Mathematics questions, chapter 8

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Measures of Dispersion for Ungrouped Data

45 marks. Use Practice to check each answer as you go, or Exam to answer everything first. Show your working for Section B. Standard deviation here means the population standard deviation (divide by N), as in the textbook.

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

A set of data has a very small range. What does this tell you about its dispersion?

The values are widely scattered.
The values are close together.
The mean must be small.
The standard deviation must be large.

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A small range means the largest and smallest values are close, so the dispersion is small. The size of the mean says nothing about dispersion.

For the teacher: Remembering, low difficulty, standard 8.1.1

Textbook p. 212

Find the range of 17, 9, 31, 22, 8, 25.

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Range = largest − smallest = 31 − 8 = 23. 22 comes from 31 − 9, which is not the smallest value.

For the teacher: Remembering, low difficulty, standard 8.2.1

Textbook p. 219

Diagram 1 shows a dot plot of the number of books read by 10 pupils in a month. What is the difference between the largest and the smallest value?

Dot plot on a number line from 3 to 11: one dot at 4, two at 5, three at 6, two at 7, one at 8 and one at 10 — Diagram 1

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The smallest value is 4 and the largest is 10. 10 − 4 = 6. The answer 7 comes from 10 − 3, using the start of the scale instead of the smallest dot.

For the teacher: Understanding, low difficulty, standard 8.1.1

Textbook p. 213

Find the interquartile range of 3, 5, 6, 8, 9, 12, 14.

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n = 7, the median is 8. The lower part is 3, 5, 6 so Q1 = 5. The upper part is 9, 12, 14 so Q3 = 12. IQR = 12 − 5 = 7. 11 is the range.

For the teacher: Applying, low difficulty, standard 8.2.1

Textbook p. 220

The table shows the scores of 40 pupils in a quiz. Find the interquartile range.

Score	1	2	3	4	5
Number of pupils	3	5	12	14	6

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Cumulative frequencies: 3, 8, 20, 34, 40. Q1 is the 10th value = 3. Q3 is the 30th value = 4. IQR = 4 − 3 = 1. The range is 4.

For the teacher: Applying, medium difficulty, standard 8.2.1

Textbook p. 221

Find the variance of 2, 4, 6, 8.

20
2.24
6.67
5

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Mean = 5. Variance = (9 + 1 + 1 + 9) ÷ 4 = 5. 2.24 is the standard deviation. 6.67 comes from dividing by 3 instead of 4.

For the teacher: Applying, low difficulty, standard 8.2.1

Textbook p. 221

The values 12, 14, 15, 15, 17, 18, 90 contain an outlier. Which measure of dispersion is the most suitable to describe the data?

Range
Interquartile range
Mean
Median

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The interquartile range ignores the extreme values, so it is not affected much by the outlier 90. The range depends on 90. Mean and median are not measures of dispersion.

For the teacher: Understanding, low difficulty, standard 8.2.2

Textbook p. 224

A set of data has a standard deviation of 2.5. Each value is multiplied by 4 and then 3 is subtracted. What is the new variance?

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Subtracting 3 changes nothing. Multiplying by 4 makes the standard deviation 10, so the variance is 10² = 100. 10 is the new standard deviation.

For the teacher: Applying, medium difficulty, standard 8.2.4

Textbook p. 229

Diagram 2 shows a box plot of the times taken, in seconds, by a group of pupils. What is the interquartile range?

Box plot on a number line from 10 to 50: minimum 12, Q1 20, median 26, Q3 35, maximum 48 — Diagram 2

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The box runs from Q1 = 20 to Q3 = 35. IQR = 35 − 20 = 15. 36 is the range, 48 − 12.

For the teacher: Understanding, low difficulty, standard 8.2.3

Textbook p. 227

Five numbers have a sum of 60 and a sum of squares of 800. What is the standard deviation?

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Mean = 12. Variance = 800 ÷ 5 − 12² = 160 − 144 = 16. Standard deviation = √16 = 4. 16 is the variance.

For the teacher: Applying, high difficulty, standard 8.2.6

Textbook p. 236

Amir and Bala each took 5 tests. Both have a mean mark of 68. The standard deviation of Amir's marks is 3.1 and Bala's is 8.4. Which statement is correct?

Bala is more consistent because his standard deviation is larger.
Amir is more consistent because his standard deviation is smaller.
Both are equally consistent because the means are the same.
Amir scored higher marks because his standard deviation is smaller.

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With equal means, the smaller standard deviation shows marks closer to the mean, which is more consistent. The standard deviation does not say who scored higher.

For the teacher: Analysing, low difficulty, standard 8.2.5

Textbook p. 234

A set of data has a range of 15, an interquartile range of 10.5 and a standard deviation of 5.3. Each value is decreased by 7. What are the new range, interquartile range and standard deviation?

8, 3.5, 5.3
15, 3.5, 5.3
15, 10.5, 5.3
15, 10.5, −1.7

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Adding or subtracting the same number shifts the data but does not change how spread out it is. All three measures stay the same.

For the teacher: Understanding, low difficulty, standard 8.2.4

Textbook p. 228

Section B: structured questions

33 marks

Answer all questions. Show your working.

The ages, in years, of 12 members of a swimming club are 15, 22, 18, 30, 25, 18, 20, 27, 24, 16, 21, 19.

[2 marks] (a)Arrange the ages in ascending order and state the range.

