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4 Measures of Dispersion


Review Exercise 4 (p. 4.5)1. (a) Amount ($) Class mark ($) Frequency

50.5 149.5 100 4150.5 249.5 200 10250.5 – 349.5 300 7350.5 – 449.5 400 14450.5 – 549.5 500 5

(b) Class interval $350.5 $449.5 has the highest frequency.

(c) The required percentage of students

2. (a) From the graph, 14 programmes have the viewing rate of 10% or below.∴ 14 programmes will be subject to review.

(b) For the programmes which have more than 1 million viewers, the viewing rate is greater than

i.e. 20%.

From the graph, 46 programmes have the viewing rate of 20% or below.∴ Number of programmes having more than

1 million viewers

3. (a) Mean income

Arrange the monthly incomes in ascending order,i.e. $6200, $6700, $7400, $7600, $8000, $8500,

$9200, $9600, $25 700Median income

(b) The median can better represent the average monthly income of the employees since the extreme data($25 700) will greatly affect the value of the mean.

4. (a) From the graph, the median score 57

(b) From the graph, the median score

5. (a) Mean of data in set Y

Let x be the datum added in set Y.

∴ The required datum is 18.(b) Arrange the set of data in ascending order,

i.e. 10, 2, 4, 4, 5, 12, 13, 20, 26.Median 5If the median remains unchanged, the required datum should equal to the median.∴ The required datum is 5.

Activity

Activity 4.1 (p. 4.7)1. Mean Median Mode

Athlete A 7.5 7.5 7.5Athlete B 7.5 7.5 7.5

1

NSS Mathematics in Action 6A Full Solutions

2


2. Since the measures of central tendency for the two athletes are all the same, they performed equally well.

3. From the dot-plots, we observe that the scores of athlete A are more spread out.Hence, the performance of athlete B is more stable.

Activity 4.2 (p. 4.27)1. (a) Set A Set B

Range 9 9Inter-quartile range 5 5

(b) No

2. No. By inspection, set A is more dispersed.

Activity 4.3 (p. 4.40)1. English

2. (a) 6(b) 5

3. No. We don’t know the distributions of the marks./ We don’t know how many students got higher marks than Mary in each subject. (or any other reasonable answers)

4. The dispersion of the marks of the class in each subject.

Activity 4.4 (p. 4.53)1. Measure of

dispersionData set

AData set

BData 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 those of data set A.

3. The measures of dispersion of data set C are twice those of data set A.

Let’s Discuss

Let’s Discuss (p. 4.30)No, the result of their suggested formula is always zero.

Let’s Discuss (p. 4.55)(a) 40, 90 (or any other reasonable answers)(b) 65, 66 (or any other reasonable answers)

Classwork

Classwork (p. 4.12)Data Lower half of

the dataUpper half of

the data(a) 2, 5, 6, 8, 12, 13 2, 5, 6 8, 12, 13(b) 1, 5, 7, 9, 11, 14,

15 1, 5, 7 11, 14, 15

(c) 3, 1, 0, 2, 2, 4, 6, 7 3, 1, 0, 2 2, 4, 6, 7

(d) 18, 30, 10, 34, 45, 18, 24, 8, 28 8, 10, 18, 18 28, 30, 34, 45

(e) 4, 4, 1, 0, 1, 0, 3, 2, 5, 2

5, 4, 4, 3, 2 2, 1, 1, 0, 0

Data Q1 Q3IQR

(Q3 Q1)(a) 2, 5, 6, 8, 12, 13 5 12 7

(b) 1, 5, 7, 9, 11, 14, 15

5 14 9

(c) 3, 1, 0, 2, 2, 4, 6, 7 0.5 5 5.5

(d) 18, 30, 10, 34, 45, 18, 24, 8, 28 14 32 18

(e) 4, 4, 1, 0, 1, 0, 3, 2, 5, 2 4 1 3

Classwork (p. 4.20)(a) Median of A median of B

Range of A > range of BIQR of A < IQR of B

(b) Median of A > median of BRange of A < range of BIQR of A IQR of B

Classwork (p. 4.29)

(a)

= 3.08

(b)

=

(c)

=

(d)

=

Classwork (p. 4.31)

Data (cor. to 3 sig. fig.)

(a) {69, 72, 65, 63, 59, 56, 52, 52} 7.04(b) {68, 71, 58, 42, 48, 52, 53, 51} 9.22(c) {64, 63, 70, 72, 57, 69, 60, 71} 5.19(d) {40, 40, 40, 21, 21, 21, 55, 55, 55, 55} 14.1

Classwork (p. 4.36)(a) It is easy to calculate.

It is easy to understand by most people.The maximum and the minimum temperatures in a day are important data for people.

(b) It takes all data into account.(c) The inter-quartile range can be conveniently found

from a cumulative frequency polygon.It is not affected by extreme values.

(d) It is not affected by extreme values.

