How can a median be greater than the mean?

affizupdaftf3opg
2022-05-07
Answered

How can a median be greater than the mean?

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notemilyu1208

Answered 2022-05-08
Author has **15** answers

Explanation:

The median of a set of numbers is the value that is in the middle (In a set with an odd number of values, it's the middle value. In a set with an even number of values, it's the mean of the two middle values).

The mean is the generally understood "average", where the sum of the values is divided by the number of values (sometimes referred to as the count of the values).

How can we set up a set of values so that the median is higher than the mean? We can do it by taking a set of numbers and skewing the values to be very low below the median and just above the median. For instance, if I take a set of five numbers and set the middle value as 10, I can place the two lower values at 1 and 2 and the higher values at 11:

1,2,10, 11, 11

The median = 10

The mean $=\frac{1+2+10+11+11}{5}=\frac{35}{5}=7$

In fact, the mean will be lower than the median in any distribution where the values "fall off", or decrease from the middle value faster than they increase from the middle value.

The median of a set of numbers is the value that is in the middle (In a set with an odd number of values, it's the middle value. In a set with an even number of values, it's the mean of the two middle values).

The mean is the generally understood "average", where the sum of the values is divided by the number of values (sometimes referred to as the count of the values).

How can we set up a set of values so that the median is higher than the mean? We can do it by taking a set of numbers and skewing the values to be very low below the median and just above the median. For instance, if I take a set of five numbers and set the middle value as 10, I can place the two lower values at 1 and 2 and the higher values at 11:

1,2,10, 11, 11

The median = 10

The mean $=\frac{1+2+10+11+11}{5}=\frac{35}{5}=7$

In fact, the mean will be lower than the median in any distribution where the values "fall off", or decrease from the middle value faster than they increase from the middle value.

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