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© Boardworks Ltd 2005 1 of 49 D2 Averages and range KS4 Mathematics

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© Boardworks Ltd 2005 2 of 49 A A A A A D2.1 The mode D2 Averages and range Contents D2.5 Comparing data D2.2 The mean D2.3 Calculating the mean from frequency tables D2.4 The median

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© Boardworks Ltd 2005 3 of 49 The three averages and range There are three different types of average: M EDIAN middle value The range is not an average, but tells you how the data is spread out: R ANGE largest value – smallest value M ODE most common M EAN sum of values number of values

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© Boardworks Ltd 2005 4 of 49 This graph shows pupils’ favourite athletics events. Favourite athletics event 0 5 10 15 20 Sprint Long distance running Hurdles High jump Long jump Triple jump Shot Discus Javelin Frequency Which is the most popular event? How do you know?

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© Boardworks Ltd 2005 5 of 49 The mode is the item that occurs the most often in a data set. The mode The most common item is called the mode. In the graph the mode is sprint because it is represented by the highest bar. We could also say “The modal athletic event is sprint.” Is it possible to have more than one modal value? Is it possible to have no modal value? Yes

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© Boardworks Ltd 2005 6 of 49 The mode We could write out all the results in a list. The list would begin: How many words (items) would there be in the list altogether? How could we work out the mode from the list if we didn’t have the graph? Can we tell how many pupils took part in the survey?

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© Boardworks Ltd 2005 7 of 49 14151513121415011 13141116141591012 The mode Here are the attendance figures at a weekly school athletics club for Year 11. What is the modal number of pupils attending? Are there any unusual results in the data set? This result is called an outlier. Can you think of any possible reasons for the outlier? If the data set were very long, what would be the best way to find the mode? Discuss : Over how many weeks were the results collected?

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© Boardworks Ltd 2005 8 of 49 0 2 4 6 8 10 12 14 16 18 20 Sprint Long distance running Hurdles High jump Long jump Triple jump Shot Discus Javelin Frequency Compare this graph to the previous one. Favourite athletics event What conclusions can you draw? Which two groups of pupils could be represented by the two graphs?

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© Boardworks Ltd 2005 9 of 49 How many sports do you play? How many pupils play more than two sports? A group of pupils were asked how many sports they played. This graph shows the results. What is the modal number of sports played? How many pupils took part in the survey? 0 2 4 6 8 10 12 14 0123456 Numbers of sports played Frequency

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© Boardworks Ltd 2005 10 of 49 This graph represents Year Ten girls’ times for a 100m sprint race. Grouped data 0 2 4 6 8 10 Frequency Times in seconds 121314151617181920 What is the modal time interval? How many girls are in this interval?

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© Boardworks Ltd 2005 11 of 49 When the mode is not appropriate 94 35 103 6 2 1 0 Numbers of sports played 2 15 17 20 Frequency Another survey is carried out among university students. The results are represented in this table: A newspaper reporter writes: “You may be surprised to learn that the average number of sports played by university students is 0.” Should the reporter say which average has been used? Why is the mode a misleading average in this example? Do you think this is a fair representation of the data?

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© Boardworks Ltd 2005 12 of 49 Data that is heavily weighted towards one end of the data set is said to be skewed. When data is skewed, the mode is not an appropriate average. Data that is heavily weighted towards one end of the data set is said to be skewed. When data is skewed, the mode is not an appropriate average. Skewed data 0 5 10 15 20 25 1234567 Numbers of sports played Frequency Negatively skewed data 0 2 4 6 8 10 12 14 1234567 Numbers of sports played Frequency Positively skewed data

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© Boardworks Ltd 2005 13 of 49 Contents A A A A A D2.2 The mean D2.5 Comparing data D2.1 The mode D2 Averages and range D2.3 Calculating the mean from frequency tables D2.4 The median

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© Boardworks Ltd 2005 14 of 49 St Clement Danes School holds an inter-form athletics competition for Year 10. Each class must select their five best boys and five best girls for each event. Comparing data Here are the times in seconds for 100 metres sprint for the two best classes. Which class should win and why? 13.116.514.315.4 13.116.413.815.4 12.815.913.415.3 12.215.212.914.7 12.014.911.512.8 10C boys10C girls10B boys10B girls