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15, 16, 18, 18, 19, 20, 21, 22, 24, 25, 27, 30 [1]1
Range = 30 − 15 = 15 [1]1

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[2 marks] (b)Find the first quartile and the third quartile.

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Q1 = (18 + 18) ÷ 2 = 18 [1]1
Q3 = (24 + 25) ÷ 2 = 24.5 [1]1

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[1 mark] (c)Find the interquartile range.

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24.5 − 18 = 6.5 [1]1

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For the teacher: Applying, low difficulty, standard 8.2.1

Textbook p. 220

A set of data is 6, 8, 9, 11, 16.

[1 mark] (a)Calculate the mean.

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(6 + 8 + 9 + 11 + 16) ÷ 5 = 50 ÷ 5 = 10 [1]1

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[2 marks] (b)Calculate the variance.

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Σx² = 36 + 64 + 81 + 121 + 256 = 558 [1]1
558 ÷ 5 − 10² = 11.6 [1]1

Using squared differences: (16 + 4 + 1 + 1 + 36) ÷ 5 = 58 ÷ 5 = 11.6.

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[1 mark] (c)Calculate the standard deviation, correct to 3 decimal places.

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√11.6 = 3.406 [1]1

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For the teacher: Applying, low difficulty, standard 8.2.1

Textbook p. 221

The table shows the number of goals scored by a school team in 20 matches.

Number of goals	0	1	2	3	4
Number of matches	4	7	5	3	1

[1 mark] (a)Calculate the mean number of goals.

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Σfx = 0 + 7 + 10 + 9 + 4 = 30; mean = 30 ÷ 20 = 1.5 [1]1

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[3 marks] (b)Calculate the variance and the standard deviation.

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Σfx² = 0 + 7 + 20 + 27 + 16 = 70 [1]1
Variance = 70 ÷ 20 − 1.5² = 1.25 [1]1
Standard deviation = √1.25 = 1.118 [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 8.2.1

Textbook p. 222

Diagram 3 shows a box plot of the time, in minutes, that 15 pupils spent on homework in one evening.

Box plot on a number line from 0 to 45: minimum 4, Q1 12, median 20, Q3 30, maximum 41 — Diagram 3

[2 marks] (a)State the median and the third quartile.

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Median = 20 minutes [1]1
Q3 = 30 minutes [1]1

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[2 marks] (b)Find the range and the interquartile range.

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Range = 41 − 4 = 37 [1]1
IQR = 30 − 12 = 18 [1]1

Tick each point you got.

For the teacher: Understanding, low difficulty, standard 8.2.3

Textbook p. 227

Diagram 4 shows dot plots of the marks of 10 pupils from Class P and 10 pupils from Class Q in the same test.

Two dot plots on one scale from 3 to 13. Class P: one dot at 6, two at 7, four at 8, two at 9, one at 10. Class Q: one dot each at 4, 5, 6, 8, 9, 10, 11, 12 and two at 7 — Diagram 4

[2 marks] (a)Which class has the greater dispersion? Give a reason.

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Class Q [1]1
Its marks range from 4 to 12 (difference 8) while Class P ranges from 6 to 10 (difference 4) [1]1

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[1 mark] (b)Both classes have a mean mark close to 8. Which class is more consistent in this test? Explain.

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Class P: its marks are closer together around the mean [1]1

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For the teacher: Analysing, medium difficulty, standard 8.1.2

Textbook p. 217

A coach must choose one of two athletes for a 100 m race. The times, in seconds, of five trials are shown.

Athlete A	12.6	12.4	12.9	12.5	12.6
Athlete B	12.2	13.1	12.5	12.9	12.8

[2 marks] (a)Calculate the mean time of each athlete.

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A: 63 ÷ 5 = 12.6 [1]1
B: 63.5 ÷ 5 = 12.7 [1]1

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[3 marks] (b)Calculate the standard deviation of each athlete's times, correct to 3 decimal places.

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A: Σx² = 793.94; variance = 793.94 ÷ 5 − 12.6² = 0.028 [1]1
σ(A) = √0.028 = 0.167 [1]1
σ(B) = √0.1 = 0.316 [1]1

For B: Σx² = 806.95, so variance = 806.95 ÷ 5 − 12.7² = 161.39 − 161.29 = 0.1.

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[1 mark] (c)Which athlete is more consistent? Give a reason.

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Athlete A, because the standard deviation is smaller [1]1

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For the teacher: Evaluating, high difficulty, standard 8.2.5

Textbook p. 235

A set of 10 numbers has Σx = 80 and Σx² = 700.

[2 marks] (a)Calculate the mean and the variance.

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Mean = 80 ÷ 10 = 8 [1]1
Variance = 700 ÷ 10 − 8² = 6 [1]1

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[2 marks] (b)A number p is added to the set and the mean becomes 9. Find the value of p.

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(80 + p) ÷ 11 = 9 [1]1
p = 99 − 80 = 19 [1]1

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[2 marks] (c)Find the standard deviation of the new set of 11 numbers.

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Σx² = 700 + 19² = 1 061; variance = 1 061 ÷ 11 − 9² = 15.45 [1]1
σ = √15.45 = 3.931 [1]1

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[1 mark] (d)State whether the standard deviation became larger or smaller. Explain.

largersmaller

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larger

p = 19 is far from the old mean 8, so the standard deviation increases from 2.449 to 3.931.

For the teacher: Evaluating, medium difficulty, standard 8.2.6

Textbook p. 236

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Section B: structured questions

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19