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erastova [34]
4 years ago
15

HELP? PLEASE HELP BY TOMORROW

Mathematics
1 answer:
kherson [118]4 years ago
8 0
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

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Step-by-step explanation:

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