2006-12-31 00:00:47 · 6 answers · asked by laura r 2 in Science & Mathematics ➔ Mathematics
I suppose the mean you are talking of is the arithmetic one (it's the most common). The arithmetic mean is the sum of your numbers divided by the occurence of numbers.
The median is the middle value of your numbers.
Example:
you have the numbers 1, 2, 3, 4, 5, 6 and 7. The mean is (1+2+3+4+5+6+7)/7 = 4
The median is also 4, as it is the middle value of this sequence.
BUT, if you change the numbers a bit: 1, 2, 3, 4, 7, 9, 15 have the mean (1+2+3+4+7+9+15)/7 = 5.8571
However, the median is still 4, as it is still the middle value of the sequence.
The median is more or less equal to the mean when you have a normal distribution of your numbers (bell-curve).
2006-12-31 00:24:07 · answer #1 · answered by Dr. Zaius 4 · 0⤊ 1⤋
Basically I agree with De_Sepp, but let me explain this a bit more.
The median is the middle number when you order your data values (increasing order). If you have an odd number of values n, then the median will be equal to the data value located at (n+1)/2. For example, the median of 1, 4, 6, 8, 11 is 6, the (5+1)/2=3rd value. If you have an even number of data values n, the median is still located at (n+1)/2. However, there is no data value there. For example, the median of 1, 4, 6, 8 is located in the (4+1)/2=2.5th position. Conventionally, we average the two middle numbers. In our example, we average the 2nd and 3rd values to get (4+6)/2 = 5 for our median. In a histogram, the median is the value with 50% of the area to the left and 50% of the area to the right.
The mean, on the other hand, is the arithmetic mean (a traditional average). The mean will be affected by extremely high (or extremely low) values. In your situation, one mean is higher than the other, so I know that there is an extreme value on the high side. Let's use the original example of 1, 4, 6, 8, 11. The median of this data set is 6 (see above). The mean of this data set is (1+4+6+8+11)/5 = 30/5 = 6. (Here the mean and median happen to be equal because I picked values that are exactly symmetric.) Now, suppose that the 11 is 21 instead. The median of 1, 4, 6, 8, 21 is still 6. However, the mean is now (1+4+6+8+21)/5 = 40/5 = 8.
The median is not affected by any changes in the "tails" of the distribution, but the mean is, since all of the data values are involved in the calculation. In the case where you have one mean higher than the other with the medians the same, the data set with the higher mean is more skewed to the right (long right tail).
Hope this helps!
2006-12-31 03:06:22 · answer #2 · answered by jbm616 2 · 0⤊ 0⤋
Short version of jbm's answer: outliers pull the mean. So if a mean is larger, it has more outliers in that direction.
2007-01-03 12:11:19 · answer #3 · answered by a_math_guy 5 · 0⤊ 0⤋
It means that the distributions of one or both of the samples are skewed.
2006-12-31 00:19:48 · answer #4 · answered by Anonymous · 0⤊ 0⤋
The other one didn't smoke enough.Serious!
2006-12-31 00:06:19 · answer #5 · answered by one10soldier 6 · 0⤊ 1⤋
um...the other numbers are different???
2006-12-31 00:08:55 · answer #6 · answered by practicalwizard 6 · 0⤊ 1⤋