# 21 Measures of Dispersion Case Study

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21 Measures of Dispersion Case Study
21.1 Range and Inter-quartile Range 21.2 Box-and-whisker Diagrams 21.3 Standard Deviation 21.4 Applications of Standard Deviation 21.5 Effects on the Dispersion with a Change in Data Chapter Summary

Case Study Both divers get the same total mark. But it seems that A’s performance is less consistent. Is that true? You need to know the meaning of dispersion first. Although both drivers got the same total mark, it does not mean that both of them had a consistent performance for all ten dives. If we plot a broken line graph for both divers, we find that the marks of diver A fluctuate more than those of diver B. In the above example, we consider the spread of the data. In this chapter, we will learn how to represent this by statistical methods.

21.1 Range and Inter-quartile Range
In junior forms, we learnt three measures of central tendency of a set of data, namely mean, median and mode. However, these measures tell us only limited information about the data. Consider two boxes of apples A and B. The weight (in g) of each apple in each box is given below: Box A: 100, 100, 100, 104, 110, 110 Box B: 85, 95, 100, 104, 116, 124 The mean and the median of the weights for both boxes of apples is 104 g and 102 g respectively, and the weights of the apples in Box B are more widely spread than those in Box A. The spread or variability of data is called the dispersion of the data. In this chapter, we are going to learn the following measures of dispersion: 1. Range 2. Inter-quartile range 3. Standard deviation

21.1 Range and Inter-quartile Range
A. Range The range is a simple measure of the dispersion of a set of data. 1. For ungrouped data, the range is the difference between the largest value and the smallest value in the set of data. Range  Largest value – Smallest value 2. For grouped data, the range is the difference between the highest class boundary and the lowest class boundary Range  Highest class boundary – Lowest class boundary

Example 21.1T 21.1 Range and Inter-quartile Range Solution: A. Range
The weights (in g) of eight pieces of meat are given below: 210, 230, 245, 180, 220, 240, 175, 195 (a) Find the range of the weights. (b) If the meat is sold at \$3 per 100 g, find the range of the prices of the meat. Solution: (a) Range  (245  175) g (b) Range of the prices

Example 21.2T 21.1 Range and Inter-quartile Range Solution: A. Range
The following table shows the weights of the boys in S6A. (a) Write down the upper class boundary of the class 70 kg – 74 kg. (b) Write down the lower class boundary of the class 50 kg – 54 kg. (c) Hence find the range of the weights. Weight (kg) 50 – 54 55 – 59 60 – 64 65 – 69 70 – 74 Frequency 3 6 8 5 2 Solution: (a) kg (b) kg (b) Range  (74.5  49.5) kg

21.1 Range and Inter-quartile Range
B. Inter-quartile Range When a set of data is arranged in ascending order of magnitude, the quartiles divide the data into four equal parts. Full set of data arranged in order of magnitude 25% of data Q Q Q3 Q1, Q2 and Q3 are also called the first, the second and the third quartiles respectively. Q1 : lower quartile  25% of data less than it Q2 : median  50% of data less than it Q3 : upper quartile  75% of data less than it The inter-quartile range is defined as the difference between the upper quartile and the lower quartile of the set of data. Inter-quartile range  Q3 – Q1

Example 21.3T 21.1 Range and Inter-quartile Range Solution:
B. Inter-quartile Range Example 21.3T The marks of 13 boys in a Chinese test are recorded below: (a) Arrange the marks in ascending order. (b) Find the median mark. (c) Find the range and the inter-quartile range. Solution: (a) Arrange the marks in ascending order: 60, 62, 62, 65, 68, 69, 70, 72, 78, 78, 80, 81, 84 (b) Median (c) Range  Inter-quartile range

Example 21.4T 21.1 Range and Inter-quartile Range Solution:
B. Inter-quartile Range Example 21.4T The cumulative frequency polygon shows the lifetimes (in hours) of 80 bulbs. (a) Find the range of the lifetimes. (b) Find the median lifetime. (c) Find the inter-quartile range. Solution: (a) Range  (780  140) hours (b) From the graph, Median (c) Q1  240 hours, Q3  600 hours  Inter-quartile range  (600  240) hours

Example 21.5T 21.1 Range and Inter-quartile Range Solution:
B. Inter-quartile Range Example 21.5T Consider the ages of passengers in two mini-buses. Mini-bus A: 18, 24, 25, 19, 12, 10, 34, 39, 45, 23, 34, 40, 24, 28 Mini-bus B: 23, 26, 28, 32, 38, 34, 19, 26, 29, 32, 35, 30, 29, 22 By comparing the ranges and the inter-quartile ranges of the ages, determine which group of passengers has a larger dispersion of ages. Solution: For Mini-bus A, For Mini-bus B, the range  45  10  35 the range  38  19  19 For Mini-bus A, For Mini-bus B, the inter-quartile range  34  19 the inter-quartile range  32  26  15  6 Since the range and the inter-quartile range of the ages of passengers of mini-bus A are larger, the passengers on mini-bus A have a larger dispersion.

