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Averages and range

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This document provides information about different types of averages (mean, median, mode) and the range, and how to calculate them from raw data and frequency tables. It discusses when each average is most appropriate to use and how outliers can affect calculations. Examples are provided of calculating averages and range from sets of data, as well as estimating the mean from grouped data. The median is defined as the middle number when values are in order. Outliers are identified as values significantly higher or lower than others.

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D2 Averages and range KS4 Mathematics
D2.1 The mode Contents A A A A A D2 Averages and range D2.5 Comparing data D2.2 The mean D2.3 Calculating the mean from frequency tables D2.4 The median
The three averages and range There are three different types of average : The range is not an average, but tells you how the data is spread out: M EDIAN middle value R ANGE largest value – smallest value M ODE most common M EAN sum of values number of values
Favourite athletics event This graph shows pupils’ favourite athletics events. Which is the most popular event? How do you know? 0 5 10 15 20 Sprint Long distance running Hurdles High jump Long jump Triple jump Shot Discus Javelin Frequency
The mode The mode is the item that occurs the most often in a data set. 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 Yes
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?
The mode 14 15 15 13 12 14 15 0 11 13 14 11 16 14 15 9 10 12 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?
Favourite athletics event Compare this graph to the previous one. What conclusions can you draw? Which two groups of pupils could be represented by the two graphs? 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
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 0 1 2 3 4 5 6 Numbers of sports played Frequency
Grouped data This graph represents Year Ten girls’ times for a 100m sprint race. What is the modal time interval? How many girls are in this interval? 0 2 4 6 8 10 Frequency Times in seconds 12 13 14 15 16 17 18 19 20
When the mode is not appropriate 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? 9 4 3 5 10 3 6 2 1 0 Numbers of sports played 2 15 17 20 Frequency
Skewed data 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. 0 5 10 15 20 25 1 2 3 4 5 6 7 Numbers of sports played Frequency Negatively skewed data 0 2 4 6 8 10 12 14 1 2 3 4 5 6 7 Numbers of sports played Frequency Positively skewed data
Contents D2.2 The mean A A A A A D2.5 Comparing data D2.1 The mode D2 Averages and range D2.3 Calculating the mean from frequency tables D2.4 The median
Comparing data 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. Here are the times in seconds for 100 metres sprint for the two best classes. Which class should win and why? 13.1 16.5 14.3 15.4 13.1 16.4 13.8 15.4 12.8 15.9 13.4 15.3 12.2 15.2 12.9 14.7 12.0 14.9 11.5 12.8 10C boys 10C girls 10B boys 10B girls
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. For example, the mean time for Class 10B girls is: 14.72 Mean = Sum of values Number of values 12.8 + 14.7 + 15.3 + 15.4 + 15.4 5 = 73.6 5 =
The mean Calculate the mean times for the other three groups. 14.72 13.18 15.78 12.64 Now calculate means for Class 10 B and Class 10 C (with girls and boys combined). 13.95 14.21 Based on these results, who should win? mean time 10C boys 10C girls 10B boys 10B girls mean time Class 10C Class 10B
Calculating the mean
Calculating a missing data item
Outliers and their effect on the mean The school athletics team take part in an inter-schools competition. James’s shot results (in metres) are below. 9.46 9.25 8.77 10.25 10.35 9.59 4.02 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?
Outliers and their effect on the mean Here are some 1500 metre race results in minutes. It may be appropriate in research or experiments to remove an outlier before carrying out analysis of results. Discuss: 6.26 6.28 6.30 6.39 5.38 4.54 10.59 6.35 7.01 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?
