16 Measures of Dispersion
16 Measures of Dispersion
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16 Measures of Dispersion
Activity
Activity 16.1 (p. 77)
(a) Measure of dispersion: range
Reason: The highest and the lowest temperature of a day can be easily obtained and the calculation is simple.
(b) Measure of dispersion: inter-quartile range
Reason: Extreme data would not affect the result.
(c) Measure of dispersion: inter-quartile range
Reason: Can be easily obtained from statistical graph.
(d) Measure of dispersion: standard deviation
Reason: All data are taken into account.
Activity 16.2 (p. 81)
1. / Measure of dispersion / Data set A / Data set B / Data set CRange / 3 / 3 / 6
Inter-quartile range / 1.5 / 1.5 / 3
Standard deviation / 1 / 1 / 2
2. The measures of dispersion of data set B are the same as that of data set A.
3. The measures of dispersion of data set C are twice that of data set A.
Follow-up Exercise
p. 52
1. (a) ∵ Largest datum = 9
Smallest datum = –1
∴ Range = 9 – (–1)
(b) ∵ Largest datum = 77
Smallest datum = 32
∴ Range = 77 – 32
(c) ∵ Largest datum = 3.2
Smallest datum = –5.4
∴ Range = 3.2 – (–5.4)
(d) ∵ Largest datum = 123
Smallest datum = 66
∴ Range = 123 – 66
2. The highest class boundary = 10.5 min
The lowest class boundary = 0.5 min
∴ Range of the waiting time
p. 55
1. (a) ∵ Q1 = 4, Q3 = 9
∴ Inter-quartile range = Q3 – Q1 = 9 – 4
(b) ,
∴ Inter-quartile range = Q3 – Q1 = 20 – 7.5
(c) Arrange the data in ascending order: 77, 95, 100, 105, 121, 144, 153
Q1 = 95, Q3 = 144
∴ Inter-quartile range = Q3 – Q1 = 144 – 95
(d) Arrange the data in ascending order: 48, 54, 61, 66, 70, 71, 77, 89, 90, 92
Q1 = 61, Q3 = 89
∴ Inter-quartile range = Q3 – Q1 = 89 – 61
2. (a) From the graph,
Q1, Q3
(b) Inter-quartile range
3. / (a) / Number of customers less than / Cumulative frequency0 / 0
10 / 6
20 / 14
30 / 26
40 / 36
50 / 40
(b)
(c) From the graph,
median
range = 50.5 – 0.5
Q1 = 15, Q3 = 34
∴ Inter-quartile range = Q3 – Q1 = 34 – 15
p. 63
1. (a)
(b)
(c)
2. (a) Median
(b) Maximum IQ = 115
Minimum IQ = 92.5
∴ Range = 115 – 92.5 =
Q1 = 97.5, Q3 = 105
∴ Inter-quartile range = 105 – 97.5 =
(c) The median IQ is closer to the lower quartile than the upper quartile. In other words, the distribution of the IQs of the group in the upper half is more dispersed than that in the lower half.
3. (a) >, >, <
(b) =, >, <
(c) >, <, =
p. 70
1. For set A,
mean
standard deviation
For set B,
mean
standard deviation
For set C,
mean
standard deviation
2. set B
p. 73
1. Mean
Standard deviation
2. (a) Mean volume
(b) Standard deviation
p. 74
1. (cor. to 3 sig. fig.)
2. (cor. to 3 sig. fig.)
3. (cor. to 3 sig. fig.)
4. (cor. to 3 sig. fig.)
p. 87
(a) ∵ Each datum is increased by 2.
∴ The mean will increase by 2.
The median will increase by 2.
The inter-quartile range and the standard deviation will remain unchanged.
∴ The new mean = 10 + 2 =
The new median = 9 + 2 =
The new inter-quartile range =
The new standard deviation =
(b) ∵ Each datum is decreased by 3.
∴ The mean will decrease by 3.
The median will decrease by 3.
The inter-quartile range and the standard deviation will remain unchanged.
∴ The new mean = 10 – 3 =
The new median = 9 – 3 =
The new inter-quartile range =
The new standard deviation =
(c) ∵ Each datum is tripled.
∴ The mean, the median, the inter-quartile range and the standard deviation will be tripled.
∴ The new mean = 10 ´ 3 =
The new median = 9 ´ 3 =
The new inter-quartile range = 5 ´ 3 =
The new standard deviation = 1 ´ 3 =
(d) ∵ Each datum is increased by 10%.
∴ The mean, the median, the inter-quartile range and the standard deviation will increase by 10%.
