both sets have the same mean and median of 14, but P has a much smaller range (8 compared with 24), so the values in P are more consistent.
11. A student scored 70 on paper 1, which is worth 40% of the final mark. Paper 2 is worth 60%. She needs an overall mark of 76. What must she score on paper 2?
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Paper 1 contributes 0.4 × 70 = 28. She needs 76 − 28 = 48 from paper 2, which has weight 0.6.
Paper 2 mark = 48 ÷ 0.6 = 80.
Check: 28 + 0.6 × 80 = 28 + 48 = 76.
12. A journalist surveys 30 members of a chess club. Twenty-four say maths is their favourite subject. She writes, “80% of the students in the school like maths the most.” Comment on the conclusion.
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24 ÷ 30 = 0.8, so 80% of the sample chose maths. The arithmetic is correct.
The conclusion is not supported. Chess club members are not typical of the whole school, so the sample is biased, and 30 students is small compared with the whole population. The data supports “80% of the 30 chess club members asked chose maths”.
If you got these wrong
- Questions 1 to 4, wrong average or no reason: revisit choosing an average for a stated question.
- Questions 5 to 7 and 11, averages of averages or missing weights: go back to calculating a weighted mean.
- Questions 8 and 9, exact values claimed from a table: read interpreting grouped data again.
- Question 10, comparing only the centre: use comparing spread and centre together.
- Question 12, accepting a claim because the percentage is right: revisit identifying an unsupported conclusion.
The mistake log and retest queue helps you record each error type and come back to it with a fresh question a few days later. The non-calculator working trainer is useful for the larger multiplications.
If the same type of error keeps coming back, online one-to-one Mathematics tuition gives you an experienced teacher who can follow your written working and help you fix the habit behind it.