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Classwork (p. 4.42)

1. (a)

(b)

(c)

(d)

2. (a) Test 1(b) Test 3

Classwork (p. 4.45)Interval Percentage

of data(a)

between and

(b) between and

(c)greater

than

Classwork (p. 4.55)For the 1st row:

For the 2nd row:

For the 3rd row:

For the 4th row:

Quick Practice

Quick Practice 4.1 (p. 4.9)(a) Range of salaries of team A

Range of salaries of team B

(b) Since the range of salaries of team A > the range of salaries of team B, the salaries of team A are more dispersed.

Quick Practice 4.2 (p. 4.10)Range of waiting times at Airline X check-in counter (20.5 – 0.5) min 20 minRange of waiting times at Airline Y check-in counter (30.5 – 0.5) min 30 min

Since the range of waiting times at Airline X check-in counter < the range of waiting times at Airline Y check-in counter, the waiting times at Airline X check-in counter are less dispersed.

Quick Practice 4.3 (p. 4.13)Inter-quartile range of set A 23 – 15 8Inter-quartile range of set B

∵ IQR for set B > IQR for set A∴ Based on the inter-quartile ranges, set B has a greater

dispersion.

Quick Practice 4.4 (p. 4.14)(a) ∵ The lower quartile is the value corresponding to the

cumulative frequency of 50.∴ From the graph, the lower quartile∵ The upper quartile is the value corresponding to the

cumulative frequency of 150.∴ From the graph, the upper quartile

(b) Inter-quartile range

Quick Practice 4.5 (p. 4.20)∵ Smallest datum 13

Largest datum 82Median 28Lower quartile 18Upper quartile 56

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∴ The required box-and-whisker diagram is:

Quick Practice 4.6 (p. 4.21)

(a) Range of the data set

IQR of the data set

(b) ∵ Upper quartile 195

∴ The total number of data

Quick Practice 4.7 (p. 4.22)(a) The minimum, median, upper quartile and maximum of

the number of visitors to the Modern Art Gallery are all higher than that to the Chinese Art Gallery. Therefore, the Modern Art Gallery has more visitors in general.

(b) Since the box for the Chinese Art Gallery is shorter, the inter-quartile range of the number of visitors to the Chinese Art Gallery is smaller. Hence, the Chinese Art Gallery has a more stable number of visitors.

(c) No. The lower-quartile of the number of visitors to the Modern Art Gallery is lower than that to the Chinese Art Gallery. Therefore, there are at least 25% of the days in which the visitors to the Chinese Art Gallery are more than that to the Modern Art Gallery.

Quick Practice 4.8 (p. 4.32)(a) Standard deviation of the height cleared by Athlete X (X)

Standard deviation of the height cleared by Athlete Y (Y)

(b) Since Y < X, the heights cleared by Athlete Y are less dispersed. Therefore, Athlete Y has a more stable performance.

Quick Practice 4.9 (p. 4.33)Standard deviation of the number of countries visited

Quick Practice 4.10 (p. 4.34)Standard deviation of the distances (cor. to 3 sig. fig.)

Quick Practice 4.11 (p. 4.34)(a) Standard deviation of the salaries of the employees in

Company A (cor. to 3 sig. fig.)

Standard deviation of the salaries of the employees in Company B

(cor. to 3 sig. fig.)

(c) Since the standard deviation of the salaries of the employees in Company B is smaller, Company B offers more uniform salaries.

Quick Practice 4.12 (p. 4.43)Standard score of the height cleared by Jack

Standard score of the height cleared by Mary

Therefore, the height cleared by Jack is 0.938 standard deviation above the mean and the height cleared by Mary is 0.435 standard deviation above the mean. Jack’s recorded height is relatively higher in his own group.

Quick Practice 4.13 (p. 4.43)(a) For practice 1,

mean 30.5, standard deviation 3.2404 (cor. to 4 d.p.)John’s standard score

For practice 2,mean 19.625,standard deviation 2.6897 (cor. to 4 d.p.)

John’s standard score

(cor. to 3 sig.

fig.)

(b) ∵ Standard score in practice 1 > standard score in practice 2

∴ John performs better in practice 1.

Quick Practice 4.14 (p. 4.47)(a) ∵ 167 cm [175 – 2(4)] cm

183 cm [175 + 2(4)] cm ∴ The required percentage is 95%.

(b) The required number of players

Quick Practice 4.15 (p. 4.48)(a) ∵ 45 min [35 + 2(5)] min

∴ The percentage of delivery times greater than 45 min

∴ Number of food coupons given out

(b) ∵ 45 min [30 + 3(5)] min ∴ The percentage of delivery times greater than 45 min

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Quick Practice 4.16 (p. 4.56)(a) The new mean

The new standard deviation = 4

(b) The new mean

The new standard deviation = 4(1 5%)= 3.8

(c) Because originally the student’s score is equal to the mean, changing it to 0 will make the distribution of data less concentrated about the mean. As a result, the standard deviation will increase.