∴ The new mean = 10 ´ (1 + 10%) =
The new median = 9 ´ (1 + 10%) =
The new inter-quartile range = 5 ´ (1 + 10%) =
The new standard deviation = 1 ´ (1 + 10%) =
(e) ∵ Each datum is decreased by 20%.
∴ The mean, the median, the inter-quartile range and the standard deviation will decrease by 20%.
∴ The new mean = 10 ´ (1 – 20%) =
The new median = 9 ´ (1 – 20%) =
The new inter-quartile range = 5 ´ (1 – 20%) =
The new standard deviation = 1 ´ (1 – 20%) =
Exercise
Exercise 16A (p. 57)
Level 1
1. (a) ∵ Largest datum = 25
Smallest datum = 8
∴ Range
(b) ∵ Largest datum = 1.2
Smallest datum = –0.9
∴ Range
(c) ∵ Largest datum = 8
Smallest datum = 1
∴ Range
2. The highest class boundary = $2999.5
The lowest class boundary = $1999.5
∴ Range = $(2999.5 – 1999.5)
=
3. (a) Q1 = –1, Q3 = 11
∴ Inter-quartile range = Q3 – Q1 = 11 – (–1) =
(b) ,
∴ Inter-quartile range = Q3 – Q1 = 2 – (–2) =
(c) ,
∴ Inter-quartile range = Q3 – Q1 = 15 – 7 =
4. (a) For athlete A:
∵ Highest score = 8.9
Lowest score = 7.7
∴ Range = 8.9 – 7.7 =
Q1 = 7.7, Q3 = 8.8
∴ Inter-quartile range = Q3 – Q1 = 8.8 – 7.7 =
For athlete B:
∵ Highest score = 9.2
Lowest score = 8.3
∴ Range = 9.2 – 8.3 =
Q1 = 8.4, Q3 = 9.2
∴ Inter-quartile range = Q3 – Q1 = 9.2 – 8.4 =
(b) ∵ Range and inter-quartile range of athlete B are smaller than that of athlete A.
∴ Scores of athlete B have a smaller dispersion.
5. (a) From the graph,
Q1 = 157 cm, Q3 = 175 cm
(b) Range = (180 – 150) cm =
Inter-quartile range = Q3 – Q1
= (175 – 157) cm
=
6. (a) From the graph,
Q1, Q3
(b) Range = 100 – 10 =
Inter-quartile range = Q3 – Q1 = 72 – 30 =
Level 2
7. (a) For class A,
median
Q1, Q3
For class B,
median
Q1, Q3
(b) For class A,
range = 35 – 5 =
inter-quartile range = Q3 – Q1 = 31.5 – 10 =
For class B,
range = 40 – 15 =
inter-quartile range = Q3 – Q1 = 30 – 20 =
(c) ∵ Range and inter-quartile range of class A are
greater than that of class B.
∴ The marks of class A have a greater dispersion.
8. / (a) / Amount of money less than ($) / Cumulative frequency0 / 0
10 / 5
20 / 20
30 / 28
40 / 38
50 / 40
(b)
(c) From the graph,
median
, Q3 = $32
∴ Inter-quartile range
9. / (a) / Number of credit card less than / Cumulative frequency1 / 6
2 / 14
3 / 26
4 / 34
5 / 40
(b)
(c) From the graph,
median
Q1 = 1.5, Q3 = 3.5
inter-quartile range = Q3 – Q1 = 3.5 – 1.5 =
10. (a) Median =
Range = (185 – 170) cm
Q1 = 175 cm, Q3 = 180 cm
Inter-quartile range = Q3 – Q1 = (180 – 175) cm =
(b) Median, quartiles and inter-quartile range are not affected since median = the mean of the 6th and 7th data which are unchanged. Similarly, Q1 = the mean of the 3rd and 4th data and Q3 = the mean of 9th and 10th data which are all unchanged. As a result, inter-quartile range = Q3 – Q1 is unchanged.
Exercise 16B (p. 65)
Level 1
1. (a)
(b)
(c)
(d)
2. (a) Median
(b) Maximum height = 182.5 cm
Minimum height = 160 cm
∴ Range = (182.5 – 160) cm =
Q1 = 165 cm, Q3 = 172.5 cm
∴ Inter-quartile range = (172.5 – 165) cm =
(c) The median height is closer to the upper quartile than the lower quartile. In other words, the distribution of heights of the boys in the lower half is more dispersed than that in the upper half.
3. (a) ∵ The length of the box of test 2 is larger
∴ Test 2 has a larger inter-quartile range of marks.
(b) ∵ The distance between two ends of whiskers of test 2 is smaller.
∴ Test 2 has a smaller range of marks.
(c) ∵ The orange bar in the box of test 1 is on the right of test 2.
∴ Test 1 has a higher median mark.
