Download 16 Measures of Dispersion

16
Measures of Dispersion
16 Measures of Dispersion
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• • • •
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Activity
• •
• • • •
1.
(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.
Data set
A
3
1.5
Data set
B
3
1.5
Data set
C
6
3
1
1
2
• •
• •
(a) ∵
∴
Q1 = 4, Q3 = 9
Inter-quartile range = Q3 – Q1 = 9 – 4  5
69
19  21
 7.5 , Q3 
 20
2
2
∴ Inter-quartile range = Q3 – Q1 = 20 – 7.5  12.5
(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  49
(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  28
2.
(a) From the graph,
Q1  10 min , Q3  25 min
(b) Inter-quartile range  Q3  Q1
 (25  10) min
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.
 15 min
3.
(a)
Follow-up Exercise
p. 52
1.
• • • •
(b) Q1 
Activity 16.2 (p. 81)
Measure of
dispersion
Range
Inter-quartile
range
Standard
deviation
• •
p. 55
Activity 16.1 (p. 77)
1.
• • • •
(a) ∵ Largest datum = 9
Smallest datum = –1
∴ Range = 9 – (–1)  10
Number of customers
less than
0
10
20
30
40
50
Cumulative
frequency
0
6
14
26
36
40
(b)
(b) ∵ Largest datum = 77
Smallest datum = 32
∴ Range = 77 – 32  45
(c) ∵ Largest datum = 3.2
Smallest datum = –5.4
∴ Range = 3.2 – (–5.4)  8.6
(d) ∵ Largest datum = 123
Smallest datum = 66
∴ Range = 123 – 66  57
2.
The highest class boundary = 10.5 min
The lowest class boundary = 0.5 min
∴ Range of the waiting time  (10.5  0.5) min
 10 min
21
Certificate Mathematics in Action Full Solutions 5B
(c) From the graph,
median  25
For set B,
1  0  10  9  (8)
2
mean 
5
standard deviation
range = 50.5 – 0.5  50
Q1 = 15, Q3 = 34
∴ Inter-quartile range = Q3 – Q1 = 34 – 15  19
(1  2) 2  (0  2) 2  (10  2) 2  (9  2) 2  (8  2) 2
5
 6.72 (cor. to 3 sig. fig.)

p. 63
1.
For set C,
11  10  19  17  26  13
 16
mean 
6
standard deviation
(a)
(11  16) 2  (10  16) 2  (19  16) 2  (17  16) 2
 (26  16) 2  (13  16) 2
6
 5.48 (cor. to 3 sig. fig.)

(b)
2.
set B
p. 73
(c)
2.
1.
7  4  12 10  17  6
 12.5
4  10  6
Standard deviation
Mean 
(7  12.5) 2  4  (12  12.5) 2  10  (17  12.5) 2  6
4  10  6
 3.5
(a) Median  100
(b) Maximum IQ = 115
Minimum IQ = 92.5
∴ Range = 115 – 92.5 = 22.5
2.
(a) Mean volume
343  3  346  8  349  12  352  2
mL
3  8  12  2
 347.56 mL

Q1 = 97.5, Q3 = 105
∴ Inter-quartile range = 105 – 97.5 = 7.5
(c)
3.
(b) Standard deviation
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.
(343  347.56) 2  3  (346  347.56) 2  8
 (349  347.56) 2  12  (352  347.56) 2  2
mL
3  8  12  2
 2.42 mL (cor. to 3 sig. fig.)

(a) >, >, <
(b) =, >, <
(c) >, <, =
p. 74
p. 70
1.
7.04 (cor. to 3 sig. fig.)
1.
2.
9.22 (cor. to 3 sig. fig.)
3.
5.19 (cor. to 3 sig. fig.)
4.
0.347 m (cor. to 3 sig. fig.)
For set A,
2  5  1  7  9
4
mean 
5
standard deviation

