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A) The range for a given set of data is MAX-MIN=6743-5139=1604
B) The middle 50% would consist of a one standard deviation difference from the median in both directions. So our range would be [5914-399.37,5888.84+399.37]=[5514.63,6313.37]. If the range were larger, e.g. 90%, we would use two standard deviations.
C) C. Typically median and IQR (interquartile range) are good summary statistics. If you look at part E, you can use context to determine the correct response.
D) The distribution is skewed to the left and multimodal. The center of the data is 3914 yards and the distribution has a spread of 624.5 yards...
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