Measures of Dispersion for Ungrouped Data · Form 4
Measures of Dispersion for Ungrouped Data: Practice Questions
Original SPM-style practice questions for Measures of Dispersion for Ungrouped Data, each with full worked solutions, Paper 1 multiple-choice and Paper 2 structured.
Original practice questions for Measures of Dispersion for Ungrouped Data, in the style of Mathematics Paper 1 (Objective) and Mathematics Paper 2 (Subjective). Each answer is worked so you can check your method, not just the result.
Multiple-choice (Paper 1 style)
Question 1
The data set below shows the number of goals scored by a team in 6 matches: 12, 7, 19, 4, 15, 9. What is the range of this data set?
- A. 12
- B. 15
- C. 19
- D. 23
Question 2
The lowest temperatures (in °C) recorded on six nights were −3, 5, −8, 2, 7, −1. Find the range of the temperatures.
- A. 8
- B. 12
- C. 15
- D. 16
Question 3
The standard deviation of a set of data is 0. Which statement is true about the data?
- A. All the values are equal
- B. All the values are zero
- C. The mean of the data is zero
- D. The range of the data is the largest possible
Question 4
A set of 5 data values has Σx = 40 and Σx² = 370. Find the variance of the data.
- A. 8
- B. 10
- C. 64
- D. 74
Question 5
Find the standard deviation of the data 2, 4, 6, 8, 10. (Give your answer correct to 2 decimal places.)
- A. 2.00
- B. 2.83
- C. 6.00
- D. 8.00
Question 6
The data set 3, 5, 7, 9 has a mean of 6. Calculate its variance.
- A. 4
- B. 5
- C. 6
- D. 20
Question 7
Every value in a data set with range 18 is increased by 6. What is the range of the new data set?
- A. 6
- B. 18
- C. 24
- D. 108
Question 8
The standard deviation of a set of data is 4. If every value is multiplied by 3, what is the new standard deviation?
- A. 4
- B. 7
- C. 12
- D. 36
Question 9
The standard deviation of variable x is 5. A new variable is defined by y = 2x + 3.
Find the standard deviation of y.
- A. 5
- B. 10
- C. 13
- D. 25
Structured (Paper 2 style)
Question 1 (6 marks)
The masses (in kg) of 5 parcels are 5, 7, 8, 9 and 11. (a) Calculate the mean mass of the parcels.
(b) Calculate the variance of the masses. (c) Hence, find the standard deviation of the masses.
- Mean x̄ = Σx ÷ N = (5 + 7 + 8 + 9 + 11) ÷ 5 = 40 ÷ 5 = 8 kg.
- Find each deviation from the mean: 5−8=−3, 7−8=−1, 8−8=0, 9−8=1, 11−8=3.
- Variance = Σ(x − x̄)² ÷ N = [(−3)² + (−1)² + 0² + 1² + 3²] ÷ 5 = (9 + 1 + 0 + 1 + 9) ÷ 5 = 20 ÷ 5 = 4 kg².
- Standard deviation = √(variance) = √4 = 2 kg.
Question 2 (7 marks)
The marks obtained by 7 students in a Mathematics test are 9, 11, 13, 15, 17, 19 and 21. (a) Find the range of the marks.
(b) Calculate the mean mark. (c) Using the formula variance = Σx²/N − x̄², calculate the variance and hence the standard deviation of the marks.
- Range = largest − smallest = 21 − 9 = 12 marks.
- Mean x̄ = Σx ÷ N = (9 + 11 + 13 + 15 + 17 + 19 + 21) ÷ 7 = 105 ÷ 7 = 15 marks.
- Σx² = 9² + 11² + 13² + 15² + 17² + 19² + 21² = 81 + 121 + 169 + 225 + 289 + 361 + 441 = 1687.
- Variance = Σx²/N − x̄² = 1687/7 − 15² = 241 − 225 = 16 marks².
- Standard deviation = √16 = 4 marks.
Question 3 (5 marks)
A set of five numbers 6, 9, x, 12 and 8 has a mean of 9. (a) Find the value of x.
(b) Using this value, calculate the standard deviation of the five numbers.
- Mean = Σx ÷ N, so (6 + 9 + x + 12 + 8) ÷ 5 = 9.
- 35 + x = 45, therefore x = 10.
- The five numbers are 6, 9, 10, 12, 8. Deviations from the mean 9: −3, 0, 1, 3, −1.
- Variance = [(−3)² + 0² + 1² + 3² + (−1)²] ÷ 5 = (9 + 0 + 1 + 9 + 1) ÷ 5 = 20 ÷ 5 = 4.
- Standard deviation = √4 = 2.
Question 4 (6 marks)
A data set of the monthly rainfall (in mm) over 6 months has a mean of 120 mm and a standard deviation of 15 mm. (a) It is later found that every reading was 8 mm too low.
State the corrected mean and the corrected standard deviation. (b) The original readings are instead converted to centimetres by multiplying each value by 0.1.
State the mean and standard deviation in centimetres. (c) Explain why the standard deviation changes in (b) but not in (a).
- (a) Adding 8 mm to every reading increases the mean by 8 but does not change the spread. Corrected mean = 120 + 8 = 128 mm; corrected standard deviation = 15 mm.
- (b) Multiplying every value by 0.1 multiplies both the mean and the standard deviation by 0.1. Mean = 120 × 0.1 = 12 cm; standard deviation = 15 × 0.1 = 1.5 cm.
- (c) Adding a constant shifts all data equally, so each distance from the mean is unchanged; multiplying by a constant scales every distance from the mean, so the spread (standard deviation) is scaled too.
Source:SPM: Format Pentaksiran mulai 2021, Matematik (1449)
Frequently asked questions
How do the Paper 1 and Paper 2 questions in this practice set differ?
Paper 1 questions are multiple-choice, testing quick recall of range, variance, or standard deviation for an ungrouped data set. Paper 2 questions are structured, asking you to show full working for variance and standard deviation, then interpret which of two data sets is more consistent.
Why attempt a question fully before looking at the worked solution?
Working through the calculation yourself reveals exactly where you lose track, at finding the mean, the deviations, or the squaring step. Jumping straight to the solution can look like understanding without building the habit of laying out variance and standard deviation working the way Paper 2 rewards.
What mistakes are common in ungrouped dispersion questions?
Students often forget to square each deviation before averaging, or stop halfway and report variance instead of taking the square root for standard deviation. Another common slip is comparing two data sets by mean alone, forgetting that a smaller standard deviation shows the data is more consistent.