4
22
(2  4)  (5  4)  (1  4)  (7  4)  (9  4)
5
2
2
2
2
2
16
p. 87
Measures of Dispersion
Exercise
(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 = 12
The new median = 9 + 2 = 11
Exercise 16A (p. 57)
Level 1
1. (a) ∵ Largest datum = 25
Smallest datum = 8
∴ Range  25  8  17
(b) ∵ Largest datum = 1.2
Smallest datum = –0.9
∴ Range  1.2  (0.9)  2.1
The new inter-quartile range = 5
The new standard deviation = 1
(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 = 7
(c)
2.
The highest class boundary = $2999.5
The lowest class boundary = $1999.5
∴ Range = $(2999.5 – 1999.5)
= $1000
3.
(a) Q1 = –1, Q3 = 11
∴ Inter-quartile range = Q3 – Q1 = 11 – (–1) = 12
The new median = 9 – 3 = 6
The new inter-quartile range = 5
The new standard deviation = 1
(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 = 30
(3)  (1)
1 3
 2 , Q3 
2
2
2
∴ Inter-quartile range = Q3 – Q1 = 2 – (–2) = 4
(b) Q1 
The new median = 9  3 = 27
The new inter-quartile range = 5  3 = 15
(c)
The new standard deviation = 1  3 = 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%) = 11
∵ Largest datum = 8
Smallest datum = 1
∴ Range  8  1  7
4.
The new median = 9  (1 + 10%) = 9.9
68
13  17
 7 , Q3 
 15
2
2
∴ Inter-quartile range = Q3 – Q1 = 15 – 7 = 8
Q1 
(a) For athlete A:
∵ Highest score = 8.9
Lowest score = 7.7
∴ Range = 8.9 – 7.7 = 1.2
Q1 = 7.7, Q3 = 8.8
∴ Inter-quartile range = Q3 – Q1 = 8.8 – 7.7 = 1.1
The new inter-quartile range = 5  (1 + 10%) = 5.5
The new standard deviation = 1  (1 + 10%) = 1.1
For athlete B:
∵ Highest score = 9.2
Lowest score = 8.3
∴ Range = 9.2 – 8.3 = 0.9
(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%) = 8
Q1 = 8.4, Q3 = 9.2
∴ Inter-quartile range = Q3 – Q1 = 9.2 – 8.4 = 0.8
The new median = 9  (1 – 20%) = 7.2
The new inter-quartile range = 5  (1 – 20%) = 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.
The new standard deviation = 1  (1 – 20%) = 0.8
5.
(a) From the graph,
Q1 = 157 cm, Q3 = 175 cm
(b) Range = (180 – 150) cm = 30 cm
Inter-quartile range = Q3 – Q1
= (175 – 157) cm
= 18 cm
23
Certificate Mathematics in Action Full Solutions 5B
6.
(a) From the graph,
Q1  30 , Q3  72
1
Q1  $13 , Q3 = $32
3
∴ Inter-quartile range  Q3  Q1
(b) Range = 100 – 10 = 90
1

