Corrigé du 12 P. 441

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a. \[\begin{aligned} \operatorname E(B) &= np = 60\times 0,37 = 22,2.& \\ \operatorname V(B) &= np(1-p) = 60\times 0,37 \times 0,63 = 13,986& \\ \implies \sigma(B) &= \sqrt{13,986}\approx 3,7398. \end{aligned}\]

b. \[\begin{aligned} \operatorname E(B') &= np = 35\times 0,61 = 21,35.& \\ \operatorname V(B') &= np(1-p) = 35\times 0,61\times 0,39 = 8,3265& \\ \implies \sigma(B') &= \sqrt{8,3265}\approx 2,8856. \end{aligned}\]

c. \[ \operatorname E(B+B') = \operatorname E(B) + \operatorname E(B') =22,2 + 21,35 = 43,55. \] Puisque $B$ et $B'$ sont indépendantes \[ V(B+B') = V(B) + V(B') =13,986 + 8,3265 = 22,3125. \] Donc $\sigma(B+B') = \sqrt{22,3125} \approx 4,7236$.

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