# Lesson starter 1: Lesson starter 1: Calculate the Mean of…… (a) 3, 4, 7, 3, 5, 2, 6, 10 (b) 8, 10, 12, 14, 7, 16, 5, 7, 9,11 (c) 64, 66, 65, 61, 67, 61,

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Lesson starter 1: Lesson starter 1: Calculate the Mean of…… (a) 3, 4, 7, 3, 5, 2, 6, 10 (b) 8, 10, 12, 14, 7, 16, 5, 7, 9,11 (c) 64, 66, 65, 61, 67, 61, 57

Answers (a) Mean = 5 (b) Mean = 9.9 (c) Mean = 63

Lesson starter 2: What is the range of each set? (a) 3, 4, 7, 3, 5, 2, 6, 10 (b) 8, 10, 12, 14, 7, 16, 5, 7, 9,11 (c) 64, 66, 65, 61, 67, 61, 57

Answers (a) Range = 8 (b) Range = 11 (c) Range = 10

Objective: Mean of large data sets A businessman is deciding whether to open a new shop selling CD’s in Market Harborough. He conducts a survey in the town and manages to ask 250 shoppers how many CD’s they buy a month. He could add up all of the answers and divide by the number surveyed, but this is a lot of data! What would be a better way to organise this data to make it useful to him?

Number of CD’sTally 0||||| ||||| 1||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| |||| 2||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||| 3||||| ||||| ||||| ||||| ||||| || 4||||| ||||| ||||| ||||| ||| 5||||| ||||| | 6||||| ||| 7||||

Number of CD’s TallyFrequency 0||||| ||||| 20 1||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| |||| 49 2||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||| 108 3||||| ||||| ||||| ||||| ||||| ||27 4||||| ||||| ||||| ||||| |||23 5||||| ||||| |11 6||||| |||8 7||||4 Totals250

Calculating the Mean The final column completes the table and allows him to calculate the mean number of CD’s bought per person. Can you suggest what might go in this column? Complete the column and calculate mean number of CD’s bought per person.

Number of CD’s TallyFrequencyFrequency x CD’s 0||||| ||||| 20 0  20 = 0 1||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| |||| 49 1  49 = 49 2||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||||| ||| 108 2  108 = 216 3||||| ||||| ||||| ||||| ||||| ||27 3  27 = 81 4||||| ||||| ||||| ||||| |||23 4  23 = 92 5||||| ||||| |11 5  11 = 55 6||||| |||8 6  8 = 48 7||||4 7  4 = 28 Totals250569 Mean = 569 / 250 = 2.276

Questions How will the information about the mean number of CD’s help him? Can you buy 2.276 CD’s or would another measure of the average have given him a clearer answer? Can we find the mode? What about the median?

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