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© Boardworks Ltd 2005 15 of 49 The mean The mean is the most commonly used average. To calculate the mean of a set of values we add together the values and divide by the total number of values. Mean = Sum of values Number of values For example, the mean time for Class 10B girls is: 12.8 + 14.7 + 15.3 + 15.4 + 15.4 5 = 14.72 73.6 5 =

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© Boardworks Ltd 2005 16 of 49 The mean Calculate the mean times for the other three groups. mean time 10C boys10C girls10B boys10B girls 14.7213.1815.7812.64 Now calculate means for Class 10B and Class 10C (with girls and boys combined). mean time Class 10CClass 10B 13.9514.21 Based on these results, who should win?

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© Boardworks Ltd 2005 17 of 49 Calculating the mean

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© Boardworks Ltd 2005 18 of 49 Calculating a missing data item

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© Boardworks Ltd 2005 19 of 49 The school athletics team take part in an inter-schools competition. James’s shot results (in metres) are below. Outliers and their effect on the mean 9.469.258.7710.2510.359.594.02 A data item that is significantly higher or lower than the other items is called an outlier. Outliers affect the mean, by reducing or increasing it. Discuss: What is the mean throw? Is this a fair representation of James’s ability? Explain. What would be a fair way for the competition to operate?

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© Boardworks Ltd 2005 20 of 49 Here are some 1500 metre race results in minutes. Outliers and their effect on the mean It may be appropriate in research or experiments to remove an outlier before carrying out analysis of results. Discuss: 6.266.286.306.395.384.5410.596.357.01 Are there any outliers? Will the mean be increased or reduced by the outlier? Calculate the mean with the outlier. Now calculate the mean without the outlier. How much does it change?

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© Boardworks Ltd 2005 21 of 49 Contents A A A A A D2.3 Calculating the mean from frequency tables D2.5 Comparing data D2.1 The mode D2 Averages and range D2.2 The mean D2.4 The median

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© Boardworks Ltd 2005 22 of 49 Calculating the mean from a frequency table Numbers of sports played Frequency 020 117 215 310 49 53 62 Here are the results of a survey carried out among university students. If you were to write out the whole list of results, what would it look like? What do you think the mean will be?

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© Boardworks Ltd 2005 23 of 49 Calculating the mean from a frequency table 26 3 9 10 15 17 20 FrequencyNumber of sports × frequency 4 5 3 2 1 0 Numbers of sports played T OTAL 0 × 20= 0 1 × 17= 17 2 × 15= 30 3 × 10= 30 4 × 9= 36 5 × 3= 15 6 × 2= 12 Mean = 140 ÷ 76 = 14076 2 sports (to the nearest whole)

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© Boardworks Ltd 2005 24 of 49 Because the data is grouped, we do not know individual scores. It is not possible to add up the scores. Grouped data Javelin distances in metres Frequency 5 ≤ d < 101 10 ≤ d < 158 15 ≤ d < 2012 20 ≤ d < 2510 25 ≤ d < 303 30 ≤ d < 351 35 ≤ d < 401 36 Here are the Year Ten boys’ javelin scores. How could you calculate the mean from this data? How is the data different from the previous examples you have calculated with?

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© Boardworks Ltd 2005 25 of 49 It is possible to find an estimate for the mean. This is done by finding the midpoint of each group. To find the midpoint of the group 10 ≤ d < 15: 10 + 15 = 25 25 ÷ 2 = Midpoints 12.5 m Javelin distances in metres Frequency 5 ≤ d < 101 10 ≤ d < 158 15 ≤ d < 2012 20 ≤ d < 2510 25 ≤ d < 303 30 ≤ d < 351 35 ≤ d < 401 Find the midpoints of the other groups.