 $ 32  13 
3

2
 $18
3
Inter-quartile range = Q3 – Q1 = 72 – 30 = 42
Level 2
7. (a) For class A,
median  25
Q1  10 , Q3  31.5
9.
(a)
For class B,
median  23
Q1  20 , Q3  30
(b) For class A,
range = 35 – 5 = 30
inter-quartile range = Q3 – Q1 = 31.5 – 10 = 21.5
Number of credit
card less than
1
2
3
4
5
Cumulative
frequency
6
14
26
34
40
(b)
For class B,
range = 40 – 15 = 25
inter-quartile range = Q3 – Q1 = 30 – 20 = 10
(c) ∵ Range and inter-quartile range of class A are
greater than that of class B.
∴ The marks of class A have a greater dispersion.
8.
(a)
Amount of money
less than ($)
0
10
20
30
40
50
(b)
Cumulative
frequency
0
5
20
28
38
40
(c) From the graph,
median  2.5
Q1 = 1.5, Q3 = 3.5
inter-quartile range = Q3 – Q1 = 3.5 – 1.5 = 2
10. (a) Median = 177 cm
Range = (185 – 170) cm  15 cm
Q1 = 175 cm, Q3 = 180 cm
Inter-quartile range = Q3 – Q1 = (180 – 175) cm = 5 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.
(c) From the graph,
median  $20
24
16
(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.
Exercise 16B (p. 65)
Level 1
1.
Measures of Dispersion
(a)
5.
(b)
or
(c)
(or any other reasonable answers)
(d)
6.
2.
(a) Median  170 cm
(b) Maximum height = 182.5 cm
Minimum height = 160 cm
∴ Range = (182.5 – 160) cm = 22.5 cm
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)
Q1 = 165 cm, Q3 = 172.5 cm
∴ Inter-quartile range = (172.5 – 165) cm = 7.5 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.
4.
(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.
(a) 25%
(b) 50%
(c) Q1 = 70 km/h, Q3 = 76 km/h
Inter-quartile range = (76 – 70) km/h = 6 km/h
(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.
25
Certificate Mathematics in Action Full Solutions 5B
8.
(a) For group A,
maximum = 10
Q3 = 56
minimum = 2
For school B,
Q1 = 44
median = 7
median = 60
Q1 = 5
Q3 = 90
Q3 = 8
(b)
For group B,
maximum = 10
minimum = 3
median = 6
Q1 = 5
Q3 = 9
(c) School A has a less dispersed mark distribution since
it has a smaller inter-quartile range.
(b)
Exercise 16C (p. 75)
Level 1
1.
Standard deviation = 3
2.
Standard deviation = 4
3.
Standard deviation = 2.73 (cor. to 3 sig. fig.)
(a) For company A,
Q1 = $5500
4.
Standard deviation = 2.77 (cor. to 3 sig. fig.)
median = $6500
5.
(a) For group A,
mean time = 15.85 s
(c) (i) group A
(ii) group B
(iii) group A
9.
Q3 = $7500
For group B,
mean time = 15.6 s
For company B,
Q1 = $6000
(b) For group A,
standard deviation = 0.415 s (cor. to 3 sig. fig.)
median = $6500
For group B,
standard deviation = 1.34 s (cor. to 3 sig. fig.)
Q3 = $7000
(b)
6.
(a) Mean waiting time = 5.7 min
(b) Standard deviation = 2.76 min (cor. to 3 sig. fig.)
(c) Company A has a more dispersed salary distribution
since it has a larger inter-quartile range.
10. (a) For school A,
Q1 = 40
median = 50
26
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)
16
(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 = 20
Level 2
9.
( x  3)  ( x  1)  ( x  4)  x  ( x  3)
 x  0.6
Mean 
5
Standard deviation
[( x  3)  ( x  0.6)]2  [( x  1)  ( x  0.6)]2
Standard deviation = 0.82  4 = 3.28
 [( x  4)  ( x  0.6)]2  [ x  ( x  0.6)]2

 [( x  3)  ( x  0.6)]2
5
2.
∵
∴
∴
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 = 10.9 (cor. to 3 sig. fig.)
(3  0.6) 2  (1  0.6) 2  (4  0.6) 2
 (0.6) 2  (3  0.6) 2
5
 2.58 (cor. to 3 sig. fig.)
Measures of Dispersion
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
= 11.2 m

a  (a  3d )  (a  5d )  (a  7d )
 a  3.75d
4
Standard deviation
10. Mean 
[a  (a  3.75d )]2  [(a  3d )  (a  3.75d )]2

 [(a  5d )  (a  3.75d )]2  [(a  7 d )  (a  3.75d )]2
4
(3.75d ) 2  (3d  3.75d ) 2  (5d  3.75d ) 2

