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F.7 數學方差問題 2
發問:
某校內考試的及格分數定為40分,共有200名考生.學生所得分數(x) x >= 40的,有160人,其平均值為64,標準差為6 x = 40的, [50 + (5/6)(x - 40)]=y x
最佳解答:
a) mean(y1)=50+(5/6)[mean(x1)-40] =50+(5/6)[64-40] =50+20 =70 mean(y2)=(5/4)mean(x2) =(5/4)32 =40 sum(y1)=mean(y1)*160=70*160=11200 sum(y2)=mean(y2)*40=40*40=1600 (according to this formula:http://upload.wikimedia.org/math/f/0/e/f0e09b2eb8e009f470ddbfe0df4cb2c6.png) sum(x^2)=[sd(x)^2+mean(x)^2]*N sum(x1^2)=[sd(x1)^2+mean(x1)^2]*160 =(6*6+64*64)*160 =661120 sum(x2^2)=[sd(x2)^2+mean(x2)^2]*40 =(4*4+32*32)*40 =41600 sum(y1^2)=sum[50+(5/6)(x1-40)]^2 =sum{[5x/6+(50-200/6)]^2} =sum{[5x/6+(100/6)]^2} =(25/36)sum(x1^2)+(1000/36)sum(x1)+(10000/36)*160 =(25/36)*661120+(1000/36)*64*160+(10000/36)*160 =788000 sum(y2^2)=[sum(5x2/4)]^2 =(25/16)*sum(x2^2) =25*41600/16 =65000 sd(y1)={[sum(y1^2)]/160-mean(y1)^2}^0.5 ={788000/160-mean(y1)^2}^0.5 ={4925-4900}^0.5 =5 sd(y2)={[sum(y2^2)]/40-mean(y2)^2}^0.5 ={65000/40-mean(y2)^2}^0.5 ={1625-1600}^0.5 =5 sum(y)=sum(y1)+sum(y2) =11200+1600 =12800 mean(y)=[mean(y1)*N1+mean(y2)*N2]/(N1+N2) =(70*160+40*40)/200 =64 sd(y)^2=sum(y^2)/N-mean(y)^2 =[sum(y1^2)+sum(y2^2)]/N-mean(y)^2 =(788000+65000)/200-64*64 =4265-4096 =169 13 ...oh my, 我路過架咋，無諗住禁煩．．． Can I give up.? 2009-10-30 00:39:37 補充： b) 因為名次並無改變，所以中位數依然是同一個人． 原本是52，即mod(x)=52>40 [50 + (5/6)(52 - 40)] =[50 + (5/6)(12)] =60 part2 易好多喎... part1 禁複雜, 唔知係因為yahoo唔方便打formula... 可能有簡單d的方法, 但係放低太耐唔記得咯...
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