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disper
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA00167623/o/701205090089413873404950.jpg plz explain it with step
最佳解答:
Let x1, x2, x3, ..., x30 (allin kg) be the weights of the 30 students. Original mean : (x1 + x2 + ...... + x30)/30 = 53 x1 + x2 + ...... + x30 = 1590 New mean = (x1 + x2 + ...... + x30 + 53)/31 kg = (1590 + 53)/31 kg = 53 kg (unchanged) Original standard deviation : √{[(x1 - 53)2 + (x2 - 53)2 + ... + (x30 -53)2]/29} = 6 [(x1 - 53)2 + (x2 - 53)2 + ... + (x30 -53)2]/29 = 36 (x1 - 53)2 + (x2 - 53)2 + ... + (x30 - 53)2 = 1044 New standard deviation : = √{[(x1 - 53)2 + (x2 - 53)2 + ... + (x30 -53)2 + (53 - 53)2]/30} kg = √[(1044 + 0)/30] kg = 5.90 kg
其他解答:
new weight=mean weight of 30 students =>addition of the new weight does not affect the original mean =>new mean = original mean = 53kg new weight = mean => deviation of new weight from mean = 0 => sum of squares of deviations from mean of 30 weights = sum of squares of deviations from mean of 31 weights Standard deviation, to a certain extent, is the average of sum of squares of deviations. With the addition of a new weight, the new standard deviation is smaller.
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