21.2 Box-and-whisker Diagrams
A box-and-whisker diagram is a statistical diagram that provides a graphical summary of the set of data by showing the quartiles and the extreme values of the data. The difference between the two end-points of the line is the range. The length of the box is the inter-quartile range. A box-and-whisker diagram shows the greatest value, the least value, the median, the lower quartile and the upper quartile of a set of data.

Example 21.6T 21.2 Box-and-whisker Diagrams Solution:
The following box-and-whisker diagram shows the number of family members of a class of students. (a) Find the median and the range of the number of family members. (b) Find the inter-quartile range. Solution: (a) Median Maximum value  6 and minimum value  1 Range (b) Q1  2 and Q3  4 Inter-quartile range

Example 21.7T 21.2 Box-and-whisker Diagrams Solution:
The following shows the measurements of the waists (in inches) of the students in a class. Girls: Boys: (a) Find the median, the lower quartile and the upper quartile of the waists for both boys and girls. (b) Draw box-and-whisker diagrams of their waist measurements on the same graph paper. Solution: (a) Arrange the measurements in ascending order: Girls: 22, 23, 24, 25, 25, 25, 26, 26, 26, 27, 28, 28, 29, 30, 32 Boys: 25, 26, 27, 28, 28, 28, 28, 29, 29, 30, 31, 32, 32, 32, 34 For girls, median  For boys, median 

Example 21.7T 21.2 Box-and-whisker Diagrams Solution:
The following shows the measurements of the waists (in inches) of the students in a class. Girls: Boys: (a) Find the median, the lower quartile and the upper quartile of the waists for both boys and girls. (b) Draw box-and-whisker diagrams of their waist measurements on the same graph paper. Solution: (b) For boys, minimum  25 inches maximum  34 inches For girls, minimum  22 inches maximum  32 inches Refer to the figure on the right.

21.3 Standard Deviation A. Standard Deviation for Ungrouped Data
Standard deviation describes how the spread out of the data are around the mean. It is usually denoted by . Consider a set of ungrouped data x1, x2, …, xn. Standard deviation,    , where is the mean and n is the total number of data. (xi – x) is the deviation of the ith data from the mean. _ Notes: The quantity  2 is called the variance of the data.

Example 21.8T 21.3 Standard Deviation Solution:
A. Standard Deviation for Ungrouped Data Example 21.8T Six students joined the inter-school cross-country race. The times taken (in min) to complete the race are recorded below: 45, 46, 49, 50, 52, y If the mean time is 49.5 min, find (a) the value of y and (b) the standard deviation of the times taken. (Give the answer correct to 3 significant figures.) Solution: (a) (b) Standard deviation (cor. to 3 sig. fig.)

Example 21.9T 21.3 Standard Deviation Solution:
A. Standard Deviation for Ungrouped Data Example 21.9T The following table shows the marks of Eric in five tests of two subjects. Find the standard deviations of the marks of each subject. (Give the answers correct to 3 significant figures.) (b) In which subject is his performance more consistent? Test 1 Test 2 Test 3 Test 4 Test 5 Chinese 65 70 76 68 78 English 72 81 85 90 80 Solution: For Chinese, mean Standard deviation (cor. to 3 sig. fig.)

Example 21.9T 21.3 Standard Deviation Solution:
A. Standard Deviation for Ungrouped Data Example 21.9T The following table shows the marks of Eric in five tests of two subjects. Find the standard deviations of the marks of each subject. (Give the answers correct to 3 significant figures.) (b) In which subject is his performance more consistent? Test 1 Test 2 Test 3 Test 4 Test 5 Chinese 65 70 76 68 78 English 72 81 85 90 80 Solution: For English, mean Standard deviation (cor. to 3 sig. fig.)

