有3種battery (A,B 同C), 佢地都有exponential distribution with mean lifetime 200,250.270 小時. 在出售的battery中, 49.97%系type A, 24.97% 系type B, 25.06% 系type C.
如果一粒出售的battery可以維持330小時, 咁e粒battery系type A的機會有幾大呢?
Statistic
2008-11-28 4:21 pm
回答 (2)
2008-11-28 8:17 pm
✔ 最佳答案
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參考: My Maths knowledge
2008-11-29 7:38 am
Pr (battery type = A | battery life = 330hrs)
= Pr (battery type = A & battery life = 330hrs) / Pr (battery life = 330hrs)
= {0.4997 x [exp (-330 / 200) / 200]} / {0.4997 x [exp (-330 / 200) / 200] + 0.2497 x [exp (-330 / 250) / 250] + 0.2506 x [exp (-330 / 270) / 270]}
= [4.7984 x (10 ^ -4)] / {[4.7984 x (10 ^ -4)] + [2.6681 x (10 ^ -4)] + [2.7341 x (10 ^ -4)]}
= 0.4704
2008-11-28 23:42:41 補充：
飛天魏國大將軍張遼's answer is not correct
From the question, "如果一粒出售的battery可以維持330小時", but not 最少維持330小時
so, we should use f(x), but not S(x) = 1 - F(x)
2008-11-28 23:43:20 補充：
i.e. x = 330hrs, but not x => 330hrs
= Pr (battery type = A & battery life = 330hrs) / Pr (battery life = 330hrs)
= {0.4997 x [exp (-330 / 200) / 200]} / {0.4997 x [exp (-330 / 200) / 200] + 0.2497 x [exp (-330 / 250) / 250] + 0.2506 x [exp (-330 / 270) / 270]}
= [4.7984 x (10 ^ -4)] / {[4.7984 x (10 ^ -4)] + [2.6681 x (10 ^ -4)] + [2.7341 x (10 ^ -4)]}
= 0.4704
2008-11-28 23:42:41 補充：
飛天魏國大將軍張遼's answer is not correct
From the question, "如果一粒出售的battery可以維持330小時", but not 最少維持330小時
so, we should use f(x), but not S(x) = 1 - F(x)
2008-11-28 23:43:20 補充：
i.e. x = 330hrs, but not x => 330hrs
收錄日期: 2021-04-23 18:14:26
原文連結 [永久失效]:
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