 (7 d  3.75d ) 2
4
 2.59d (cor. to 3 sig. fig.)
11. (a) Mean weight  199.8 g (cor. to 4 sig. fig.)
(b) Standard deviation  0.9171 g (cor. to 4 sig. fig.)
12. (a) (i)
Level 2
6. (a) Data set A
(b) For data set A,
range = 7 – 3 = 4
Mean mark  49.95
Q1 = the 8th datum = 4
Q3 = the 23th datum = 6
∴ Inter-quartile range = 6 – 4 = 2
(b) (ii) Standard deviation  12.48 (cor. to 4 sig. fig.)
(c) ∵ Standard deviation of school A > standard
deviation of school B
∴ School A’s result is more dispersed.
For data set B,
range = 5 – 1 = 4
Q1 = the 8th datum = 2
Q3 = the 23th datum = 4
∴ Inter-quartile range = 4 – 2 = 2
Exercise 16D (p. 87)
(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.
Level 1
1. (a) ∵ Each datum is increased by 3.
∴ The range and the standard deviation will
remain unchanged.
∴ Range = 5
Standard deviation = 0.82
(b) ∵ Each datum is multiplied by 4.
∴ The range and the standard deviation will be
multiplied by 4.
∴ Range = 5  4 = 20
Standard deviation = 0.82  4 = 3.28
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.
(b) Standard deviation = 0.110 kg (cor. to 3 sig. fig.)
(c) The machine does not function properly since the
standard deviation is greater than 0.1 kg.
27
Certificate Mathematics in Action Full Solutions 5B
8.
(a) Standard deviation = 13.5 (cor. to 3 sig. fig.)
(b) (i) The range increases.
(ii) New standard deviation = 35.7 (cor. to 3 sig.
fig.)
The standard deviation increases.
9.
(a) ∵ The salary of each employee is increased by
$70.
∴ The mean will be increased by $70.
The standard deviation will remain unchanged.
∴ The new mean = $(7000 + 70) = $7070
The new standard deviation = $2000
(b) ∵ The salary of each employee is increased by 1%.
∴ The mean and the standard deviation will both
increased by 1%.
∴ The new mean = $7000  (1 + 1%) = $7070
The new standard deviation = $2000  (1 + 1%)
= $2020
(c) The standard deviation will increase. Since those
employees have salaries equal to the mean, deleting
them leads the data less concentrated about the mean.
10. (a) For group A,
14 1  15  7  16 14  17  7  18 1
mean 
1  7  14  7  1
 16
the 15th datum  the 16th datum
2
16  16

2
 16
median 
range = largest datum – smallest datum
= 18 – 14
= 4
inter-quartile range = 23rd datum – the 8th datum
= 17 – 15
= 2
For group B,
14  3  15  7  16 10  17  7  18  3
mean 
3  7  10  7  3
 16
the 15th datum  the 16th datum
median 
2
16  16

2
 16
range = largest datum – smallest datum
= 18 – 14
= 4
28
inter-quartile range = the 23rd datum – the 8th datum
= 17 – 15
= 2
(b) Group B
(c) (i) The standard deviation will increase in group A
since the leaving member has age equal to the
mean.
(ii) The standard deviation will decrease in group B
since the joining member has age equal to the
mean.
(d) The standard deviation will increase since the new
member’s age is not close to the mean. The data are
relatively less concentrated about the mean.
(e) The standard deviation will decrease since the two
leaving members have ages not close to the mean.
The data are relatively more concentrated about the
mean.
Revision Exercise 16 (p. 93)
Level 1
1. (a) Arrange the data in ascending order: –3, 2, 4, 5, 5
Range = 5 – (–3) = 8
2  (3)
 0.5
2
55
Q3 
5
2
Inter-quartile range = Q3 – Q1
= 5 – (–0.5)
= 5.5
Q1 
(b) Arrange the data in ascending order:
–3.30, –2.50, –1.72, 1.77, 2.34, 3.54, 4.10, 5.03
Range = 5.03 – (–3.30)
= 8.33
(1.72)  (2.50)
2
 2.11
3.54  4.10
Q3 
2
 3.82
Inter-quartile range = Q3 – Q1
= 3.82 – (–2.11)
= 5.93
Q1 
(c) Arrange the data in ascending order:
2 1 2 5 3
, , , , ,1
7 3 5 8 4
2 5
Range  1  
7 7
Q1 
1
3
, Q3 
3
4
16
Inter-quartile range  Q3  Q1
(c) ∵ The orange bar in the box of class B is on the
right of class A.
∴ Class B has a higher median height.
(d) The median of class A is closer to the lower quartile
than to the upper quartile. In other words, the
distribution of the heights is more dispersed in the
upper half than that in the lower half.
3 1
 
