Sam NicholsSTAT 311 HW 1 Textbook Exercises: 37) I would report a trimmed mean of 11.46 as a representative value of October snow cover for this period. My choice was prompted by a particularly low outlier and a particularly high outlier, and when trimmed, the mean is more representative of the typical snow cover. 44) a) Sample range = 182.6 - 180.3 = 2.3 b) First, I obtained the average/mean as 181.4083. Then, I added together the squares of each data point subtracted by the mean. Finally, dividing that all by (12 - 1) 11 gives a sample variance of s2= 0.5245 c) The sample standard deviation is the square root of the sample variance, which equals 0.7242 d) Using the shortcut, I summed the values, both on their own (2176.9) and after squaring each value (394913.57). Then I did the calculation = 5.7692. Finally, dividing that94913.57 3− 122176.92number by n - 1 (11) yields an s2of 0.5245 using the shortcut 52) Because the sum over all deviations from the mean must equal 0, a simple algebra equation reveals that the fifth deviation must be -3.5. An example sample.3 0.9 1.0 1.3 x00+ + + + = for which these are the five deviations from the mean is [3,8, 4.4, 4.5, 4.8, 0].