Example 21.9T 21.3 Standard Deviation Solution:
A. Standard Deviation for Ungrouped Data Example 21.9T The following table shows the marks of Eric in five tests of two subjects. Find the standard deviations of the marks of each subject. (Give the answers correct to 3 significant figures.) (b) In which subject is his performance more consistent? Test 1 Test 2 Test 3 Test 4 Test 5 Chinese 65 70 76 68 78 English 72 81 85 90 80 Solution: (b) For Chinese, standard deviation  4.88 For English, standard deviation  5.95 The standard deviation of Chinese is smaller than that of English, so Eric’s performance in Chinese is more consistent.

Example 21.10T 21.3 Standard Deviation Solution:
A. Standard Deviation for Ungrouped Data Example 21.10T The numbers of students in five classes are given as: y – 15, y + 6, y + 9, y – 20, y + 15 (a) Find the mean and the standard deviation. (Give the answers correct to 3 significant figures if necessary.) (b) Find the range and the median if the mean is 34. Solution: (a) Mean Standard deviation (b) ∵ y  1  34 ∴ y  35 The five numbers are 20, 41, 44, 15 and 50. (cor. to 3 sig. fig.) ∴ Range and Median

21.3 Standard Deviation B. Standard Deviation for Grouped Data
For a set of grouped data, we have to consider the frequency of each group. Standard deviation,  = = , where fi and xi are the frequency and the class mark of the ith class interval respectively, is the mean and n is the total number of class marks.

Example 21.11T 21.3 Standard Deviation Solution:
B. Standard Deviation for Grouped Data Example 21.11T The following table shows the ages of 50 workers in a company. (a) Find the mean age of the workers. (b) Find the standard deviation of the ages. (Give the answer correct to 3 significant figures.) Age 11 – 20 21 – 30 31 – 40 41 – 50 51 – 60 Number of workers 4 24 12 8 2 Solution: (a) Class mark 15.5 25.5 35.5 45.5 55.5 Frequency 4 24 12 8 2 Mean age (b) Standard deviation (cor. to 3 sig. fig.)

21.3 Standard Deviation C. Finding Standard Deviation by a Calculator
In actual practice, it is quite difficult to calculate the standard deviation if the amount of data is very large. In such circumstances, a calculator can help us to find the standard deviation. In order to use a calculator, we have to set the function mode of the calculator to standard deviation ‘SD’. We also have to clear all the previous data in the ‘SD’ mode. For grouped data, use the class mark to represent the entire group. For both ungrouped and grouped data, we can use a calculator to find the mean and the standard deviation.

21.3 Standard Deviation The following table summarizes the advantages and disadvantages of the three different measures of dispersion. Measure of dispersion Advantage Disadvantage 1. Range Only two data are involved, so it is the easiest one to calculate. Only extreme values are considered which may give a misleading impression. 2. Inter-quartile range It only focuses on the middle 50% of data, thus avoiding the influence of extreme values. Cannot show the dispersion of the whole group of data. 3. Standard deviation It takes all the data into account that can show the dispersion of the whole group of data. Difficult to compute without using a calculator.

21.4 Applications of Standard Deviation
A. Standard Scores Standard score is used to compare data in relation with the mean and the standard deviation . The standard score z of a given value x from a set of data with mean and standard deviation  is defined as: Notes: The standard score may be positive, negative or zero. A positive standard score means the given value is z times the standard deviation above the mean while a negative standard score means the given value is z times the standard deviation below the mean.

Example 21.12T 21.4 Applications of Standard Deviation Solution:
A. Standard Scores Example 21.12T Ryan sat for a mathematics examination which consisted of two papers. The following table shows his marks as well as the means and the standard deviations of the marks for the whole class in these papers. (a) Find his standard scores in the two papers. (Give the answers correct to 3 significant figures.) (b) In which paper did he perform better? Paper I Paper II Marks 66 71 Mean 60.8 62.4 Standard deviation 4.2 6.4 Solution: (a) Paper I: (cor. to 3 sig. fig.) Paper II: (cor. to 3 sig. fig.) (b) Since 1.34  1.24, Ryan performed better in Paper II than in Paper I.