4 3
5

12
2.
(a) Standard deviation = 3
8.
(b) Standard deviation = 2
(c) Standard deviation = 1
3.
Measures of Dispersion
(a) From the graph,
Q1 = $8999
Q3 = $12 499
(b) Inter-quartile range = Q3 – Q1
= $(12 499 – 8999)
= $3500
(a) Median = 100
(b) Range = 106 – 93 = 13
Inter-quartile range = 102 – 96 = 6
(d) The median IQ is closer to the upper quartile than to
the lower quartile. In other words, the distribution of
IQs of students in the lower half (93 – 100) is more
dispersed than that in the upper half (100 – 106).
4.
Standard deviation = 7.18 cm (cor. to 3 sig. fig.)
5.
(a)
9.
(a) Range = 9 – (–2)  11
(2)  1
 0.5
2
79
Q3 
8
2
Inter-quartile range = Q3 – Q1
= 8 – (–0.5)
= 8.5
Q1 
Standard deviation = 4
(b) (i)
∵
∵
∴
(b)
(c)
x is added to each datum.
The standard deviation will remain
unchanged.
The standard deviation = 4
(ii) ∵
∴
∴
Each datum is doubled.
The standard deviation will be doubled.
The standard deviation = 4  2  8
(iii) ∵
∴
Each datum is divided by 4.
The standard deviation will be divided by
4.
The standard deviation = 4  4 = 1
∴
10. Standard deviation, since it can be applied in further
statistical calculations and analysis and all data are taken
into account.
6.
(a) Range = 90 – 30 = 60
(b) Inter-quartile range = 80 – 50 = 30
(c) Median = 60
7.
(a) ∵ The distance between two ends of the whiskers
of class B is longer.
∴ Class B has the heights with a larger range.
(b) ∵ The length of the box of class A is larger.
∴ Class A has the heights with a larger
inter-quartile range.
11. Standard deviation, since it can be applied in further
statistical calculations and analysis and all data are taken
into account.
12. Data set A: 1, 2, 3, 3, 4, 5
Data set B: 0, 1, 2, 3, 3, 4, 5, 6 or
Data set A: 1, 3, 3, 5
Data set B: 0, 1, 3, 3, 5, 6 (or any other reasonable answers)
13. (a) 1, 2, 3, 4, 5 or 2, 4, 6, 8
(or any other reasonable answers)
(b) 0, 0, 0, 0, 1 or 0, 0, 0, 3, 4
(or any other reasonable answers)
29
Certificate Mathematics in Action Full Solutions 5B
Level 2
14. (a) Range = 6 – 2 = 4
the 20th datum  the 21st datum
2
44

2
4
(c) From the graph,
median  13
Q1 = 10
Q3 = 17
Inter-quartile range = Q3 – Q1 = 17 – 10 = 7
Median 
the10th datum  the 11th datum
2
3 4

2
 3.5
the 30th datum  the 31th datum
Q3 
2
55

2
5
Inter-quartile range = Q3 – Q1= 5 – 3.5  1.5
(d)
Q1 
(b)
15. (a)
No. of time less than Cumulative frequency
0
0
5
4
10
7
15
19
20
27
25
29
30
30
16. (a) For class A,
median  the 16th datum
 28
Q1 = the 8th datum
= 10
Q3 = the 23rd datum
= 34
For class B,
median  the 15th datum
 26
the 7th datum  the 8th datum
Q1 
2
14  17

2
 15.5
the 22nd datum  the 23rd datum
Q3 
2
34  34

2
 34
(b)
(b) (i) class A
(ii) class B
(iii) class A
17. (a) For Mathematics,
median  48
Q1  40
Q3  63
For Chinese,
median  40
Q1  22
Q3  60
30
16
20. (a)
Measures of Dispersion
0  2  1 7  2 12  3  7  4  2
30
2
M ean 
the 15th datum  the16th datum
2
22

2
2
Median 
Inter-quartile range  the 23rd datum  the 8th datum
(b) Chinese has a more dispersed distribution since it has
a larger inter-quartile range.
18. (a) (i)
For battery A,
Q1 = 6.5 h, Q3 = 30 h
Inter-quartile range  Q3  Q1
 (30  6.5) h
 23.5 h
For battery B,
Q1 = 10 h, Q3 = 35 h
Inter-quartile range  Q3  Q1
 (35  10) h
 25 h
(ii) battery B
(b) Yes, since it is less affected by extreme data.
19. (a) (i)
 3 1
2
Standard deviation  1
(b) The inter-quartile range will remain unchanged since
the new and original lower and upper quartiles are the
same. The standard deviation will increase since the
removed data are all equal to the mean, relatively less
data are concentrated about the mean.
(c) The inter-quartile range will remain unchanged since
the new and original lower and upper quartiles are the
same. The standard deviation will increase since the
new datum is not close to the mean, relatively less
data are concentrated about the mean.
21. (a) ∵
Range  16  13
3
∴
the12th datum  the13 datum
2
14  14

