5bch16(measures Of Dispersion)

16 Measures of Dispersion

16 Measures of Dispersion • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • =(10 .5 −0.5) m in

Activity

=10 m in

Activity 16.1 (p. 77)

p. 55

(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.

1.

(a) ∵ ∴

Measure of dispersion Range Inter-quartile range Standard deviation

Data set A 3 1.5

Data set B 3 1.5

Data set C 6 3

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

∴

(a) ∵ ∴ (b) ∵ ∴ (c) ∵ ∴ (d) ∵ ∴

2.

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 =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, in , Q3 =25 m in Q1 =10 m = Q3 −Q1 (b) Inter-quartile range = ( 25 −10 ) min =15 min

3.

(a)

Largest datum = 9 Smallest datum = –1 Range = 9 – (–1) =10 Largest datum = 77 Smallest datum = 32 Range = 77 – 32 =45 Largest datum = 3.2 Smallest datum = –5.4 Range = 3.2 – (–5.4) =8.6 Largest datum = 123 Smallest datum = 66 Range = 123 – 66 =57

The highest class boundary = 10.5 min The lowest class boundary = 0.5 min ∴ Range of the waiting time

6 +9 19 + 21 = 7.5 , Q3 = = 20 2 2

=12 .5

p. 52 1.

Q1 =

(b)

Activity 16.2 (p. 81) 1.

Q1 = 4, Q3 = 9 Inter-quartile range = Q3 – Q1 = 9 – 4 =5

Number of customers less than 0 10 20 30 40 50

(c) From the graph, median =25 range = 50.5 – 0.5 =50 Q1 = 15, Q3 = 34 (b)

Cumulative frequency 0 6 14 26 36 40

Certificate Mathematics in Action Full Solutions 5B

∴ Inter-quartile range = Q3 – Q1 = 34 – 15 =19

standard deviation

( −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

For set C, mean = 1.

11 +10 +19 +17 + 26 +13 =16 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)

1.

Mean =

7 ×4 +12 ×10 +17 ×6 =12 .5 4 +10 + 6

Standard deviation 2.

3.

(a) Median =100 (b) Maximum IQ = 115 Minimum IQ = 92.5 ∴ Range = 115 – 92.5 = 22 .5 Q1 = 97.5, Q3 = 105 ∴ Inter-quartile range = 105 – 97.5 = 7.5 (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.

(7 −12 .5) 2 ×4 +(12 −12 .5) 2 ×10 +(17 −12 .5) 2 ×6 4 +10 +6 = 3. 5

2.

(a) Mean volume 343 ×3 +346 ×8 +349 ×12 +352 ×2 mL 3 +8 +12 + 2 = 347 .56 mL =

(b) Standard deviation

(a) >, >, < (b) =, >, < (c) >, <, =

(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.)

p. 70

1.

For set A,

mean =

−2 + 5 +1 + 7 + 9 =4 5 p. 74

standard deviation

=

( −2 − 4) + (5 − 4) + (1 − 4) + (7 − 4) + (9 −1.4) 2 7.04 (cor. to 3 sig. fig.) 9.22 (cor. to 3 sig. fig.) 5 2. 2

2

2

=4

2

3. 4.

For set B, mean =

−1 + 0 +10 + 9 + ( −8) =2 5

5.19 (cor. to 3 sig. fig.) m (cor. to 3 sig. fig.)

0.347

16 Measures of Dispersion

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 = 12

Exercise Exercise 16A (p. 57) Level 1 1. (a) ∵ ∴ (b) ∵

The new median = 9 + 2 = 11 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

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.

(a) Q1 = –1, Q3 = 11 ∴ Inter-quartile range = Q3 – Q1 = 11 – (–1) = 12 (b)

The new median = 9 × 3 = 27

Q1 = ∴

The new inter-quartile range = 5 × 3 = 15 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 = 25 Smallest datum = 8 Range =25 −8 =17 Largest datum = 1.2 Smallest datum = –0.9 0.9) =2.1 Range =1.2 −( − Largest datum = 8 Smallest datum = 1 Range =8 −1 =7

(c)

4.

