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    There is a diagram of a cumulative frequency graph, and also a histogram showing the same data.
    It was later found that times recorded within each interval were in fact at the upper class boundary for that interval. state with a reason, how this information would affect your answers to part iii
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    what was part iii?
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    (Original post by CrystalSalvatore)
    what was part iii?
    It was to calculate the mean and standard deviation
    p.s: team stefan or damon
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    (Original post by imaan2121)
    It was to calculate the mean and standard deviation
    p.s: team stefan or damon
    so the mean you be adding the difference between the midvalue and the upper boundary because to calculate it you probably used the misvalue like in a group of 2-4 you would use 3 to calculate the mean so in this case i think the mean would increase by 1 (not sure you may need to calculate it to make sure)
    the standard deviation is the spread of the date, since the spread is the same,( as the results just shifted by 1 or whatever the real number is) the standard deviation is the same
    (and ofc team Damon <3)
 
 
 
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