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© Boardworks Ltd 2005 26 of 49 135 ≤ d < 40 1 3 10 12 8 1 FrequencyMidpoint 30 ≤ d < 35 Frequency × midpoint 25 ≤ d < 30 20 ≤ d < 25 15 ≤ d < 20 10 ≤ d < 15 5 ≤ d < 10 Javelin distances in metres Estimating the mean from grouped data 1 × 7.5= 7.5 8 × 12.5= 100 12 × 17.5= 210 10 × 22.5= 225 3 × 27.5= 82.5 1 × 32.5 Estimated mean = 695 ÷ 36 1 × 37.5 = 32.5 7.5 12.5 17.5 22.5 27.5 32.5 37.5= 37.5 36695T OTAL = 19.3 m (to 1 d.p.)

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© Boardworks Ltd 2005 27 of 49 How accurate is the estimated mean? 35.0031.0528.8925.6025.3324.1123.5021.8221.78 21.7721.6021.0020.7020.2020.0019.50 18.82 17.3517.3116.6415.7915.7515.6915.5215.2515.00 14.5012.8012.5012.00 11.8510.009.50 Here are the javelin distances thrown by Year 10 before the data was grouped. Work out the mean from the original data above and compare it with the estimated mean found from the grouped data. How accurate was the estimated mean? The estimated mean is 19.3 metres (to 1 d.p.). The actual mean is18.7 metres (to 1 d.p.).

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© Boardworks Ltd 2005 28 of 49 A A A A A Contents D2.4 The median D2.5 Comparing data D2.1 The mode D2 Averages and range D2.2 The mean D2.3 Calculating the mean from frequency tables

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© Boardworks Ltd 2005 29 of 49 6.266.286.306.395.384.5410.59 6.35 7.01 The median is the middle number when all numbers are in order. The median Calculate the median of the 1500 m results. Why is this a more appropriate average than the mean for these results? 4.545.386.266.286.30 6.35 6.397.01 10.59 Write the results in order and find the middle value:

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© Boardworks Ltd 2005 30 of 49 Choosing the most appropriate average What are the mean and median for these sets of attendance figures for three lunchtime activities? Explain your answers. To decide which of the three activities is the most popular, which average is a better one to use? Why? 23222120191817Choir 20 Drama club 292825201918 Orchestra

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© Boardworks Ltd 2005 31 of 49 Outliers and the median and mean

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© Boardworks Ltd 2005 32 of 49 If there are two middle numbers, you need to find what is halfway between them. 2.15 2.21 2.40 2.55 2.80 3.32 3.46 3.63 3.83 4.74 When there are two middle numbers Here are 10B girls’ long jump results in metres. How could you work out the median jump? 2.80 m + 3.32 m = 6.12 m 6.12 m ÷ 2 = 3.06 m If the numbers are far apart, a quick way to find the middle of those two numbers is to add them up and divide by two.

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© Boardworks Ltd 2005 33 of 49 Finding halfway between two numbers

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© Boardworks Ltd 2005 34 of 49 One or two middle numbers? 12346678911 If there are 9 numbers in a list, will there be 1 or 2 middle numbers? 2346678911 If there are 10 numbers in a list, will there be 1 or 2 middle numbers? If there is an even number of numbers in a list, there will be two middle numbers. If there is an odd number of numbers in a list, there will be one middle number.

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© Boardworks Ltd 2005 35 of 49 To find out where a middle number in a very long list, call the number of numbers n. Then the middle number is then When there are two middle numbers ( n + 1) ÷ 2 101 ÷ 2 = 50.5 th number in the list (halfway between the 50 th and the 51 st ). 38 ÷ 2 = 19 th number in the list. For example, There are 100 numbers in a list. Where is the median? There are 37 numbers in a list. Where is the median?

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© Boardworks Ltd 2005 36 of 49 Where is the median?

