點畫standard deviation bar??10點

For the set of 5 temperatures:i) mean (of these temps) = (20 +23 + 30 + 40 +50) ÷ 5 = 32.6 (deg C)ii)variance(of these temps) = 〔(20-32.6)^2 +(23-32.6)^2 + (30-32.6)^2+(40-32.6)^2 +(50-32.6)^2 〕÷ 5 = 123.04 ∴ standard deviation (of these temps) = √123.04 = 11.09 (deg C) ＊ • ＊ • ＊ • ＊ • ＊ • ＊ • ＊ • ＊ • ＊ • ＊ Note :(a)i)A x-y graph can be drawn if there are 2 sets of nos. x,y , which exist in form ofordered pairs : ( x1 , y1 ), ( x2 , y2), (x3, y3 ), ….ii)A linear graph can drawn if there are 2 sets of nos. x,y, where y is a function of x (i.e. y is formed from x)e.g.y=2x+1, where x,y Î R

(b) For the set of 5 temps 20, 23,30, 40, 50： there exists only 1 mean temp (平均温度) & also 1 S.D. (標準差)∴ there is NO GRAPH to be drawn between the set of 5 temps & their S.D.(11.09)！

(c)Also,obviously, your 3rd set of S.D. are NOT FORMED from the 1stset of temps. That is,there is no relationship between the 1st & 3rd set of nos.∴ there is no 1 - 1 correspondence between the 2 set of nos.Thus,you should not pair the 2 sets up one by one to plot a graph!! Did you make this problem up? ?