Contents D2.3 Calculating the mean from frequency tables A A A A A D2.5 Comparing data D2.1 The mode D2 Averages and range D2.2 The mean D2.4 The median
Calculating the mean from a frequency table 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? 9 4 3 5 10 3 6 2 1 0 Numbers of sports played 2 15 17 20 Frequency
Calculating the mean from a frequency table 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 = 140 76 2 sports (to the nearest whole) 2 6 3 9 10 15 17 20 Frequency Number of sports × frequency 4 5 3 2 1 0 Numbers of sports played
Grouped data Because the data is grouped, we do not know individual scores. It is not possible to add up the scores. 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? 1 35 ≤ d < 40 36 1 3 10 12 8 1 Frequency 30 ≤ d < 35 25 ≤ d < 30 20 ≤ d < 25 15 ≤ d < 20 10 ≤ d < 15 5 ≤ d < 10 Javelin distances in metres
Midpoints 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 = 12.5 m Find the midpoints of the other groups. 1 35 ≤ d < 40 1 3 10 12 8 1 Frequency 30 ≤ d < 35 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 36 695 T OTAL = 19.3 m (to 1 d.p.) 1 35 ≤ d < 40 1 3 10 12 8 1 Frequency Midpoint 30 ≤ d < 35 Frequency × midpoint 25 ≤ d < 30 20 ≤ d < 25 15 ≤ d < 20 10 ≤ d < 15 5 ≤ d < 10 Javelin distances in metres
How accurate is the estimated mean? 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 is 18.7 metres (to 1 d.p.). 9.50 10.00 11.85 12.00 12.00 12.00 12.50 12.80 14.50 15.00 15.25 15.52 15.69 15.75 15.79 16.64 17.31 17.35 18.82 19.50 19.50 20.00 20.20 20.70 21.00 21.60 21.77 21.78 21.82 23.50 24.11 25.33 25.60 28.89 31.05 35.00
Contents D2.4 The median 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
The median 6.26 6.28 6.30 6.39 5.38 4.54 10.59 6.35 7.01 The median is the middle number when all numbers are in order. Calculate the median of the 1500 m results. Why is this a more appropriate average than the mean for these results? 4.54 5.38 6.26 6.28 6.30 6.35 6.39 7.01 10.59 Write the results in order and find the middle value :
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? 23 22 21 20 19 18 17 Choir 20 20 20 20 20 20 20 Drama club 29 28 25 20 19 18 18 Orchestra
Outliers and the median and mean
When there are two middle numbers 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 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.
Finding halfway between two numbers
One or two middle numbers? 1 2 3 4 6 6 7 8 9 11 If there are 9 numbers in a list, will there be 1 or 2 middle numbers? 2 3 4 6 6 7 8 9 11 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.
When there are two middle numbers To find out where a middle number in a very long list, call the number of numbers n . Then the middle number is then ( 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?
Where is the median?
Contents D2.5 Comparing data A A A A A D2.1 The mode D2 Averages and range D2.2 The mean D2.3 Calculating the mean from frequency tables D2.4 The median
The range Here are the high jump scores for two girls in metres. 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 1.30 1.30 1.41 1.45 1.59 Kirsty 1.15 1.20 1.35 1.41 1.62 Joanna
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.
The range 1.35 m 0.47 m 1.41 m 0.29 m 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. Range Mean Kirsty Joanna 1.30 1.30 1.41 1.45 1.59 Kirsty 1.15 1.20 1.35 1.41 1.62 Joanna
Calculating the mean, median and range
Comparing sets of data 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. Range Mean 0.9 seconds 25.0 seconds Rob 1.4 seconds 24.8 seconds Chris
Comparing sets of data 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? Range Mean 0.9 seconds 25.0 seconds Rob 1.4 seconds 24.8 seconds Chris 24.4 24.5 24.4 24.5 24.6 24.9 25.0 25.0 25.1 25.8 24.3 24.9 25.0 25.0 25.0 25.1 25.1 25.1 25.2 25.2 Here is the original data for Chris and Rob.
Comparing hurdles scores Year 9 12.1 14 .0 15.3 15.4 15.4 15.6 15.7 15.7 16.1 16.7 17 .0 Year 10 12.3 13.7 15.5 15.5 15.6 15.9 16.0 16.1 16.1 17.1 22 . 9 15.4 16.1 4.9 10.6 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? Range Mean Year 10 Year 9
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
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 15.4 15.6 15.7 15.7 16.1 16.7 17 .0 Year 10 12.3 13.7 15.5 15.5 15.6 15.9 16.0 16.1 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?