The new median = 9 × (1 + 10%) = 9.9 The new inter-quartile range = 5 × (1 + 10%) = 5.5 The new standard deviation = 1 × (1 + 10%) = 1.1 (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

Inter-quartile range = Q3 – Q1 = 2 – (–2) = 4

Q1 = ∴

( −3) + ( −1) 1+3 = −2 , Q3 = =2 2 2 6 +8 13 +17 = 7 , Q3 = = 15 2 2

Inter-quartile range = Q3 – Q1 = 15 – 7 = 8

(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

For athlete B: ∵ Highest score = 9.2 Lowest score = 8.3 ∴ Range = 9.2 – 8.3 = 0.9 Q1 = 8.4, Q3 = 9.2 ∴ Inter-quartile range = Q3 – Q1 = 9.2 – 8.4 =

The new median = 9 × (1 – 20%) = 7.2 The new inter-quartile range = 5 × (1 – 20%) = 4 The new standard deviation = 1 × (1 – 20%) = 0.8

0.8

(b) ∵ ∴ 5.

Range and inter-quartile range of athlete B are smaller than that of athlete A. Scores of athlete B have a smaller dispersion.

(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

Certificate Mathematics in Action Full Solutions 5B

6.

= Q3 − Q1

(a) From the graph, Q1 =30 , Q3 =72 (b) Range = 100 – 10 = 90

∴

Inter-quartile range = Q3 – Q1 = 72 – 30 = 42



= $18

Level 2 7. (a) For class A, median =25

9.

(a)

Q1 =10 , Q3 =31 .5 For class B, median =23 Q1 =20 , Q3 =30 (b) For class A, range = 35 – 5 = 30

1

Inter-quartile range = $32 −13 3   

Number of credit card less than 1 2 3 4 5

2 3

Cumulative frequency 6 14 26 34 40

inter-quartile range = Q3 – Q1 = 31.5 – 10 = 21 .5 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

Cumulative frequency 0 5 20 28 38 40

(b) (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.

(b) (c) From the graph, median =$20

Q1 = $13

1 , Q3 = $32 3

16 Measures of Dispersion

Exercise 16B (p. 65)

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.

Level 1

1.

(a) 5. or (b)

(c)

(or any other reasonable answers)

6.

(d)

Level 2

2.

(a) Median =170 cm (b) Maximum height = 182.5 cm Minimum height = 160 cm ∴ Range = (182.5 – 160) cm = 22 .5 cm 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.

(a) ∵ ∴ (b) ∵ ∴ (c) ∵ ∴

4.

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)

The length of the box of test 2 is larger Test 2 has a larger inter-quartile range of marks. The distance between two ends of whiskers of test 2 is smaller. Test 2 has a smaller range of marks. 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 (d) The median speed is closer to the lower quartile than

7.

(a) (b) (ii) (iii) (c)

(i) Mathematics English English 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.

Certificate Mathematics in Action Full Solutions 5B

8.

(a) For group A, maximum = 10

Q1 = 44

minimum = 2

Q3 = 90

median = 60

median = 7 Q1 = 5 Q3 = 8 For group B, maximum = 10 minimum = 3 median = 6 Q1 = 5

(b)

Q3 = 9

(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 = 3

2.

Standard deviation = 4

(c) (i) group A (ii) group B (iii) group A

3.

Standard deviation = 2.73 (cor. to 3 sig. fig.)

4.

Standard deviation = 2.77 (cor. to 3 sig. fig.)

(a) For company A, Q1 = $5500

5.

(a) For group A, mean time = 15 .85 s For group B, mean time = 15 .6 s (b) For group A, standard deviation = 0.415 s (cor. to 3 sig. fig.) For group B, standard deviation = 1.34 s (cor. to 3 sig. fig.)

6.

(a) Mean waiting time = 5.7 min

(b)

9.

median = $6500 Q3 = $7500 For company B, Q1 = $6000 median = $6500 Q3 = $7000

(b) Standard deviation = 2.76 min 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)

(b) (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 Q3 = 56 For school B,

(cor. to 3 sig.

16 Measures of Dispersion

Level 2

9.

(b) ∵ ∴

Mean

=

( x − 3) + ( x +1) + ( x − 4) + x + ( x + 3) = x − 0.6 5

Standard deviation [( x −3) −( x −0.6)] 2 +[( x +1) −( x −0.6)] 2 +[( x −4) −( x −0.6)] 2 +[ x −( x −0.6)] 2

(c) ∵ ∴

+[( x +3) −( x −0.6)] 2 5

=

∴

∴

(− 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.) =

Standard deviation = 0.82 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 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 Standard deviation = 0.82 × 4 = 3.28

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 = 11 .2 m

10. Mean a + (a + 3d ) + ( a + 5d ) + (a + 7 d ) = = a + 3.753.d Range, because the highest price and the lowest price are 4 the most important information. Standard deviation

=

=

4.

(a) ∵

5.

(a) Standard deviation should be used since all data are taken into account. (b) Standard deviation = 10 .9 (cor. to 3 sig. fig.)

