The median is least affected by an extreme outlier. Mean and standard deviation ARE affected by extreme outliers.
You cannot because the standard deviation is not related to the median.
Karl Pearson simplified the topic of skewness and gave us some formulas to help. The first is the Pearson mode or first skewness coefficient. It is defined by the (mean-median)/standard deviation. So in this case the Pearson mode is: (8-6)/2 =1 There is also the Pearson Median. This is also called second skewness coefficient. It is defined as 3(mean-median)/standard deviation which in this case is 6/2 =3 hence the distribution is positive skewed
No, it is not
I believe the standard deviations are measured from the median, not the mean.1 Standard Deviation is 34% each side of median, so that is 68% total.2 Standard Deviations is 48% each side of median, so that is 96% total.
In the same way that you calculate mean and median that are greater than the standard deviation!
characteristics of mean
mean | 32 median | 32 standard deviation | 4.472 ========================================================================
The mean, median, and mode of a normal distribution are equal; in this case, 22. The standard deviation has no bearing on this question.
msd 0.560
mean | 30 median | 18 standard deviation | 35.496
The median is least affected by an extreme outlier. Mean and standard deviation ARE affected by extreme outliers.
You cannot because the standard deviation is not related to the median.
mean
Mean: 26.33 Median: 29.5 Mode: 10, 35 Standard Deviation: 14.1515 Standard Error: 5.7773
You cannot because the median of a distribution is not related to its standard deviation.
When using the mean: the variance or standard deviation. When using the median: the range or inter-quartile range.