Quick Practice 4.17 (p. 4.58) (a) Range of Tom’s marks

Standard deviation of Tom’s marks= 5.54 (cor. to 3 sig. fig.)

(b) (i) Since Tom’s mark in Test 5 is lower than the lowest mark of the first 4 tests, the range of his marks will increase.

(ii) Standard deviation of Tom’s marks in the 5 tests (cor. to 3 sig. fig.)

Hence, the standard deviation of Tom’s marks increases when compared with that in (a).

Further Practice

Further Practice (p. 4.15)1. (a) (i) Range of the starting salaries for Group A

Range of the starting salaries for Group B

(ii) Since the range of the starting salaries for Group A > the range of the starting salaries for Group B, the starting salaries for Group A are more dispersed.

(b) (i) Inter-quartile range of the starting salaries for Group A

Inter-quartile range of the starting salaries for Group B

(ii) ∵ IQR for Group A < IQR for Group B∴ Based on the inter-quartile ranges of the

starting salaries of the graduates, graduates of Group B have a greater dispersion in their starting salaries.

2. (a)

(b) From the graph, the median

From the graph, the range

∵ The lower quartile is the value corresponding to the cumulative frequency of 10.

∴ From the graph, the lower quartile .∵ The upper quartile is the value corresponding to

the cumulative frequency of 30.∴ From the graph, the upper quartile .

Inter-quartile range

Further Practice (p. 4.23)1. (a) Since the box for Mathematics test is the longest, the

inter-quartile range of marks in Mathematics test is the largest.

(b) Since the length of the whole box-and-whisker diagram for Chinese test is the shortest, the range of marks in Chinese test is the smallest.

(c) From reading the bars inside the boxes in the diagrams, we can see that the median mark in Mathematics test is the highest.

(d) In Mathematics test, Peter’s result is below the upper quartile. In Chinese test, Peter’s result is equal to the upper quartile. In English test, Peter’s result is above the upper quartile. Therefore, Peter performs the best in English test relative to his classmates.

2. ∵ Smallest datum 20Largest datum 100Median 72

The lower quartile is the value corresponding to the cumulative frequency of 12.5.From the graph, the lower quartile 60The upper quartile is the value corresponding to the cumulative frequency of 37.5.From the graph, the upper quartile 84

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∴ The required box-and-whisker diagram is:

3. (a) From the diagram, we can see that Paul’s score is equal to the upper-quartile of Group A. Therefore, Paul’s score is higher than the scores of 75% of the students in his own group. Similarly, Mary’s score is equal to the median of Group B. Therefore, Mary’s score is higher than the scores of 50% of the students in her own group. Hence, Paul’s performance is relatively better.

(b) Let x be the number of students in each group.Number of students having marks lower than 60 in Group A 0.75xNumber of students having marks lower than 60 in Group B 0.5x∴ The required percentage

Further Practice (p. 4.35)1. (a) Standard deviation

(b) (i) Data 1 – 10 11 – 20 21 – 30 31 – 40Frequency 3 6 3 4

(ii) Standard deviation

2. (a) He should choose Thomas. It is because the best time given by Thomas is 53.4 s, which is fast enough to break the even record.

(b) He should choose Mike. Since the standard deviation of time taken by Mike is less than that by Thomas, Mike is a more reliable swimmer. Also, his mean time is 58.8 s, which is fast enough for him to be qualified.

3. (a) Mean

(b) Standard deviation

Further Practice (p. 4.44)1. (a) ∵ Standard score of May’s time

∴ The time taken for May to complete the 100 m event is greater than the mean.

∴ May doesn’t perform better than the average.

(b)

∴ The mean time of all the contestants is 13.8964 s.

(c) Eve’s standard score

2. (a) Mean of the marks

Standard deviation of the marks

(b) Let x be the passing mark.

∴ The passing mark is 3.∴ 11 students pass the quiz.

Further Practice (p. 4.48)1. (a) ∵ 5 years

9 years∴ The percentage of LEDs with lifetimes between

5 and 9 years 95%∴ The required number of LEDs

(b) ∵

∴ 2.5% or more of the LEDs have lifetimes less than or equal to

∴ The required time =

2. (a) ∵ 3 min

∴ The required probability

(b) ∵∴ The percentage of customers who do not need to

wait more than 108 s before the call is answered

∴ It is not true that over 85% of the customers will wait less than 100 s for the service.

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Further Practice (p. 4.58)1. (a) Range

Standard deviation

(b) RangeStandard deviation

(c) RangeStandard deviation

(d) Range

Standard deviation

2. If the datum in the interval 61 min – 75 min is moved to the interval 1 min – 15 min, the distribution will be more concentrated. Therefore, the standard deviation of the travelling time of the class will decrease.

ExerciseExercise 4A (p. 4.15)

Level 11. (a) Range

(b) Range(c) Range

2. (a) For A,

For B,

For C,

(b) For A,

For B,

For C,

3. (a) Range

(b)

4. (a) Range of pulse rates of teenagers

Range of pulse rates of senior citizens

(b) Range of pulse rates of teenagers <range of pulse rates of senior citizens∴ The group of senior citizens has more dispersed

pulse rates.

