推 loveliver:P(Z>1.96)=0.025? 03/19 15:59
→ hahakelly:謝謝^^" 03/19 16:06
推 MousePads:Y1-Y2 = 3.92 :P 03/19 16:24
※ 編輯: hahakelly 來自: 61.58.85.190 (03/19 16:35)
→ hahakelly:一直打錯T^T 謝謝樓上兩位 03/19 16:35
※ 引述《xiang1989215 (let it be)》之銘言：
: Suppose the relationship between applied stress x and time-to-failure Y is
: described by the simple linear regression model with true regression line
: E(y)=100-3.92x and σ2 =2. Let Y1 denotes an observation on time-to-failure
: made with x=15 and Y2 denotes an independent observation made with x=14, find
: the probability that Y1 exceed Y2, i.e.,P(Y1>Y2)=?
: 請問上述問題該如何解?
Y1 ~ N(100-3.92*15 = 41.2 , 2)
Y2 ~ N(100-3.92*14 = 45.12, 2)
Y1-Y2 ~N(41.2-45.12 = -3.92 , 2+2 = 4 )
P(Y1>Y2) = P(Y1-Y2>0) = P(Z> 3.92/2 =1.96 ) = 0.025
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