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How can two data sets with different numbers have the same mean median and mode?
(10,10,30,30,30,50,50) (20,20,30,30,30,40,40) These two sets have the same mean, median and mode.
Is it possible for the mean meadian and mode of a data set to be the same number?
Yes, it is. One easy way for this to happen is if every number in a data set is the same: then it's the mean, median, and mode at the same time. That's not the only way for it to happen, of course. For example, if the data set is 1, 2, 2, 3 then the mean, median, and mode is 2.
How do you get the mean median and mode all the same number?
Have all the numbers be the same. Example: 2,2,2,2,2,2,2 Mean=2 Median=2 Mode=2 There's probably another way but you can figure it out just try.
In any data set are the median and mean usually very similar in value?
yes they are if you have 0 and 10 the mean is 5 and so is the median. The mean and the median can in fact be the same value. But basically to answer your question, One possible way is that if the values are ascending by 1 in the data set, then the number of values left to the median should be the same as the number of values right to the median. e.g. 6+7+8+9+10 6,7 = 2 terms 9,10 = 2 terms median =8 mode = 8
Can the mean median mode and range all the same number?
They quick answer is YES!Here is an example.Before we begin let quickly recap what the we mean by "mean", "median", "mode" and "range":[MEAN] - The sum of all the values, divided by the total number of values.[MEDIAN] - The middle value when the data is arranged in numerical order.[MODE] - The most common value in a data set.[RANGE] - The difference between the highest and lowest values in the set.If we had the following numbers 1, 2, 2, 2, 3,The [MEAN] would be: TWO= 1+2+2+2+3/5 = 10/5 = (2)The [MEDIAN] would be: TWO= 1 2 (2) 2 3 = (2)The [MODE] would be: TWOThe most common value is (2)The [RANGE] would be: TWOrange = (highest - lowest) = (3-1) = (2)Therefore; Mean, Median, Mode and Range = (2)So the Mean, Median, Mode and Range can all be the same number![Answered by F:A:W:B:Y] - (As always, glad to help)