Example 21.13T 21.4 Applications of Standard Deviation Solution:
A. Standard Scores Example 21.13T Given that the standard scores of Doris’s marks in Art and Music are –2.3 and 1.4 respectively, find (a) her mark in Art if the mean and the standard deviation of the marks are 30 and 2 respectively; (b) the mean mark of Music if Doris got 41.5 marks and the standard deviation of the marks is 3.5. Solution: (a) Doris’s mark in Art  (2.3)  (b) The mean mark of Music  41.5  1.4  3.5

21.4 Applications of Standard Deviation
B. Normal Distribution For a large number of the frequency distributions we meet in our daily life, their frequency curves have the shape of a bell: The bell can have different shapes. This bell-shaped frequency curve is called the normal curve and the corresponding frequency distribution is called the normal distribution. For a normal distribution, the mean, median and the mode of the data lie at the centre of the distribution. In a normal distribution, mean  median  mode. Therefore, the normal curve is symmetrical about the mean, i.e., the axis of symmetry for the normal curve is x  .

21.4 Applications of Standard Deviation
B. Normal Distribution In addition, we can tell the percentage of the data lie within a number of standard deviations from the mean: 1. About 68% of the data lie within one standard deviation from the mean, i.e., –  and . 2. About 95% of the data lie within two standard deviations from the mean, i.e., – 2 and . 3. About 99.7% of the data lie within three standard deviations from the mean, i.e., – 3 and .

Example 21.14T 21.4 Applications of Standard Deviation Solution:
B. Normal Distribution Example 21.14T The heights of some soccer players are normally distributed with a mean of 180 cm and a standard deviation of 8 cm. Find the percentage of players (a) whose heights are between 172 cm and 188 cm, (b) whose heights are greater than 188 cm. Solution: Given  180 and s  8. (a) 172  180  8  34% of the players’ heights lie between ( ) cm and cm. 188   34% of the players’ heights lie between cm and ( ) cm. Percentage of players whose heights are between 172 cm and 188 cm  34% + 34% (b) 188   Percentage of players whose heights are greater than 188 cm  50%  34%

Example 21.15T 21.4 Applications of Standard Deviation Solution:
B. Normal Distribution Example 21.15T The weights of 2000 children are normally distributed with a mean of kg and a standard deviation of 6.2 kg. (a) How many children have weights between 50.2 kg and 68.8 kg? (b) How many children are heavier than 50.2 kg? Solution: Given  56.4 and s  6.2. (a) 50.2  56.4  6.2 68.8   6.2 81.5% of children have weights between ( ) kg and ( ) kg.  Number of children  2000  81.5% (b) 50.2  56.4  6.2 Percentage of children who are heavier than 50.2 kg  50% + 34%  84%  Number of children  2000  84%

21.5 Effects on the Dispersion with a Change in Data
A. Removal of the Largest or Smallest item from the Data In junior forms, we learnt that if we remove a datum greater than the mean of the data set, then the mean will decrease. Similarly, if we remove a datum less than the mean of the data set, then the mean will increase. We can deduce that: If the greatest or the least value (assuming the removed datum is unique) in a data set is removed, then 1. the range will decrease; 2. the inter-quartile range may increase, decrease or remain unchanged; 3. the standard deviation may increase or decrease.

21.5 Effects on the Dispersion with a Change in Data
B. Adding a Common Constant to the Whole Set of Data We have the following conclusion: If a constant k is added to each datum in a set of data, then the following measures of dispersion will not change: If a constant k is added to each datum, then the mean, median and the mode will also increase by k. 1. the range, 2. the inter-quartile range and 3. the standard deviation

21.5 Effects on the Dispersion with a Change in Data
C. Multiplying the Whole Set of Data by a Common Constant We have the following conclusion: The range, the inter-quartile range and the standard deviation will be k times the original values if each datum in a set of data is multiplied by a constant k. Notes: If the quartiles are not members of the data set, the conclusion will be the same.

Example 21.16T 21.5 Effects on the Dispersion with a Change in Data
C. Multiplying the Whole Set of Data by a Common Constant Example 21.16T Consider the nine different numbers: 20, 45, 25, 30, 32, 28, 35, 51, 40 (a) Find the range, the inter-quartile range and the standard deviation. (b) Find the new range, the new inter-quartile range and the new standard deviation of the new set of data if (i) 10 is subtracted from each data; (ii) each number is halved; (iii) the datum 45 is removed. (Give the answers correct to 3 significant figures if necessary.) Solution: (a) Range Inter-quartile range Standard deviation (cor. to 3 sig. fig.)