The location of quartiles in an ordered data set When there are n values in an ordered data set: The interquartile range = the upper quartile – the lower quartile The lower quartile = n + 1 4 th value The median = n + 1 2 th value The upper quartile = 3( n + 1) 4 th value
Finding the interquartile range
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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Averages and range

  • 1. D2 Averages and range KS4 Mathematics
  • 2. D2.1 The mode Contents A A A A A D2 Averages and range D2.5 Comparing data D2.2 The mean D2.3 Calculating the mean from frequency tables D2.4 The median
  • 3. The three averages and range There are three different types of average : The range is not an average, but tells you how the data is spread out: M EDIAN middle value R ANGE largest value – smallest value M ODE most common M EAN sum of values number of values
  • 4. Favourite athletics event This graph shows pupils’ favourite athletics events. Which is the most popular event? How do you know? 0 5 10 15 20 Sprint Long distance running Hurdles High jump Long jump Triple jump Shot Discus Javelin Frequency
  • 5. The mode The mode is the item that occurs the most often in a data set. 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 Yes
  • 6. 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?
  • 7. The mode 14 15 15 13 12 14 15 0 11 13 14 11 16 14 15 9 10 12 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?
  • 8. Favourite athletics event Compare this graph to the previous one. What conclusions can you draw? Which two groups of pupils could be represented by the two graphs? 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
  • 9. 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 0 1 2 3 4 5 6 Numbers of sports played Frequency
  • 10. Grouped data This graph represents Year Ten girls’ times for a 100m sprint race. What is the modal time interval? How many girls are in this interval? 0 2 4 6 8 10 Frequency Times in seconds 12 13 14 15 16 17 18 19 20
  • 11. When the mode is not appropriate 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? 9 4 3 5 10 3 6 2 1 0 Numbers of sports played 2 15 17 20 Frequency
  • 12. Skewed data 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. 0 5 10 15 20 25 1 2 3 4 5 6 7 Numbers of sports played Frequency Negatively skewed data 0 2 4 6 8 10 12 14 1 2 3 4 5 6 7 Numbers of sports played Frequency Positively skewed data
  • 13. Contents D2.2 The mean A A A A A D2.5 Comparing data D2.1 The mode D2 Averages and range D2.3 Calculating the mean from frequency tables D2.4 The median
  • 14. Comparing data 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. Here are the times in seconds for 100 metres sprint for the two best classes. Which class should win and why? 13.1 16.5 14.3 15.4 13.1 16.4 13.8 15.4 12.8 15.9 13.4 15.3 12.2 15.2 12.9 14.7 12.0 14.9 11.5 12.8 10C boys 10C girls 10B boys 10B girls
  • 15. 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. For example, the mean time for Class 10B girls is: 14.72 Mean = Sum of values Number of values 12.8 + 14.7 + 15.3 + 15.4 + 15.4 5 = 73.6 5 =
  • 16. The mean Calculate the mean times for the other three groups. 14.72 13.18 15.78 12.64 Now calculate means for Class 10 B and Class 10 C (with girls and boys combined). 13.95 14.21 Based on these results, who should win? mean time 10C boys 10C girls 10B boys 10B girls mean time Class 10C Class 10B
  • 19. Outliers and their effect on the mean The school athletics team take part in an inter-schools competition. James’s shot results (in metres) are below. 9.46 9.25 8.77 10.25 10.35 9.59 4.02 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?
  • 20. Outliers and their effect on the mean Here are some 1500 metre race results in minutes. It may be appropriate in research or experiments to remove an outlier before carrying out analysis of results. Discuss: 6.26 6.28 6.30 6.39 5.38 4.54 10.59 6.35 7.01 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?
