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Td Stats 3.2 et 4.1 - 3.2 1 X N(200 40 A P X 250 =...

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3.2 1. X N(200 , 40) A P ( X ≥ 250) = p ( (X-200)/40) ≥ p ((250 – 200)/40) = p (T≥1,25) = 1 - F(1,25) = 1 – 0,8944 = 0,1056 B P(X<100) = ( p (X-200)/40) < p (100 – 200) /40 )) = p (T < - 2,5 ) = 1 – F (2,5) = 1 – 0,9938 = 0,0062 B’. P ( X ≥ 180) = p ( (X-200)/40) ≥ p ((180 – 200)/40) = p (T≥ - 0,5) = 1 – F(-0,5) = 1 – 1 + F(0,5) = F(0,5) = 0,6915 C p (140 < X < 260 ) = p ( X < 260 ) – p ( X < 140) = ( p (X-200)/40) < p (260-200)/40) ) – ( p (X-200)/40) < p (140-200)/40) = ( p (T< 1,5) – ( p (T< - 1,5)) = (F(1,5)) – (1 – F(1,5)) = 0,9332 – 1 + 0,9332 = 0,8664 D p (150 < X < 180 ) = p ( X < 180 ) – p ( X < 150) = ( p (X-200)/40) < p (180-200)/40) ) – ( p (X-200)/40) < p (150-200)/40) = ( p (T< (-20/40) – ( p (T< (-50/40)) = (1 - F(0,5)) – (1 – F(1,25)) = (1 – 0,6915) – (1-0,8944) = 0,2029 2.
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P(X≥x 0 ) = 0,04 P(X< x 0 ) = 1 – 0,04 = 0,96 F((x 0 – 200)/40) = 0,96 = F(t) F(1,75) t = 1,75 Donc (x 0 – 200)/40 = 1,75 x 0 = (1,75 * 40) + 200 = 270 unités en stock en début de semaine 3. P(x1<X<x2) = P (X<x2) – P(X<x1) = F((x2-200)/40) – F((x1-200)/40)= 0,8 F(t) – F(-t)=0,8 2F(t) – 1 = 0,8 F(1,28)=0,8997 F(1,29)=0,9015 On choisi :
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