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One or two middle numbers? 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? 1 2 3 4 6 6 7 8 9 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. 11 Averages (The Median) The median is the middle value of a set of data once the data has been ordered. Example 1. Robert hit 12 balls at Grimsby driving range. The recorded distances of his drives, measured in yards, are given below. Find the median distance for his drives. 85, 125, 130, 65, 100, 70, 75, 50, 140, 135, 95, 70 50, 65, 70, 70, 75, 85, 95, 100, 125, 130, 135, 140 Two middle values so take the mean. Ordered data Median drive = 90 yards 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 For example, There are 100 numbers in a list. Where is the median? 101 ÷ 2 = 50.5th number in the list (halfway between the 50th and the 51st). There are 37 numbers in a list. Where is the median? 38 ÷ 2 = 19th number in the list. Questions Answers Mean from Frequency table LO:- to be able to calculate the mean from grouped data. 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 Numbers of Frequency whole list of results, what sports played would it look like? 0 20 1 17 2 15 3 10 4 9 5 3 6 2 What do you think the mean would be? Calculating the mean from a frequency table Numbers of Frequency sports played 0 1 2 3 4 5 6 TOTAL 20 17 15 10 9 3 2 76 Number of sports × frequency 0 × 20 1 × 17 2 × 15 3 × 10 4×9 5×3 6×2 =0 = 17 = 30 = 30 = 36 = 15 = 12 140 Mean = 140 ÷ 76 = 2 sports (to the nearest whole) Grouped Data Grouped data Javelin Frequency distances in metres 5 ≤ d < 10 1 10 ≤ d < 15 8 15 ≤ d < 20 12 20 ≤ d < 25 10 25 ≤ d < 30 3 30 ≤ d < 35 1 35 ≤ d < 40 1 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? Because the data is grouped, we do not know individual scores. It is not possible to add up the scores. Midpoints Javelin Frequency distances in metres 5 ≤ d < 10 1 10 ≤ d < 15 8 15 ≤ d < 20 12 20 ≤ d < 25 10 25 ≤ d < 30 3 30 ≤ d < 35 1 35 ≤ d < 40 1 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. Estimating the mean from grouped data Javelin Frequency distances in metres 5 ≤ d < 10 1 10 ≤ d < 15 8 15 ≤ d < 20 12 20 ≤ d < 25 10 25 ≤ d < 30 3 30 ≤ d < 35 1 35 ≤ d < 40 1 TOTAL 36 Midpoint 7.5 12.5 17.5 22.5 27.5 32.5 37.5 Frequency × midpoint 1 × 7.5 8 × 12.5 12 × 17.5 10 × 22.5 3 × 27.5 1 × 32.5 1 × 37.5 = 7.5 = 100 = 210 = 225 = 82.5 = 32.5 = 37.5 695 Estimated mean = 695 ÷ 36 = 19.3 m (to 1 d.p.) How accurate is the estimated mean? Here are the javelin distances thrown by Year 10 before the data was grouped. 35.00 31.05 28.89 25.60 25.33 24.11 23.50 21.82 21.78 21.77 21.60 21.00 20.70 20.20 20.00 19.50 19.50 18.82 17.35 17.31 16.64 15.79 15.75 15.69 15.52 15.25 15.00 14.50 12.80 12.50 12.00 12.00 12.00 11.85 10.00 9.50 Work out the mean from the original data above and compare it with the estimated mean found from the grouped data. The estimated mean is 19.3 metres (to 1 d.p.). The actual mean is 18.7 metres (to 1 d.p.). How accurate was the estimated mean? Questions Answers