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© Boardworks Ltd 2005 37 of 49 Contents A A A A A D2.5 Comparing data D2.1 The mode D2 Averages and range D2.2 The mean D2.3 Calculating the mean from frequency tables D2.4 The median

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© Boardworks Ltd 2005 38 of 49 The range Here are the high jump scores for two girls in metres. Joanna1.621.411.351.201.15 Kirsty1.591.451.411.30 Find the range for each girl’s results and use this to find out who is consistently better. Joanna’s range = 1.62 – 1.15 = 0.47 Kirsty’s range = 1.59 – 1.30 = 0.29

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© Boardworks Ltd 2005 39 of 49 The range The highest and lowest scores can be useful in deciding who is more consistent. If the scores are close together then the range will be lower and the scores more consistent. The lowest score subtracted from the highest score is called the range. Remember that the range is not an average, but a measure of spread. If the scores are spread out then the range will be higher and the scores less consistent.

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© Boardworks Ltd 2005 40 of 49 The range JoannaKirsty Mean Range 1.35 m 0.47 m 1.41 m 0.29 m Joanna1.621.411.351.201.15 Kirsty1.591.451.411.30 Calculate the mean and the range for each girl. Use these results to decide which one you would enter into the athletics competition and why.

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© Boardworks Ltd 2005 41 of 49 Calculating the mean, median and range

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© Boardworks Ltd 2005 42 of 49 Comparing sets of data ChrisRob Mean24.8 seconds25.0 seconds Range1.4 seconds0.9 seconds Here is a summary of Chris and Rob’s performance in the 200 metres over a season. They each ran 10 races. Which of these conclusions are correct? Robert is more reliable. Robert is better because his mean is higher. Chris is better because his range is higher. Chris must have run a better time for his quickest race. On average, Chris is faster but he is less consistent.

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© Boardworks Ltd 2005 43 of 49 Comparing sets of data ChrisRob Mean24.8 seconds25.0 seconds Range1.4 seconds0.9 seconds 24.424.524.424.524.624.925.0 25.125.8 24.324.925.0 25.1 25.2 Here is the original data for Chris and Rob. Use the summary table above to decide which data set is Chris’s and which is Rob’s? Who has the best time? Who has the worst time?

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© Boardworks Ltd 2005 44 of 49 Year 9 12.1 14.0 15.3 15.4 15.6 15.7 16.1 16.7 17.0 Comparing hurdles scores Year 9Year 10 Mean Range Year 10 12.3 13.7 15.5 15.6 15.9 16.0 16.1 17.1 22.9 15.416.1 4.910.6 Here are the top eleven hurdles scores in seconds for Year 9 and Year 10. Work out the mean and range. Which year group do you think is better and why? Why might Year 10 feel the comparison is unfair?

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© Boardworks Ltd 2005 45 of 49 Finding the interquartile range When there are outliers in the data, it is more appropriate to calculate the interquartile range. The time of 22.9 seconds is an outlier. The interquartile range is the range of the middle half of the data. The lower quartile is the data value that is quarter of the way along the list. The upper quartile is the data value that is three quarters of the way along the list. interquartile range = upper quartile – lower quartile

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© Boardworks Ltd 2005 46 of 49 Locating the upper and lower quartiles There are 11 times in each list. 16.1 – 15.3 = 0.8 16.1 – 15.5 = 0.6 Year 9 12.1 14.0 15.3 15.4 15.6 15.7 16.1 16.7 17.0 Year 10 12.3 13.7 15.5 15.6 15.9 16.0 16.1 17.1 22.9 Interquartile range for Year 9: Interquartile range for Year 10: Where is the median in each list? Where is the lower quartile in each list? Where is the upper quartile in each list?

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© Boardworks Ltd 2005 47 of 49 The location of quartiles in an ordered data set When there are n values in an ordered data set: The lower quartile = n + 1 4 th valueThe median = n + 1 2 th valueThe upper quartile = 3( n + 1) 4 th value The interquartile range = the upper quartile – the lower quartile

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© Boardworks Ltd 2005 48 of 49 Finding the interquartile range

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© Boardworks Ltd 2005 49 of 49 Review To review the work you have covered in this topic: 2) Make up challenges involving sets of data for your partner, such as working out the mean. 3) Make a list of possible mistakes to avoid in this topic. Write out the key words on cards. Shuffle the cards. Describe the word on each card to your partner. Your partner must guess the word. Do as many as you can in one minute, then swap over. 1) Play “Guess the word”.

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