Example 21.16T 21.5 Effects on the Dispersion with a Change in Data
C. Multiplying the Whole Set of Data by a Common Constant Example 21.16T Consider the nine different numbers: 20, 45, 25, 30, 32, 28, 35, 51, 40 (a) Find the range, the inter-quartile range and the standard deviation. (b) Find the new range, the new inter-quartile range and the new standard deviation of the new set of data if (i) 10 is subtracted from each data; (ii) each number is halved; (iii) the datum 45 is removed. (Give the answers correct to 3 significant figures if necessary.) Solution: (b) (i) If 10 is subtracted from each datum, the range, the inter-quartile range and the standard deviation of the new data remain unchanged. Range Inter-quartile range Standard deviation (cor. to 3 sig. fig.)

Example 21.16T 21.5 Effects on the Dispersion with a Change in Data
C. Multiplying the Whole Set of Data by a Common Constant Example 21.16T Consider the nine different numbers: 20, 45, 25, 30, 32, 28, 35, 51, 40 (a) Find the range, the inter-quartile range and the standard deviation. (b) Find the new range, the new inter-quartile range and the new standard deviation of the new set of data if (i) 10 is subtracted from each data; (ii) each number is halved; (iii) the datum 45 is removed. (Give the answers correct to 3 significant figures if necessary.) Solution: (b) (ii) If each datum is halved, the range, the inter-quartile range and the standard deviation of the new data are multiplied by 0.5. Range Inter-quartile range Standard deviation (cor. to 3 sig. fig.)

Example 21.16T 21.5 Effects on the Dispersion with a Change in Data
C. Multiplying the Whole Set of Data by a Common Constant Example 21.16T Consider the nine different numbers: 20, 45, 25, 30, 32, 28, 35, 51, 40 (a) Find the range, the inter-quartile range and the standard deviation. (b) Find the new range, the new inter-quartile range and the new standard deviation of the new set of data if (i) 10 is subtracted from each data; (ii) each number is halved; (iii) the datum 45 is removed. (Give the answers correct to 3 significant figures if necessary.) Solution: (b) (iii) The remaining data are 20, 25, 28, 30, 32, 35, 40, 51. Range Inter-quartile range Standard deviation (cor. to 3 sig. fig.)

21.5 Effects on the Dispersion with a Change in Data
D. Insertion of Zero in the Data Set We have the following conclusion: If a zero value is inserted in a non-negative data set, then 1. the range may increase or remain unchanged; 2. the inter-quartile range may increase, decrease or remain unchanged; 3. the standard deviation may increase or decrease.

Example 21.17T 21.5 Effects on the Dispersion with a Change in Data
D. Insertion of Zero in the Data Set Example 21.17T The daily income (\$) of a hawker during the last two weeks was: 200, 220, 230, 240, 250, 320, 340, 360, 380, 400, 450, 580, 650 (a) Find the inter-quartile range and the standard deviation. (b) There was a thunderstorm last Monday and the income on that day was zero. If the income from last Monday is also considered, find the new inter-quartile range and the new standard deviation. (Give the answers correct to 1 decimal place if necessary.) Solution: (a) (b) Inter-quartile range Inter-quartile range The standard deviation The standard deviation (cor. to 1 d. p.) (cor. to 1 d. p.)

Chapter Summary 21.1 Range and Inter-quartile Range
1. The range is the difference between the largest value (highest class boundary) and the smallest value (lowest class boundary) in a set of ungrouped (grouped) data. 2. The inter-quartile range is the difference between the upper quartile Q3 and the lower quartile Q1 of a set of data.

Chapter Summary 21.2 Box-and-whisker Diagrams
A box-and-whisker diagram illustrates the spread of a set of data. It shows the greatest value, the least value, the median, the lower quartile and the upper quartile of the data.

Chapter Summary 21.3 Standard Deviation
Standard deviation  is the measure of dispersion that describes how spread out a set of data is around the mean value. 1. For ungrouped data: 2. For grouped data: Larger values for the range, the inter-quartile range and the standard deviation of the data indicate a larger dispersion and vice versa.

Chapter Summary 21.4 Applications of Standard Deviation
1. The standard score z is the number of standard deviations that a given value is above or below the mean, and is given by 2. The curve of a normal distribution is bell-shaped and is called the normal curve. In the normal distribution, different percentages of data lie within different standard deviations from the mean.