5. (a) Range

(b) Both measures are affected. Since the smallest datum is discarded, the range will be changed. Moreover, since the data are different from each other, the lower quartile and the upper-quartile will also be changed. In other words, the IQR will be changed.

6. (a) Range

(b) The range will remain unchanged. It is because the range is independent to the frequency.

7. (a) Lower quartileUpper quartile

(b) RangeIQR

8. (a) Lower quartileUpper quartile

(b) RangeIQR

9. (a) Range of the scores of athlete A

IQR of the scores of athlete A

Range of the scores of athlete B

IQR of the scores of athlete B

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(b) ∵ Both the range and the IQR of the scores of athlete B are smaller than that of athlete A.

∴ Athlete B’s scores have a smaller dispersion.

Level 210. (a) For class A,

For class B,

(b) For class A,

For class B,

(c) ∵ The range and the IQR of class A are greater than that of class B.

∴ Class A has a greater dispersion in marks.

11. (a)

(b)

12. (a) Range

(b) (i) Weight less than (kg) 1.45 1.95 2.45 2.95 3.45 3.95

Cumulative frequency 0 7 13 17 19 20

(ii) From the graph,

13. (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

14. (a) Median

Range

,

(b) None of them is affected. The data set becomes{170 cm, 174 cm, 175 cm, 175 cm, 177 cm, 177 cm, 177 cm, 180 cm, 180 cm, 180 cm, 185 cm, 186 cm}. Then, the median is still 177 cm; Q1 and Q3 are still 175 cm and 180 cm respectively. Furthermore, IQR is still 5 cm.

15. (a) Length of telephone call less than (min)

0.5 4.5 8.5 12.5 16.5 20.5 24.5

Cumulative frequency 0 21 55 70 79 79 80

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(b) Range

(c) From the graph,

(d) Inter-quartile range is a better measure of the dispersion since it is not affected by extreme values.

Exercise 4B (p. 4.24)

Level 11. (a)

(b)

(c)

(d)

2. (a) Highest daily maximum temperatureLowest daily maximum temperature

(b) Median of the daily maximum temperature(c)

3. (a) Range

(b) The required number of students

(c) No. According to the diagram, the highest IQ of a student in the class is 125. Therefore, we can’t find a student in the group with an IQ of 130.

4. (a) MedianLower quartileUpper quartile

(b)

(c) RangeIQR

5. (a) Test 2

(b) Test 2

(c) Test 1

(d) (i) Let x be the number of students in the class.

∴ There are 36 students in the class.

(ii) The required number of students

6.

(or any other reasonable answers)

Level 27. (a) The required percentage =

(b) The required percentage =(c)(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.

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8. (a)

(b) (i) Mathematics(ii) Chinese(iii) English

(c) Mary performs the best relative to her class in Chinese examination. It is because her result in Chinese examination is above the upper quartile, while the other two results in Mathematics and English examinations are below the upper quartiles.

9. (a) For group A,

For group B,

∴ The required box-and-whisker diagrams:

(b) (i) Group A(ii) Group B(iii) Group A

(c) The median score of Group A is closer to the upper quartile while that of Group B is closer to the lower quartile. In other words, the distribution of scores in the upper half is more dispersed in Group B than in Group A, and the distribution of scores in the lower half is more dispersed in Group A than in Group B.

10. (a) (i)

(ii) The middle 50% of the scores lie between the lower quartile and the upper quartile, i.e. 98 and 107.

(b) (i)

(ii) The minimum score of team A equals to the median score of team B. So, the scores of team A are higher than those of team B in at least 50% of the matches, i.e. 5 matches.

11. (a) For school A,

For school B,

(b)

(c) School A has a less dispersed mark distribution since it has a smaller inter-quartile range.

12. (a) For company A,

For company B,

(b)

(c) Company A has a more dispersed salary distribution since it has a larger inter-quartile range.

Exercise 4C (p. 4.37)

Level 1

1. (a) Standard deviation

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(b) Standard deviation(c) Standard deviation

(d) Standard deviation

2. Mean

Standard deviation

3. Mean

Standard deviation

4. (a) Mean

Standard deviation

(b) Factory X has a more dispersed distribution of salaries since the standard deviation of the salaries of the workers in factory X is higher.

5. (a) For group A,mean timeFor group B,mean time

(b) For group A,standard deviation

For group B,standard deviation

(c) Group B gives a more widely spread result since the standard deviation of the results of group B is larger.

6. (a) Mean waiting time

(b) Standard deviation of the waiting time

7. No. Consider the following two data sets.Set A = {1, 2, 3, 4, 5, 6} and set B = {1, 1, 1, 1, 1, 10}

Inter-quartile range of A > Inter-quartile range of BStandard deviation of A < Standard deviation of B(or any other reasonable answers)

Level 2

8. (a) Mean

Standard deviation

(b) Mean

Standard deviation

9. (a) Machine Y produces pins with a smaller standard deviation in diameter since the distribution of the diameters of pins produced by machine Y is more concentrated about the mean.