The curve of city A is always higher than that of city B. 2 2 ∴ A is warmer. + [(a + 5d ) − (a + 3.75d )] + [(a + 7 d ) − ( a + 3.75d )](b) RangeCity should be used since extreme temperatures are the most important information. 4 (c) Range of city A = (23 – 20)°C = 3°C Range of city B = (22 – 16)°C = 6°C (−3.75d ) 2 + (3d − 3.75d ) 2 + (5d − 3.75d ) 2 ∴ City B has a greater variation in temperature in that day. + (7 d − 3.75d ) 2

[a − ( a + 3.75d )]2 + [(a + 3d ) − (a + 3.75d )]2

4 = 2.59d (cor. to 3 sig. fig.) 11. (a) Mean weight = 1 9 9

.8 g (co r.

to 4 sig .

fig .)

(b) Standard deviation = 0.9 1 7 1

g (co r.

to 4 sig .

fig .)

12. (a) (i) Mean mark =49 .95 (b) (ii) Standard deviation = 12 .48

(c) ∵ ∴

(cor.

to 4 sig.

fig.)

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 = 5

Level 2 6. (a) Data set A (b) For data set A, range = 7 – 3 = 4 Q1 = the 8th datum = 4 Q3 = the 23th datum = 6 ∴ Inter-quartile range = 6 – 4 = 2 For data set B, range = 5 – 1 = 4 Q1 = the 8th datum = 2 Q3 = the 23th datum = 4 ∴ Inter-quartile range = 4 – 2 = 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

Certificate Mathematics in Action Full Solutions 5B

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.

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 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.

(b) ∵ ∴

10. (a) For group A, mean

14 ×1 +15 ×7 +16 ×14 +17 ×7 +18 ×1 1 +7 +14 +7 +1 =16 =

median

the 15th datum + the 16th datum 2 16 +16 = 2 =16 =

range = largest datum – smallest datum = 18 – 14 = 4 inter-quartile range = 23rd datum – the 8th datum = 17 – 15 = 2 For group B, mean

14 ×3 +15 ×7 +16 ×10 +17 ×7 +18 ×3 3 +7 +10 +7 +3 =16 =

median

16 Measures of Dispersion

the 15th datum + the 16th datum 2 16 +16 = 2 =16 =

(b) (c)

range = largest datum – smallest datum = 18 – 14 = 4 (d) (e)

inter-quartile range = the 23rd datum – the 8th datum = 17 – 15 = 2 Group B (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. 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. 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 Q1 =

Inter-quartile range = Q3 – Q1 = 5 – (–0.5) = 5.5 (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 Q1 =

Inter-quartile range = Q3 – Q1 = 3.82 – (–2.11) = 5.93 (c) Arrange the data in ascending order:

2 1 2 5 3 , , , , ,1 7 3 5 8 4 Range =1 −

Q1 =

2 5 = 7 7

1 3 , Q3 = 3 4

Certificate Mathematics in Action Full Solutions 5B

= Q3 −Q1 3

(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.

1

Inter-quartile range = 4 − 3

=

2.

5 12

(a) Standard deviation = 3

8.

(b) Standard deviation = 2

Q3 = $12 499 (b) Inter-quartile range = Q3 – Q1 = $(12 499 – 8999) = $3500

(c) Standard deviation = 1 3.

(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.

5.

Standard deviation = 7.18 cm (cor. to 3 sig. fig.)

(a) From the graph, Q1 = $8999

9.

(a) Range = 9 – (–2) =11

( −2) +1 = −0.5 2 7 +9 Q3 = =8 2 Q1 =

Inter-quartile range = Q3 – Q1 = 8 – (–0.5) = 8 .5 Standard deviation = 4 (b) (i) ∵ 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

(a)

(b)

10. Standard deviation, since it can be applied in further statistical calculations and analysis and all data are taken into account. 11. Standard deviation, since it can be applied in further statistical calculations and analysis and all data are taken into account.

(c)

6.

(a) Range = 90 – 30 = 60 (b) Inter-quartile range = 80 – 50 = 30 (c) Median = 60

7.

(a) ∵ ∴ (b) ∵ ∴

The distance between two ends of the whiskers of class B is longer. Class B has the heights with a larger range. The length of the box of class A is larger. Class A has the heights with a larger interquartile range.

