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.
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.
Yes. It can be higher or lower
it is a divison
No because the mean is the highest numeral and the median is the middle numeral of the set of numbers so it is tecnictly impossible, but if you are using decimals, the median could get pretty close to the mean, but never higher.
The Mean is the average of a given set of values. The Median is the value that has the same number of smaller values than the number of higher values, it is in the middle of them. In a symmetrical distribution the Mean is equal to the Median. In an asymmetrical distribution they have different value.
The distribution is skewed to the right.
Average price higher than median price in this situation means that there was a number of EXTREMALLY expensive homes sold, way, way, higher than the median price, which skewed the average to the upside. There is a cluster of "average" properties around the mean, but a significant number of million dollar / pound houses than pull house price averages up. The mean is also higher than the median in the UK by about 20% - see Zoopla data that shows average house price as about 200k vs median.
The mean is better than the median when there are outliers.
If the distribution is not symmetric, the mean will be different from the median. A negatively skewed distribution will have a mean hat is smaller than the median, provided it is unimodal.
Yes. If the lower values tend to be farther below the median than the highest values are above the median, the mean is smaller than the median. why are write wrong
When a distribution is skewed to the right, the mean is greater than median.