For the data9, 21, 15, 24, x, y, z, the mean and standard deviation are 15 and (√1260)/7respectively. If 15 is removed from the data, what will be the new standarddeviation?please detail steps
Ans is √30
measures of dispersion urgent!
2011-04-07 1:40 am
回答 (2)
2011-04-07 2:45 am
✔ 最佳答案
By definition of standard deviation, s. d. =sqrt {[ (15 - 9)^2 + (15 - 21)^2 + (15 - 15)^2 + (15 - 24)^2 + (15 - x)^2 + (15 - y)2 + (15 - z)^2]/7} = sqrt { [ 36 + 36 + 0 + 81 + (15 - x)^2 + (15 - y)^2 + (15 - z)^2]/7} = sqrt { [ 153 + (15 - x)^2 + (15 - y)^2 + (15 - z)^2]/7}
= (sqrt 1260)/7
so 153 + (15 - x)^2 + (15 - y)^2 + (15 - z)^2 = 1260/7 = 180
so (15 - x)^2 + (15 - y)^2 + (15 - z)^2 = 180 - 153 = 27.
When 15 is removed, the mean is still 15, the new s.d.
= sqrt { [ 36 + 36 + 81 + (15 - x)^2 + (15 - y)^2 + (15 - z)^2]/6}
= sqrt { (153 + 27)/6} = sqrt (180/6) = sqrt 30.
2011-04-07 3:18 am
收錄日期: 2021-04-20 00:25:45
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