Chapter Summary 21.5 Effects on the Dispersion with a Change in Data
1. If the greatest or the least value (assuming both are unique) in a data set is removed, then the range will decrease. However, the inter- quartile range may increase, decrease or remain unchanged and the standard deviation may increase or decrease. 2. If a constant k is added to each datum in a set of data, then the range, the inter-quartile range and the standard deviation will not change. 3. If each item in the data is multiplied by a positive constant k, then the range, the inter-quartile range and the standard deviation will be k times their original values. 4. If a zero value is inserted in a non-negative data set, then the range may increase or remain unchanged, the inter-quartile range may increase, decrease or remain unchanged and the standard deviation may increase or decrease.

Follow-up 21.1 21.1 Range and Inter-quartile Range Solution: A. Range
The results (in m) of the best eight boys in the long jump are: 5.2, 5.6, 4.8, 4.2, 5.3, 4.5, 5.0, 4.8 Find the range of the results. Solution: Range  (5.6  4.2) m

Follow-up 21.2 21.1 Range and Inter-quartile Range Solution: A. Range
The following table shows the heights of the students in S6B. (a) What is the upper class boundary of the class 171 cm – 175 cm? (b) What is the lower class boundary of the class 151 cm – 155 cm? (c) Hence find the range of the heights. Height (cm) 151 – 155 156 – 160 161 – 165 166 – 170 171 – 175 Frequency 1 3 6 2 4 Solution: (a) cm (b) cm (c) Range  (175.5  150.5) cm

Follow-up 21.3 21.1 Range and Inter-quartile Range Solution:
B. Inter-quartile Range Follow-up 21.3 The air pollution indices of a city are recorded at noon every day. The following are the indices in the last 15 days. (a) Arrange the data in ascending order. (b) Find the median. (c) Find the inter-quartile range. Solution: (a) Arrange the marks in ascending order: 56, 56, 58, 58, 59, 60, 62, 63, 63, 64, 65, 67, 69, 69, 70 (b) Median (c) Q3  67 and Q1  58 Inter-quartile range

Follow-up 21.4 21.1 Range and Inter-quartile Range Solution:
B. Inter-quartile Range Follow-up 21.4 The figure shows the cumulative frequency polygon of the heights (in cm) of 100 trees. (a) Find the lower quartile and the upper quartile of the heights. (b) Find the inter-quartile range. Solution: (a) From the graph, (b) Inter-quartile range

Follow-up 21.5 21.1 Range and Inter-quartile Range Solution:
B. Inter-quartile Range Follow-up 21.5 The following table shows the distribution of members’ ages in two sports clubs. (a) How many members are there in each of the clubs? Age 12 13 14 15 16 17 Badminton Club 10 8 6 2 Basketball Club Solution: (a) For Badminton Club, For Basketball Club, number of members number of members

Follow-up 21.5 21.1 Range and Inter-quartile Range Solution:
B. Inter-quartile Range Follow-up 21.5 The following table shows the distribution of members’ ages in two sports clubs. (b) Find the range and the inter-quartile range of the distribution of ages of the members in each club. (c) Which club has a larger dispersion of ages? Age 12 13 14 15 16 17 Badminton Club 10 8 6 2 Basketball Club Solution: (b) For Badminton Club, range  17 – 12 inter-quartile range  15 – 12 For Basketball Club, range  16 – 13 inter-quartile range  16 – 14 (c) Since the range and the inter-quartile range of the ages of the members in the Badminton Club are larger, Badminton Club has a larger dispersion of ages.

Follow-up 21.6 21.2 Box-and-whisker Diagrams Solution:
The following box-and-whisker diagram shows the lengths (in cm) of the hands of a class of students. (a) Find the median length of the hands. (b) Find the range of the lengths of the hands. (c) Find the inter-quartile range of the lengths of the hands. Solution: (a) Median (b) Maximum length  19.4 cm and minimum length  17.3 cm Range  (19.4  17.3) cm (c) Inter-quartile range  (19.0  18.2) cm

Follow-up 21.7 21.2 Box-and-whisker Diagrams Solution:
Kelvin and Johnny are comparing their results on 10 mathematics tests. The following are their marks. Kelvin’s marks: 54, 70, 67, 92, 75, 80, 84, 78, 66, 82 Johnny’s marks: 70, 74, 76, 78, 78, 79, 80, 78, 72, 75 (a) Find the median, the lower quartile and the upper quartile of the marks for each student. Solution: (a) Arrange the marks in ascending order: Kelvin’s marks: 54, 66, 67, 70, 75, 78, 80, 82, 84, 92 Median Johnny’s marks: 70, 72, 74, 75, 76, 78, 78, 78, 79, 80 Median

