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î ì ì õ } l µ } p z z Ì } Á Ç z ~ Ì x h x Ì î ì í õ u } Ì x ô ð ó u x i x x ô X K } Ì Ç ' µ i } ^ Á µ ^D^ i ' i P } h Ì v l u Ç u u Ç u l µ i WP(Z < −z) = P(Z > z) Therefore, if the sum of these probabilities has to be equal to 18, and both probabilities are equal we must have that each probabilityF(x) = P(X x) = Z x 10 10 u2 du= 10 u jx 10)F(x) = 1 10 x c We want to nd a value of X(call it p) such that P(X p) = 075 Z p 10 10 x2 dx= 075 ) 10 x jp 10 = 075 )1 10 p = 075 )p= 40 Therefore the 75 thpercentile is 40, which means P(X 40) = 075 d We rst nd the probability that a device willfunction for at least 15 hours P(X>15) = Z




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