  • 21. Contents D2.3 Calculating the mean from frequency tables A A A A A D2.5 Comparing data D2.1 The mode D2 Averages and range D2.2 The mean D2.4 The median
  • 22. Calculating the mean from a frequency table 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? 9 4 3 5 10 3 6 2 1 0 Numbers of sports played 2 15 17 20 Frequency
  • 23. Calculating the mean from a frequency table 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 = 140 76 2 sports (to the nearest whole) 2 6 3 9 10 15 17 20 Frequency Number of sports × frequency 4 5 3 2 1 0 Numbers of sports played
  • 24. Grouped data Because the data is grouped, we do not know individual scores. It is not possible to add up the scores. 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? 1 35 ≤ d < 40 36 1 3 10 12 8 1 Frequency 30 ≤ d < 35 25 ≤ d < 30 20 ≤ d < 25 15 ≤ d < 20 10 ≤ d < 15 5 ≤ d < 10 Javelin distances in metres
  • 25. Midpoints 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 = 12.5 m Find the midpoints of the other groups. 1 35 ≤ d < 40 1 3 10 12 8 1 Frequency 30 ≤ d < 35 25 ≤ d < 30 20 ≤ d < 25 15 ≤ d < 20 10 ≤ d < 15 5 ≤ d < 10 Javelin distances in metres
  • 26. 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 36 695 T OTAL = 19.3 m (to 1 d.p.) 1 35 ≤ d < 40 1 3 10 12 8 1 Frequency Midpoint 30 ≤ d < 35 Frequency × midpoint 25 ≤ d < 30 20 ≤ d < 25 15 ≤ d < 20 10 ≤ d < 15 5 ≤ d < 10 Javelin distances in metres
  • 27. How accurate is the estimated mean? 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 is 18.7 metres (to 1 d.p.). 9.50 10.00 11.85 12.00 12.00 12.00 12.50 12.80 14.50 15.00 15.25 15.52 15.69 15.75 15.79 16.64 17.31 17.35 18.82 19.50 19.50 20.00 20.20 20.70 21.00 21.60 21.77 21.78 21.82 23.50 24.11 25.33 25.60 28.89 31.05 35.00
  • 28. Contents D2.4 The median 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
  • 29. The median 6.26 6.28 6.30 6.39 5.38 4.54 10.59 6.35 7.01 The median is the middle number when all numbers are in order. Calculate the median of the 1500 m results. Why is this a more appropriate average than the mean for these results? 4.54 5.38 6.26 6.28 6.30 6.35 6.39 7.01 10.59 Write the results in order and find the middle value :
  • 30. 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? 23 22 21 20 19 18 17 Choir 20 20 20 20 20 20 20 Drama club 29 28 25 20 19 18 18 Orchestra
  • 31. Outliers and the median and mean
  • 32. When there are two middle numbers 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 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.
  • 33. Finding halfway between two numbers
  • 34. One or two middle numbers? 1 2 3 4 6 6 7 8 9 11 If there are 9 numbers in a list, will there be 1 or 2 middle numbers? 2 3 4 6 6 7 8 9 11 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.
  • 35. When there are two middle numbers To find out where a middle number in a very long list, call the number of numbers n . Then the middle number is then ( 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?
  • 36. Where is the median?
  • 37. Contents D2.5 Comparing data A A A A A D2.1 The mode D2 Averages and range D2.2 The mean D2.3 Calculating the mean from frequency tables D2.4 The median
  • 38. The range Here are the high jump scores for two girls in metres. 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 1.30 1.30 1.41 1.45 1.59 Kirsty 1.15 1.20 1.35 1.41 1.62 Joanna
  • 39. 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.
  • 40. The range 1.35 m 0.47 m 1.41 m 0.29 m 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. Range Mean Kirsty Joanna 1.30 1.30 1.41 1.45 1.59 Kirsty 1.15 1.20 1.35 1.41 1.62 Joanna
  • 41. Calculating the mean, median and range
  • 42. Comparing sets of data 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. Range Mean 0.9 seconds 25.0 seconds Rob 1.4 seconds 24.8 seconds Chris
  • 43. Comparing sets of data 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? Range Mean 0.9 seconds 25.0 seconds Rob 1.4 seconds 24.8 seconds Chris 24.4 24.5 24.4 24.5 24.6 24.9 25.0 25.0 25.1 25.8 24.3 24.9 25.0 25.0 25.0 25.1 25.1 25.1 25.2 25.2 Here is the original data for Chris and Rob.
  • 44. Comparing hurdles scores Year 9 12.1 14 .0 15.3 15.4 15.4 15.6 15.7 15.7 16.1 16.7 17 .0 Year 10 12.3 13.7 15.5 15.5 15.6 15.9 16.0 16.1 16.1 17.1 22 . 9 15.4 16.1 4.9 10.6 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? Range Mean Year 10 Year 9
  • 45. 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
  • 46. 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 15.4 15.6 15.7 15.7 16.1 16.7 17 .0 Year 10 12.3 13.7 15.5 15.5 15.6 15.9 16.0 16.1 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?
  • 47. The location of quartiles in an ordered data set When there are n values in an ordered data set: The interquartile range = the upper quartile – the lower quartile The lower quartile = n + 1 4 th value The median = n + 1 2 th value The upper quartile = 3( n + 1) 4 th value
  • 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”.