2
 14
the 36th datum  the 37th datum
Q3 
2
15  15

2
 15
Inter-quartile range  Q3  Q1
(ii) Q1 
(a  5) 2  (b  5) 2  (c  5) 2  (d  5) 2
2
4
∴ Standard deviation
(b) ∵
(a  5) 2  (b  5) 2  (c  5) 2

 (d  5) 2  ( x  5) 2
5
22  4  (5  5) 2
5
 1.79 (cor. to 3 sig. fig.)

 15  14
1
(iii) Standard deviation  1
abcd
abcd  x
 5 and
5
5
4
5 4  x
5
5
x5
22. ∵
(b) The range will increase to 4 since the highest age is
increased by 1.
The inter-quartile range will remain unchanged since
the 37th datum will be the same as before.
The standard deviation will increase to 1.05 since the
new datum is not close to the mean, relatively less
data are concentrated about the mean.
5  7  m  13
4
25  m

4
M ean 
2
2
25  m  
25  m 

5 
  7 
 
4  
4 

2
∴
25  m  
25  m 

m 
  13 

4
4 

 
4
2
3
31
Certificate Mathematics in Action Full Solutions 5B
For group B,
mean = a
inter-quartile range = 10
 25  m   25  m 
 25  m 
25  5

  49  7

 2   4 
 2 
2
 25  m 
 25  m   25  m 

  m 2  m


4


 2   4 
 25  m   25  m 
 169  13


 2   4 
4
2
standard deviation = 22
∴ Only I is true.
2
5.
Answer: A
From the diagram,
median of data set A = median of data set B
range of data set A > range of data set B
inter-quartile range of data set A < inter-quartile range of
data set B
∴ Only I and II are true.
6.
Answer: B
From the diagram, Chinese has the least inter-quartile
range.
7.
Answer: D
∵ Each datum is decreased by 5%.
∴ The mean and the standard deviation of the set of
data will also be decreased by 5%.
∴ The new mean  m  (1  5%)  0.95m
3
2
 25  m   25  m 
243  m 2  (25  m)


 2   4  9
4
972  3m 2  625  50m  144
3m 2  50m  203  0
(m  7)(3m  29)  0
m  7 or m 
29
3
Multiple Choice Questions (p. 99)
1.
Answer: B
Q1 = the 2nd datum = 4
Q3 = the 5th datum = 10
∴ Inter-quartile range  Q3  Q1  10  4  6
The new standard deviation  s  (1  5%)  0.95 s
8.
Answer: C
∵ Each datum is decreased by 2.
∴ The mean of the set of data will also be decreased by
2.
The inter-quartile range and the standard deviation
will remain unchanged.
∴ The new mean  m  2
2.
Answer: A
From the graph,
Q1 = 65, Q3 = 80
∴ Inter-quartile range  Q3  Q1  80  65  15
3.
Answer: C
The new inter-quartile range  q
( a  2d )  ( a  d )  a  ( a  d )  ( a  2d )
M ean 
5
a
∴ Standard deviation
The new standard deviation  s
[(a  2d )  a]2  [(a  d )  a]2  (a  a) 2

 [(a  d )  a]2  [(a  2d )  a]2
5

10d 2
5
 2d 2
 2d
4.
32
Answer: A
For group A,
mean = a,
inter-quartile range = 4
standard deviation = 2
9.
Answer: B
Because all the salaries of the leaving employees are equal
to the mean, removing them will not change the mean.
Moreover, the distribution of data will be less concentrated
about the mean, as a result, the standard deviation will
increase.
10. Answer: B
From the graph,
range of P = range of Q
inter-quartile range of P < inter-quartile range of Q
median of P = median of Q
∴ Only II is true.
11. Answer: C
For I, standard deviation = 14.1 (cor. to 3 sig. fig.)
For II, standard deviation = 11.0 (cor. to 3 sig. fig.)
For III, standard deviation = 16.0 (cor. to 3 sig. fig.)
∴ II < I < III