Follow-up 21.7 21.2 Box-and-whisker Diagrams Solution:
Kelvin and Johnny are comparing their results on 10 mathematics tests. The following are their marks. Kelvin’s marks: 54, 70, 67, 92, 75, 80, 84, 78, 66, 82 Johnny’s marks: 70, 74, 76, 78, 78, 79, 80, 78, 72, 75 (b) Draw box-and-whisker diagrams on the same graph to compare the results. (c) Which student performs more consistently in the tests? Solution: (b) For Kelvin, minimum  54 marks maximum  92 marks For Johnny, minimum  70 marks maximum  80 marks Refer to the figure on the right. (c) Johnny performs more consistently.

Follow-up 21.8 21.3 Standard Deviation Solution:
A. Standard Deviation for Ungrouped Data Follow-up 21.8 Emily got the following marks in five English tests. 59, 65, 76, 68, 82 (a) Find the mean mark. (b) Find the standard deviation of the marks. (Give the answer correct to 3 significant figures.) Solution: (a) Mean (b)  330 Standard deviation (cor. to 3 sig. fig.)

Follow-up 21.9 21.3 Standard Deviation Solution:
A. Standard Deviation for Ungrouped Data Follow-up 21.9 The following table shows the marks of two students in five tests. (a) Find the standard deviations of the marks of the two students. (Give the answers correct to 3 significant figures.) (b) Which student has a more consistent performance? Test 1 Test 2 Test 3 Test 4 Test 5 Winnie 80 65 82 66 78 Cherry 76 74 72 Solution: (a) For Winnie, For Cherry, standard deviation standard deviation (cor. to 3 sig. fig.) (cor. to 3 sig. fig.) (b) Since the standard deviation of Cherry’s marks is smaller than Winnie’s, Cherry has a more consistent performance.

Follow-up 21.10 21.3 Standard Deviation Solution:
A. Standard Deviation for Ungrouped Data Follow-up 21.10 The following are the ages of five people. 66 + x , x, 71 – x , 68 + x , 70 + x (a) Find the mean age in terms of x. (b) If x  4, find the standard deviation. (Give the answer correct to 3 significant figures.) Solution: (a) Mean (b) ∵ x  4  Mean   72 and the five numbers are 70, 77, 67, 72 and 74.  Standard deviation (cor. to 3 sig. fig.)

Follow-up 21.11 21.3 Standard Deviation Solution:
B. Standard Deviation for Grouped Data Follow-up 21.11 The following table shows the air pollution index recorded daily in a particular district at 5:00 p.m. in June. (a) Find the value of y. (b) Find the mean. (c) Find the standard deviation. Index 51 – 55 56 – 60 61 – 65 66 – 70 Number of days 8 10 7 y Solution: (a) y  30 807.5 30 Total 361.25 72.25 5 68 85.75 12.25 7 63 22.5 2.25 10 58 338 42.25 8 53 f x (b) Mean (c) Standard deviation (cor. to 3 sig. fig.)

Follow-up 21.12 21.4 Applications of Standard Deviation Solution:
A. Standard Scores Follow-up 21.12 The following table shows the means and the standard deviations of the marks for the whole class in two subjects, as well as Helen’s marks. (a) Find the standard scores of Helen’s marks in these subjects. (Give the answers correct to 3 significant figures.) (b) In which subject did she perform better? Subject Chinese History Helen’s mark 85 70 Mean 79 62 Standard deviation 8.2 4.3 Solution: (a) Chinese: History: (cor. to 3 sig. fig.) (cor. to 3 sig. fig.) (b) Since 1.86  0.732, Helen performed better in History than in Chinese.