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) Level 2

16 Measures of Dispersion

14. (a) Range = 6 – 2 = 4

the 20th datum + the 21st datum 2 4 +4 = 2 =4

Median =

the 10th datum + the 11th datum 2 3 +4 = 2 = 3 .5 the 30th datum + the 31th datum Q3 = 2 5 +5 = 2 =5 Q1 =

Inter-quartile range = Q3 – Q1= 5 – 3.5 =1.5

(b) (c) From the graph, median =13 Q1 = 10 Q3 = 17 Inter-quartile range = Q3 – Q1 = 17 – 10 = 7

(b) 15. (a)

No. of time less than Cumulative frequency 0 0 5 4 10 7 15 19 20 27 25 29 30 30

(d) 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 2 14 +17 = 2 = 15 .5 the 22nd datum + the 23rd datum Q3 = 2 34 + 34 = 2 = 34 Q1 =

Certificate Mathematics in Action Full Solutions 5B

the 12th datum + the 13 datum 2 14 +14 = 2 = 14

Q1 =

the 36th datum + the 37th datum 2 15 +15 = 2 = 15

Q3 = (b) (i) class A (ii) class B (iii) class A

=Q3 −Q1 Inter-quartile range =15 −14

17. (a) For Mathematics,

median = 48

=1

Q1 = 40

1 (iii) Standard deviation = (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.

Q3 = 63 For Chinese,

median = 40 Q1 = 22 Q3 = 60 20. (a)

0 ×2 +1×7 +2 ×12 +3 ×7 +4 ×2 30 =2

Mean =

Median

(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 =Q3 −Q1 Inter-quartile range =(30 −6.5) h = 23.5 h

For battery B, Q1 = 10 h, Q3 = 35 h =Q3 −Q1 Inter-quartile range = (35 −10 ) h = 25 h

(ii) battery B (b) Yes, since it is less affected by extreme data. 19. (a) (i) (ii)

Range

=16 −13 =3

the 15th datum + the 16th datum 2 2 +2 = 2 =2 =

Inter-quartile range = the 23rd datum

−the 8th datum

=3 −1 =2 1 Standard deviation =

(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

16 Measures of Dispersion

the mean, relatively less data are concentrated about the mean.

a +b +c + d = 5 and 4 a +b +c + d + x =5 5

m = 7 or m =

29 3

21. (a) ∵

Multiple Choice Questions (p. 99) 1.

5 ×4 + x =5 5 x =5

∴

∴

(b) ∵

( a − 5) 2 + (b − 5) 2 + (c − 5) 2 + ( d − 5) 2 =2 4 ∴

2.

Standard deviation

(a − 5)2 + (b − 5)2 + (c − 5)2 + (d − 5) + ( x − 5) 5 2

=

Answer: B Q1 = the 2nd datum = 4 Q3 = the 5th datum = 10

2

Answer: A From the graph, Q1 = 65, Q3 = 80 ∴

3.

5 + 7 + m +13 4 25 + m = 4

Mean =

2

2

2

2 5+ m   2 5+ m   m−  +  13 −  4   4   =3 4 2 5

( a − 2d ) + ( a − d ) + a + ( a + d ) + ( a + 2 d ) 5 =a

∴

Standard deviation

[( a − 2d ) − a ]2 + [( a − d ) − a ]2 + ( a − a ) 2

 25 + m   25 + m  5−  + 7−  + 4   4   ∴

= Q3 −Q1 = 80 −65 =15

Mean =

2 2 ×4 +(5 −5) 2 5 =1.79 (cor. to 3 sig. fig.)

2

Inter-quartile range

= Q3 −Q1 =10 −4 = 6

Answer: C

=

22. ∵

Inter-quartile range

=

+ [( a + d ) − a ]2 + [( a + 2d ) − a ]2 5

=

10 d 2 5

= 2d 2 = 2d

Answer: A For group A, mean = a, inter-quartile range = 4 + m  2 5 + m = 2 5   standard deviation 2 + + 4 9 − 7   

2 5  − 5  2

4.

  4

+ m  2



2

2 5 + m  2 5 + m  2 5 + m     + + m2 − m +       4   2   4  2

2 5 + m  2 5 + m    + 1 6 9 − 1 3  +     2   4  4 2 4 3

2 5 + m  2 5 + m   + m2 − (2 5 + m ) +    2   4 4 9 7 2 + 3m 2 − 6 2 5 − 5 0 3m 2

− 5 0 m + 2 0 3

(m − 7 ) ( 3 m − 2 9

Certificate Mathematics in Action Full Solutions 5B

For group B, mean = a inter-quartile range = 10 standard deviation = 22 ∴ Only I is true. 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%) =0.95 m 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 The new inter-quartile range =q The new standard deviation =s

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