Yes. If the predominant data are higher than the median, the mean average will be higher than the median average. For example, the median average of the numbers one through ten is five. The mean average is five and one-half.
If the mean is 6 then the 5 numbers must total 6 * 5 = 30. Median is 7, so that leaves 4 numbers totaling 23. So we need 2 numbers less than (or equal to) 7 and 2 numbers greater than (or equal to) 7 which all add up to 23 and where the highest number minus the lowest number equals 10. So how about, 1,2,7,9,11 or 1,1,7,10,11 etc
No. Not necessarily. Here are two examples with a smaller set of numbers: A) 1, 2, 3, 4 Mean: 10/4 = 2.5 Median: the mean of the two central elements - also 2.5. B) 1, 2, 3, 394 Mean: 400 / 4 = 100 Median: the mean of the two central elements = 2.5. You can extend the same principle - of having numbers on side of the center farther from the center than the numbers on the other side - to get a different mean and median, for sets of just about any size.
The median is the middle number of any given set when they are in order, least to greatest. If there is an even number of values, the median is the average of the middle two. It tells you that half the members of the set are greater than it and half are less than it.
The question will specify.
7,6,4,92,57,32
When a distribution is skewed to the right, the mean is greater than median.
Yes.
If there are only three numbers, the median MUST be the central number. Any question that claims otherwise is incorrect.
I am guessing you are asking for an example of a set of numbers with these properties. Let's start with 5 numbers, so the median will be the middle number; say 1, 2, 3, 4, 5. The median is 3, but so is the mean. Now let's replace the 5 with 10. The median is still 3, but the mean is 4. To make the mode less than 3, let us change the 2 into a 1. Now the median is still 3, the mode is 1, and the mean is 3.8. So 1, 1, 3, 4, 10 will work.
Generally, when the median is greater than the mean it is because the distribution is skewed to the left. This results in outliers or values further below the median than above the median which results in a lower mean value than median value. When a distribution is skewed left, it is generally not very symmetrical or normally distributed.
Yes, the median can be greater than the mean. It just depends on the values of the data. A simple series of 1,5,6 has 5 as the median, with a mean of 4.
In the same way that you calculate mean and median that are greater than the standard deviation!
By definition yes. The median is the number which is exactly in the middle of a set of numbers. The first quartile is the point where 25% of the numbers are to the left, and 75% of the numbers are to the right.
true
Yes. If the predominant data are higher than the median, the mean average will be higher than the median average. For example, the median average of the numbers one through ten is five. The mean average is five and one-half.
Oh, what a happy little math question we have here! To find the four numbers, we can start with the median, which is 6. We can then choose two numbers less than 6 and one number greater than 6 to balance it out, like 4, 5, 6, and 7. This way, when we find the mean by adding them up and dividing by 4, we get 5. Happy calculating!