(b) Machine Y. It is because the frequency of pins of 10 mm diameter produced by machine Y is higher than that of machine X.

10. (a) Mean age

(b) Standard deviation of the ages of the employees

(c) The new age distribution:Age 20 29 30 39 40 49 50 59Frequency 7 9 10 2

New mean ageNew standard deviation of the ages of the remaining employees

11. (a) (i) Mean mark(ii) Standard deviation of the marks

(b) ∵ Standard deviation of school A > Standard deviation of school B

∴ School A’s result is more dispersed.

12. To obtain the largest standard deviation, the data should be dispersed from the mean as much as possible. Hence, we should choose the maximum and the minimum numbers only.

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The required data set {0, 0, 0, 0, 0, 10, 10, 10, 10, 10}The largest possible value of standard deviation

13. (a) Mean waiting time

(b) Standard deviation of waiting time

(c) No. Although the mean waiting time has been reduced to 8 min, the standard deviation is relatively large. In other words, the new distribution of waiting time is more dispersed than before and extreme values may be more likely to appear.

14. (a) ∵ The temperature 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 (32 – 27)C = 5CRange of city B (27 – 20)C = 7C∴ City B has a greater variation in temperature on

that day.

15. (a) He made a good choice. The standard deviation takes all data into account and therefore, it is the most reliable measure. Moreover, it can be applied in further statistical calculations and analysis.


(c) The machine does not function property since the standard deviation is greater than 0.1 kg.

Exercise 4D (p. 4.49)

Level 1

1. (a) Let x be the standard deviation of the class.

∴ The standard deviation of the class is 6.

(b) Let y be the marks of Sam.

∴ The marks of Sam is 65.

(c) Let z be the mean of the class.

∴ The mean of the class is 60.

(d) Standard score of Sam

2. Mary’s standard score

3. Let x be the number of fish Jacky caught.

∴ Jacky caught 67 fish.

4. (a) ∵∴ The required percentage

(b) ∵

∴ The required percentage

(c) ∵


(d) ∵∴ The required percentage

(e) ∵∴ The required percentage

5. (a) ∵


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The required number of females

(b) ∵∴ The required percentage

The required number of females

6. (a) Sales of salesman A was below the mean sales of the company last month because his standard score is less than 0.

(b) Salesman A made more sales last month as his standard score is larger than salesman B’s.

7. (a) Standard score of Mary

(b) The higher standard score a competitor obtains, the longer finishing time he/she has made. Hence, a higher standard score means a worse performance.

8. (a) David’s standard score in test 1

David’s standard score in test 2

(b) David performs relatively better in test 1. It is because his standard score in test 1 is higher.

Level 2

9. (a) Let x be the standard deviation of the number of tickets sold.

∴ The standard deviation of the number of tickets sold is 5.

(b) Let y be the number of tickets Tom sold.

∴ Tom sold 89 tickets.

10.

The percentage of boxes of washing powder weighing between 4.8 kg and 4.9 kg

Let x be the total number of boxes of washing powder.

∴ There are 2000 boxes of washing powder.Percentage of boxes of washing powder weighing not less than 4.8 kg

The required number of boxes of washing powder

11. ∵

∴

∵

∴(2) – (1) :

By substituting into (2), we have

12. (a) (i) Standard score of the weight of the baby boy

(ii) Standard score of the weight of the baby girl

(b) The baby boy is comparatively heavier since the standard score of the weight of the baby boy is greater.

13. (a) (i) Mean

Standard deviation(ii) Mean

Standard deviation

14


15


(b) Adam’s standard score in geography test

Adam’s standard score in history test

(c) Adam performs better in geography test compared to his seven close friends since his standard score in geography test is higher than that in history test.

14. (a) Standard score in jumping

Standard score in dressage

Standard score in eventing

(b) (i) Tom performs the best in jumping since his standard score in jumping is the highest among all the areas.

(ii) Tom performs the worst in eventing since his standard score in eventing is the lowest among all the areas.

15. ∵

∴ Length of the warranty period

16. (a) ∵∴ The required percentage

(b) (i) ∵∴ The required probability

(ii) The required probability

(iii) Let x be the total number of bottles of candles that should be produced.

∴ The total number of bottles of candies is 3200.

17. (a) (i) ∵

∴ The required probability

(ii) ∵∴ The required probability

(b)

18. (a) Connie’s standard score

(b) Let x be Connie’s adjusted mark.

∴ Connie’s adjusted mark is 50.4.

19.

(or any other reasonable answers)

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20. (a) For machine A,

∴ Percentage of ball bearings produced with diameters between 5.88 mm and 6.12 mm 95%

17


For machine B,

∴ Percentage of ball bearings produced with diameters between 5.88 mm and 6.12 mm

∴ Machine B will produce more ball bearings with acceptable diameters.

