By tomorrow 2. the data set shows the weights of pumpkins, in pounds, that are chosen for a photograph in a farming magazine. 18 22 14 30 26 (a) what is the mean, x , of the data set? (b) what is the sum of the squares of the differences between each data value and the mean? use the table to organize your work. x x x x x 2 (c) what is the standard deviation of the data set? use the sum from part (b) and show your work. (d) a second group of pumpkins, with weights of 32, 35, 33, 34, and 36 pounds, are chosen for another photograph. will the standard deviation of these weights be greater or less than the standard deviation found in part (c)? answer this without doing a calculation and explain your reasoning

A) What is the mean of : 18, 22, 14, 30, 26 x = (18 +22+ 14+ 30+ 26)/5 = 22 (b) What is the sum of the squares of the differences between each data value and the mean: (data point - x)² or (data point - 22)² (18-22)² = 16 (22-22)² = 0 (14-22)² = 64 (30-22)² = 64 (26-22)² = 16 Sum = 16+0+64+64+16 = 160 (c) What is the standard deviation s = √[(∑(x-22)²/(n-1)], where x is the data point, 20 the mean and n =5 √[(∑(x-22)² = 160 and n = 5 (5 data points) then s = 160(5-1) = 160/4 = 40 d) 32, 35, 33, 34, and 36 Rewrite it from smaller to greater: 32, 34, 34, 35 36. We notice that the range is from 32 to 36 is only 4 (36-32), Where as the range of the 1st pumpkins is 30-14 = 16 Since the spread of the 1st one is by far larger than the 2nd, we can conclude that the standard deviation of the 1st is greater than the second

The answer is poppy pants69