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S5 Maths: Dispersion and Standard Deviation: Paper 3 (Challenge)
S5 Maths: Dispersion and Standard Deviation: Paper 3 (Challenge)
Ten items on grouped data, combining two sets and transforming variables, always via the variance.
Direct answers Items 1-3: variance 9 with standard deviation 3, then 66 and 161. Items 4-6: 11, standard deviation 8 with mean 23, mean falling from 10 to 5. Items 7-9: 4.12, 3.33, 25. Item 10: not correct.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S5 Maths: Dispersion and Standard Deviation |
| Level | Challenge |
02 Questions (10 items)
- FoundationA set of 10 values has a sum of squared deviations of 90. Find the variance and the standard deviation.
- FoundationGroup X has 10 students averaging 60 marks and group Y has 15 averaging 70. Find the combined mean.
- Foundation20 students average 155 cm in height and another 30 average 165 cm. Find the mean height of all 50 students.
- IntermediateFor grouped data, 4 students fall in 0 to 10 marks and 6 in 10 to 20 marks. Estimate the mean using class mid-points.
- IntermediateLet Y = 2X + 3. If X has mean 10 and standard deviation 4, find the mean and standard deviation of Y.
- IntermediateWhen 30 is removed from the data set 2, 4, 6, 8, 30, how do the mean and the standard deviation change?
- ChallengeTwo sets of 10 values share a mean of 10 and have standard deviations 3 and 5. Find the standard deviation of the 20 combined values (2 d.p.).
- ChallengeThe data set D is 7, 9, 10, 11, 13 with variance 4. If one more value equal to the mean, 10, is added, what is the new variance (2 d.p.)?
- ChallengeA symmetric distribution has mean 25 and standard deviation 4. Find its median.
- Error redoA student averages the two standard deviations to stand for the combined standard deviation. Is that correct?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Variance 9, standard deviation 3 | 90 / 10 = 9, and the square root of 9 is 3. |
| 2 | 66 marks | (10 x 60 + 15 x 70) / 25 = 66. |
| 3 | 161 cm | (20 x 155 + 30 x 165) / 50 = 161. |
| 4 | 11 | The mid-points are 5 and 15, so (4 x 5 + 6 x 15) / 10 = 11. |
| 5 | Mean 23, standard deviation 8 | Multiplying by 2 makes the standard deviation 8, and the mean is 2 x 10 + 3 = 23. |
| 6 | The mean falls from 10 to 5 and the standard deviation falls from 10.2 to 2.24 | 30 is an extreme value far from the rest, so removing it lowers the dispersion sharply. |
| 7 | 4.12 | The combined variance is (9 + 25) / 2 = 17, whose square root is about 4.12. |
| 8 | 3.33 | The sum of squared deviations stays at 20, and dividing by 6 gives 3.33. |
| 9 | 25 | In a symmetric distribution the mean and the median are equal. |
| 10 | Not correct | Standard deviations cannot be averaged directly; the variances must be combined first, then the square root taken. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Write the mean first, then list each squared deviation; that keeps the working clean.Tutor Registration →
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