(b) For machine A,


For machine B,


∴ Machine A will produce more ball bearings with acceptable diameters.

(c) (i) To produce at least 95% acceptable ball bearings,

(1) + (2) :

Substitute into (1).

The standard deviation should be adjusted to 0.03 mm.The minimum cost of fixing machine A

(ii) ∵∴ There is no need to adjust .To produce at least 95% acceptable ball bearings,

The mean diameter should be adjusted to5.98 mm.The minimum cost of fixing machine B

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Exercise 4E (p. 4.59)

Level 11. (a) Range : unchanged

Standard deviation : decrease(b) Range : increase

Standard deviation : increase(c) Range : unchanged

Standard deviation : increase(d) Range : unchanged

Standard deviation : decrease

2. (a) Range of set B(b) Range of set C

3. (a) (i) Mean

Standard deviation

(b) (i) New mean

New standard deviation

(ii) New mean


4. (a) (i) Range

Standard deviation

(ii) RangeStandard deviation

(b) (i) Range = 6 (5) = 11

Mean

Standard deviation

(ii) From (a), takeRangeStandard deviation

5. The new inter-quartile range

6. (a) Range

(b) Standard deviation of the ages of students

(c) New rangeNew IQRNew standard deviation (cor. to 3 sig. fig.)

7. (a) RangeIQR

(b) Since each datum is reduced by 5 min, the range and the IQR will remain unchanged.

8. (a) Mean

Standard deviation

(b) (i) Increase

(ii) New mean


∴ The standard deviation increases.

Level 29. The mean of the data set 19.2 (cor. to 3 sig. fig.)

∵ The range remains unchanged.∴ The minimum and the maximum datum are not the

possible values of x.∵ The standard deviation increases.∴ The data are more dispersed from the mean.∴ 18 and 21 are the possible values of x.

10. (a) Class B(b) If all marks are multiplied by 10, the ranges, the IQRs

and the standard deviations will be all multiplied by 10.

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11. (a)

Standard deviation

(b) ∵ More tickets are sold at the most expensive price.∴ The distribution of the data is more dispersed.∴ The mean and the standard deviation increase.

12. (a) New meanNew standard deviation

(b) New meanNew standard deviation

(c) ∵ The salaries of the ten employees are equal to the mean.

∴ The distribution of the data is less concentrated.∴ The standard deviation increases.

13. (a) RangeIQR

(b) (i) Range

IQR

(ii) Highest temperature

Lowest temperature

20


14. (a) Group B

(b) (i) For group A,

inter-quartile range

For group B,

inter-quartile range

(ii) No

(c) (i) The standard deviation will increase in group A since the leaving member has age equal to the mean. The data are relatively less concentrated about the mean.

(ii) The standard deviation will decrease in group B since the joining member has age equal to the mean. The data are relatively more concentrated about 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.

15. (a) (i) New mean


(ii) The range and the IQR will be multiplied

by . Hence, both decrease by 25%.

(b) (i) Sum of scores of the students

Correct mean

(ii) The standard deviation will increase. Since 50 is closer to the mean than 60, correcting the score will make the distribution less concentrated about the mean.

Revision Exercise 4 (p. 4.66)

Level 11. (a) Range

IQR

(b) Range

(c) Range

(d) Range

2. (a) Mean

Standard deviation

(b) Mean

Standard deviation

21


22


(c) Mean

Standard deviation

3. (a) Range(b) IQR(c) Median

4. (a) Median(b) Range

IQR

5. (a) Range(b) Median

Lower quartile

Upper quartile

6. (a) Party B(b) Party A(c) Party B

7. (a) Median

(b) RangeIQR

(c)

(d) The median is closer to the minimum than to the maximum. In other words, the distribution of the data in the lower half is less dispersed than that in the upper half.

8. (a) Median

Lower quartile

Upper quartile

(b)

(c) The required interval is between 49 m3 and 88 m3.

9. (a) City A has a population with more dispersed income.

(b) In city A, the monthly salary of the job offer is greater than the mean monthly income of a person by $500. Such difference is very small when compared with the standard deviation $3000. In other words, the monthly salary given to Lucy is not high.In city B, the monthly salary of the job offer is greater than the mean monthly income of a person by $2000. Such difference is very large when compared with the standard deviation $800. In other words, the monthly salary given to Lucy is very high.∴ Lucy should take the job offer in city B.

10. (a)Class mark ($1000) 29.5 49.5 69.5 89.5 109.5

Frequency 12 x y 15 9


11. An engineer always needs the most accurate and reliable measure. Hence, all data must be taken into account and standard deviation should be used.

12. If the statistician wants to do accurate statistical calculations or reliable analysis, standard deviation should be used as it takes all data into account. If the statistician does not want complicated calculations, inter-quartile range is good enough for the purpose as it is not affected by extreme values.