Follow-up 21.13 21.4 Applications of Standard Deviation Solution:
A. Standard Scores Follow-up 21.13 Given that the standard scores of Kelvin’s marks in the Chinese reading and written tests are 1 and –1 respectively, find (a) his mark in the reading test if the mean and the standard deviation of the marks for his whole class are 6.9 and 1.1 respectively; (b) the mean mark of the written test if his mark in the test and the standard deviation of the marks are 5.5 and 2 respectively. Solution: (a) Kelvin’s mark in the reading test (b) The mean mark of the written test

Follow-up 21.14 21.4 Applications of Standard Deviation Solution:
B. Normal Distribution Follow-up 21.14 The foot sizes of a group of children in a child care centre are normally distributed with a mean of 20 cm and a standard deviation of 2.5 cm. (a) Find the percentage of children having foot sizes between 15 cm and 22.5 cm. (b) Find the percentage of children having foot sizes less than 12.5 cm. Solution: (a) Given  20 and s  2.5. 15  20  2  2.5  47.5% of the children’s foot sizes lie between ( ) cm and cm. 22.5   34% of the children’s foot sizes lie between cm and ( ) cm. Percentage of children having foot sizes between 15 cm and 22.5 cm  47.5% + 34% (b) 12.5  20  3  2.5 Percentage of children having foot sizes less than 12.5 cm  50% – 49.85%

Follow-up 21.15 21.4 Applications of Standard Deviation Solution:
B. Normal Distribution Follow-up 21.15 The lifetimes of a pack of 1500 bulbs is normally distributed with a mean of 1200 hours and a standard deviation of 50 hours. (a) How many bulbs have a lifetime between 1100 hours and 1300 hours? (b) How many bulbs have a lifetime less than 1150 hours? Solution: Since  1200 and s  50, (a) 1100  1200  2  50 1300   50 95% of bulbs have a lifetime between ( ) hours and ( ) hours.  Number of bulbs  1500  95% (b) 1150  1200  50 Percentage of bulbs having a lifetime less than 1150 hours  50% – 34%  16%  Number of bulbs  1500  16%

Follow-up 21.16 21.5 Effects on the Dispersion with a Change in Data
C. Multiplying the Whole Set of Data by a Common Constant Follow-up 21.16 Consider the following ten numbers: (a) Find the range, the inter-quartile range and the standard deviation. (b) Find the change in the range, the inter-quartile range and the standard deviation after each of the following changes: (i) 1 is added to each datum (ii) each datum is multiplied by 5 (c) Find the range, the inter-quartile range and the standard deviation of the positive numbers. (Give the answers correct to 3 significant figures if necessary.) Solution: (a) Range Inter-quartile range Standard deviation (cor. to 3 sig. fig.)

Follow-up 21.16 21.5 Effects on the Dispersion with a Change in Data
C. Multiplying the Whole Set of Data by a Common Constant Follow-up 21.16 Consider the following ten numbers: (a) Find the range, the inter-quartile range and the standard deviation. (b) Find the change in the range, the inter-quartile range and the standard deviation after each of the following changes: (i) 1 is added to each datum (ii) each datum is multiplied by 5 (c) Find the range, the inter-quartile range and the standard deviation of the positive numbers. (Give the answers correct to 3 significant figures if necessary.) Solution: (b) (i) If 1 is added to each datum, the range, the inter-quartile range and the standard deviation of the new data remain unchanged. (ii) If each datum is multiplied by 5, the range, the inter-quartile range and the standard deviation of the new data are 5 times the original values.

Follow-up 21.16 21.5 Effects on the Dispersion with a Change in Data
C. Multiplying the Whole Set of Data by a Common Constant Follow-up 21.16 Consider the following ten numbers: (a) Find the range, the inter-quartile range and the standard deviation. (b) Find the change in the range, the inter-quartile range and the standard deviation after each of the following changes: (i) 1 is added to each datum (ii) each datum is multiplied by 5 (c) Find the range, the inter-quartile range and the standard deviation of the positive numbers. (Give the answers correct to 3 significant figures if necessary.) Solution: (c) Range Inter-quartile range Standard deviation (cor. to 3 sig. fig.)

Follow-up 21.17 21.5 Effects on the Dispersion with a Change in Data
D. Insertion of Zero in the Data Set Follow-up 21.17 The figure shows the stem-and-leaf diagram of the marks of the girls who took the same English test. (a) Find the inter-quartile range and the standard deviation of the marks. (b) Fiona was also absent, so she got a zero mark. If Fiona’s mark is also considered, find the new inter-quartile range and the new standard deviation of the marks. (Give the answers correct to 3 significant figures if necessary.) Stem (Tens digit) Leaf (Units digit) 1 2 2 4 3 1 1 Solution: (a) The inter-quartile range The standard deviation (cor. to 3 sig. fig.) (b) The total number of girls is 14. The new inter-quartile range The new standard deviation (cor. to 3 sig. fig.)