4. (a) 25%
(b) 50%
(c) Q1 = 70 km/h, Q3 = 76 km/h
Inter-quartile range = (76 – 70) km/h =
(d) The median speed is closer to the lower quartile than the upper quartile. In other words, the distribution of speeds of cars in the upper half is more dispersed than that in the lower half.
5.
or
(or any other reasonable answers)
6. No, the lengths of the whiskers only represent the distribution of the first 25% and the last 25% of data. Their lengths may not be the same. (or any other reasonable answers)
Level 2
7. (a)
(b) (i) Mathematics
(ii) English
(iii) English
(c) The median mark of Chinese is closer to the lower quartile than the upper quartile. In other words, the distribution of marks of Chinese in the upper half is more dispersed than that in the lower half.
(d) Chinese, because her mark is in the top 25% of the class.
8. (a) For group A,
maximum =
minimum =
median =
Q1 =
Q3 =
For group B,
maximum =
minimum =
median =
Q1 =
Q3 =
(b)
(c) (i) group A
(ii) group B
(iii) group A
9. (a) For company A,
Q1 =
median =
Q3 =
For company B,
Q1 =
median =
Q3 =
(b)
(c) Company A has a more dispersed salary distribution since it has a larger inter-quartile range.
10. (a) For school A,
Q1 =
median =
Q3 =
For school B,
Q1 =
median =
Q3 =
(b)
(c) School A has a less dispersed mark distribution since it has a smaller inter-quartile range.
Exercise 16C (p. 75)
Level 1
1. Standard deviation =
2. Standard deviation =
3. Standard deviation = (cor. to 3 sig. fig.)
4. Standard deviation = (cor. to 3 sig. fig.)
5. (a) For group A,
mean time =
For group B,
mean time =
(b) For group A,
standard deviation = (cor. to 3 sig. fig.)
For group B,
standard deviation = (cor. to 3 sig. fig.)
6. (a) Mean waiting time =
(b) Standard deviation = (cor. to 3 sig. fig.)
7. (a) 2, 2, 4, 4 or –2, –2, –4, –4 (or any other reasonable answers)
(b) 4, 4, 8, 8 or –4, –4, –8, –8 (or any other reasonable answers)
8. No, for example:
A: 1, 2, 3, 4, 5 and B: 1, 2, 3, 3, 8
Inter-quartile range of A > inter-quartile range of B
standard deviation of A < standard deviation of B
(or any other reasonable answers)
Level 2
9. Mean
Standard deviation
10. Mean
Standard deviation
11. (a) Mean weight
(b) Standard deviation
12. (a) (i) Mean mark
(b) (ii) Standard deviation
(c) ∵ Standard deviation of school A > standard
deviation of school B
∴ School A’s result is more dispersed.
Exercise 16D (p. 87)
Level 1
1. (a) ∵ Each datum is increased by 3.
∴ The range and the standard deviation will remain unchanged.
∴ Range =
Standard deviation =
(b) ∵ Each datum is multiplied by 4.
∴ The range and the standard deviation will be multiplied by 4.
∴ Range = 5 ´ 4 =
Standard deviation = 0.82 ´ 4 =
(c) ∵ Each datum is multiplied by 4 and then increased by 3.
∴ The range and the standard deviation will be multiplied by 4.
∴ Range = 5 ´ 4 =
Standard deviation = 0.82 ´ 4 =
2. ∵ The height of each tree is increased by 12%.
∴ The inter-quartile range will be increased by 12%.
∴ The new inter-quartile range = 10 ´ (1 + 12%) m
=
3. Range, because the highest price and the lowest price are the most important information.
4. (a) ∵ The curve of city A is always higher than that of
city B.
∴ City A is warmer.
(b) Range should be used since extreme temperatures are the most important information.
(c) Range of city A = (23 – 20)°C = 3°C
Range of city B = (22 – 16)°C = 6°C
∴ City B has a greater variation in temperature in that day.
5. (a) Standard deviation should be used since all data are taken into account.
(b) Standard deviation = (cor. to 3 sig. fig.)
Level 2
6. (a) Data set A
(b) For data set A,
range = 7 – 3 =
Q1 = the 8th datum = 4
Q3 = the 23th datum = 6
∴ Inter-quartile range = 6 – 4 =
For data set B,
range = 5 – 1 =
Q1 = the 8th datum = 2
Q3 = the 23th datum = 4
∴ Inter-quartile range = 4 – 2 =
(c) Standard deviation.
Standard deviation of data set A = 1.39
Standard deviation of data set B = 1.13
∵ Standard deviation of data set A > standard deviation of data set B
∵ Data set A is more spread out.
7. (a) He made a good choice. The standard deviation takes all data into account and can be applied in further statistical calculations and analysis.