13. (a) Range


Standard deviation

(b) (i) ∵ x is added to each datum.∴ The standard deviation will remain

unchanged.

23


∴ The standard deviation(ii) ∵ Each datum is doubled.

∴ The standard deviation will be doubled.∴ The standard deviation

(iii) ∵ Each datum is divided by 4.∴ The standard deviation will be divided by 4.∴ The standard deviation

14. (a)

(b) ∵

∴

15. (a) Mean

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

(b) The datum 2200 cc is close to the mean. So, if the manufacturer stops producing a 2200 cc model, the distribution of the data will be less concentrated about the mean. Therefore, the standard deviation will increase.

16. (a) Mean

Standard deviation

(b) Standard score

17. (a) For piano group,

mean

standard deviation

For violin group,

mean

standard deviation

For flute group,

mean

standard deviation

(b) Ben, Jess and Marcy are the best participants in their own groups. Among these 3 participants, the one with the highest standard score should be the overall champion.Standard score of Ben

Standard score of Jess

Standard score of Marcy

∴ Jess will win the overall champion.

18. (a) (i) ∵∴ The required percentage

24


25


(ii) ∵


(b) Percentage of pigs which are either heavier than 135 kg or lighter than 90 kg

The required number of pigs

19. The two groups have equal mean, equal modal class and equal median. The standard deviation for group A is greater than that of group B.

20. (a) The required set of data {1, 2, 3, 4, 5} or {2, 4, 6, 8} (or any other reasonable answers)

(b) The required set of data {0, 0, 0, 0, 1} or {0, 0, 0, 3, 4} (or any other reasonable answers)

21. (a)

(b) ∵ The data set in (b) can be formed by adding a common constant 10 to each datum of the data set in (a).

∴ The minimum, maximum, median, lower quartile and upper quartile of data set in (b) should be also increased by the same common constant 10.

∴

(c) (i)

(ii)

22. (a) ∵∴ The required probability

(b)

The probability that the weight of a tomato is greater than or equal to

which is

very low. In other words, the probability of getting a tomato with weight more than 280 g is even lower.∴ It is very unlikely that a tomato weighs more than

280 g.

Level 2

23. (a) Years of experience less than

Cumulative frequency

0 05 4

10 715 1920 2725 2930 30

(b)

(c) Median (cor. to the nearest integer)

,

(cor. to the nearest integer)

26


24. (a) For group A,median the 16th datum

28 min

27


Q1 the 8th datum 10 min

Q3 the 23rd datum 34 min

For group B,median = the 15th datum

= 26 min

The required diagrams are:

(b) (i) Group A(ii) Group B(iii) Group A

25. (a) Median

,

(b) (i)

(ii) No. The diagrams only suggest that in general the members read more books. Since the diagrams cannot show the actual number of books read by each member, the president cannot guarantee every member reads more books after joining the club.

26. (a)

(b)

(c) ∵

∴ Set X is more dispersed.

27. (a) (i) For battery A,,


For battery B,,


(ii) Battery B(b) Yes. The middle 50% of the data of battery B are

really more dispersed than that of battery A. Since the inter-quartile range focuses on the distribution of the middle 50% of given data, it is a suitable measure of dispersion in this case.

28. (a) Standard score for May

Standard score for John

(b) The lower standard score a visitor obtains, the faster he/she reacts.∴ John reacts relatively faster when compared with

other members in his own group.

29. (a) ∵ Karen’s standard score in paper I > Karen’s standard score in paper II

∴ Karen performs better in paper I compared with the scores of the whole class.

(b)

(c) Karen’s score in the overall Mathematics examination

30. Let and be the new mean and the new standard deviation of the test results respectively.

For Bob,

......(1)

For Kate,

......(2)(1) – (2), we have

28


By substituting 8 into (2), we have

∴ The new mean and the new standard deviation of the test results are 60 and 8 respectively.

31. (a) Mean price$4840 (cor. to 3 sig. fig.)Standard deviation $4360 (cor. to 3 sig. fig.)The price should be between$(4840 – 4360) and $(4840 + 4360),i.e. $480 and $9200.∴ Mr. Mok can choose watches B, C, D, E and F.

(b) The required percentage

(c) If the prices are normally distributed, there should be 68% of the data lying within one standard deviation from the mean. However, there are only 55.6% of the watches with prices lying within one standard deviation from the mean. Therefore, the prices of the watches do not follow the normal distribution.

32. (a) ∵ 69.5 mm [70 – 2(0.25)] mm 70.5 mm [70 + 2(0.25)] mm


(b) Net profit

33. (a) (i) ∵ 1.3 kg [1 + 2(0.15)] kg ∴ The percentage of chickens which weigh

over 1.3 kg

(ii) ∵ 1.3 kg (1.2 + 0.1) kg ∴ The percentage of ducks which weigh over

1.3 kg

(b) P(at least one weighs over 1.3 kg)

34. (a) ∵ 1595 lm [1700 – 3(35)] lm 1770 lm [1700 + 2(35)] lm =

∴ Percentage of light bulbs having brightness between 1595 lm and 1770 lm

∴ The required number of light bulbs

(b) Let y be the number of light bulbs which should be manufactured in total.

