The mean is affected by the skewness, whereas the median is not.
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Let's go through the answer choices to see which are true and which are false.
A. True. The mean takes account of all the values since we sum all the values and divide by the number of values. The median only looks at the middle most value. The median does not take into account any other value. So the outliers can move wherever you want and the median will not be affected.
B. False. It depends on what the context of the situation is. Both measures of center have their strengths and weaknesses. The median is only preferred if there are outliers. That way we get a sense of where the center is and know that the center isn't being pulled on by the outliers. A good example is home real estate as this problem mentions.
C. False. The mean is larger than the median if the data is skewed right. Skewed right data means we have a large outlier(s) to the right side pulling the mean toward that location. An example would be having homes in some range of say 50 to 100 thousand, and then an outlier mansion would have a price of 20 million dollars. This skews the mean to be larger than it should be (when it should be somewhere between 50 and 100 thousand). Note: sometimes a trimmed mean is used instead of the median
D. False. The mean should only be used if we don't have any outliers at all. In other words, the mean should only be used if the data is not skewed left and not skewed right either.
E. False. This computes the midrange and not the median. In my experience with many stats problems, the midrange doesn't really come up that often. Though I could just have a limited viewpoint of course.
It bisects the triangle at the angle 90° . So that , we can say that an right angle triangle has only one line of symmetry which cuts triangle to two equal halves.