∴ 10 000 light bulbs should be manufactured in total.

(c) ∵ 1595 lm [1700 – 3(35)] lm ∴ Percentage of light bulbs having brightness

lower than 1595 lm

P(more than one light bulb has brightness lower than 1595 lm) P(2 light bulbs have brightness lower than 1595

lm) + P(all light bulbs have brightness lower than 1595 lm)

35. (a) (i) If the two records are corrected, the sum of heights of the 30 students will be decreased by 20 cm. ∴ The mean will be decreased by

(ii) Median

If both incorrect data are 171 cm or 173 cm, then after the corrections, they become 161 cm or 163 cm, which are still not the 15th and the 16th datum. So, the median remains unchanged.

(iii) The original mode is 157 cm, with a frequency of 5. If both incorrect data are 171 cm or 173 cm, then after the corrections, they become 161 cm or 163 cm, with a frequency less than 5. Therefore, the mode remains unchanged.

(iv) If both incorrect data are 173 cm, then after the corrections, the largest datum is 171 cm which is less than the original one by 2 cm, and hence, the range will be decreased by 2 cm. Otherwise, the range remains unchanged.

(v) Before the corrections,Q1 155 cm, Q3 167 cmAfter the corrections,

Since the IQR will be decreased by 2 cm.

(b) Originally, the distribution of the data is concentrated about the interval 155 cm 167 cm, i.e. the middle 50% of the data.No matter what corrections will be done, the two incorrect data will certainly fall in the interval155 cm 165 cm. Precisely, from (a)(v), the IQR will

29


be decreased by 2 cm. So, the new distribution is more concentrated than before.In other words, the data are less dispersed.

36. (a) (i) Range

(ii)


(i) Standard deviation

(b) The range will be increased 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 be increased since the new datum is not close to the mean and relatively less data are concentrated about the mean.

37. (a) Mean

Median

Inter-quartile range = the 23rd datum the 8th datum= 3 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 and 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 and relatively less

data are concentrated about the mean.

38. (a) ∵ and

∴

(b) ∵

∴ Standard deviation

39. ∵ Mean

and standard deviation = 3

∴

Multiple Choice Questions (p. 4.76)

1. Answer: AArrange the heights (cm) in ascending order:

∴ Inter-quartile range of the heights

2. Answer: AFrom the graph,

30


∴ Inter-quartile range

3. Answer: B

Mean

31


Standard deviation

4. Answer: B∵ Each datum is increased by 10.∴ Median is increased by 10 while range and inter-

quartile range will remain unchanged.∴ Only I and II are correct.

5. Answer: CThe data set can be formed by adding the common constant a to each datum of data set

The data set can be formed

by multiplying each datum of by the common constant d.Obviously, the mean and the standard deviation of

are 0 and respectively. So, the

required standard deviation is

6. Answer: AFrom the diagram, median of data set A median of data set Brange of data set A > range of data set Binter-quartile range of data set A < inter-quartile range of data set B∴ Only I and II are true.

7. Answer: BFrom the diagram,∵ The length of the box for Chinese examination is the

shortest.∴ Chinese examination has the smallest inter-quartile

range of marks.

8. 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

The new standard deviation

9. Answer: BSince 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: BFrom the graph,range of P range of Qinter-quartile range of P < inter-quartile range of Qmedian of P median of Q∴ Only II is true.

11. Answer: CFor I, standard deviation 14.1 (cor. to 3 sig. fig.)For II, standard deviation 11.5 (cor. to 3 sig. fig.)For III, standard deviation 16.0 (cor. to 3 sig. fig.)∴ II < I < III

12. Answer: A

Let x marks be the standard deviation.

∴ The standard deviation is 5 marks.

13. Answer: C∵∴ The required percentage

Investigation Corner (p. 4.81)

1. Consider the following table.

Marks (x) Class MarkNumber of candidates

in 2008

Number of candidates

in 2009

1260 370

6745 3100

16 970 21 075

17 035 21 890

6710 3205

1280 360

Mean in 2008

Standard deviation in 2008

Mean in 2009

Standard deviation in 2009

2. For 2008,

passing mark

Lowest mark for a distinction

32


For 2009,

passing mark

Lowest mark for a distinction

3. Under system Y in 2008,lowest mark for ‘B’

lowest mark for ‘D’

Candidate Year Marks GradeSystem X System Y

Joanna 2008 92 A BAnthony 2009 87 B ARachel 2008 36 F ETim 2009 42 E F

4. (a) Under system Y, since the mean in 2008 and 2009 are very close but the standard deviation in 2009 is much less than that in 2008, the mark for a distinction in 2009 is much lower than that in 2008.

(b) By using the same argument in (a), the passing mark in 2009 is much higher than that in 2008.

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