05

TD 5

Exercice 29:

1

\(X \sim \mathcal{N}(200,70^{2})\)

1

\(\begin{aligned} {\mathbb{P}}(X > 250) = {\mathbb{P}}(\frac{X - 200}{70} > \frac{5}{7}) & = 1 - {\mathbb{P}}(\frac{X - 200}{70} \leq \frac{5}{7}) \\ & = 1 - \Phi(\frac{5}{7}) \end{aligned}\)
\({\mathbb{P}}(X \leq 180) = {\mathbb{P}}(X < 180) = {\mathbb{P}}(\frac{X - 200}{70} < \frac{180 - 200}{70}) = \Phi( - \frac{2}{7}) = 1 - \Phi(\frac{2}{7})\)

2

\[\begin{aligned} {\mathbb{P}}(190 < X \leq 210) & = {\mathbb{P}}(\frac{190 - 200}{70} < \frac{X - 200}{70} \leq \frac{210 - 200}{70}) = {\mathbb{P}}( - \frac{1}{7} < \frac{X - 200}{70} \leq \frac{1}{7}) \\ & = {\mathbb{P}}(\frac{X - 200}{70} \leq \frac{1}{7}) - {\mathbb{P}}(\frac{X - 200}{70} \leq - \frac{1}{7}) = \Phi(\frac{1}{7}) - \Phi( - \frac{1}{7}) = 2\Phi(\frac{1}{7}) - 1 \end{aligned}\]

2

\(X \sim \mathcal{N}(24,3^{2})\)

1

\({\mathbb{P}}(\frac{X - 24}{3} > \frac{x_{4} - 24}{3}) = 0.05\) par table on a: \(\frac{x_{4} - 24}{3} = 1.65\) d’où x_4 = 28.95.
On trouve après grace aux équations: \(\left( x_{1} + x_{4} \right) = 24,x_{1} + \frac{x_{4} - x_{1}}{3} = x_{2},\ldots\) \(x_{1} = 19.05,x_{2} = 22.35,x_{3} = 25.65\)

2

On calcule \({\mathbb{P}}(X < x_{1}) = {\mathbb{P}}(X > x_{4}),{\mathbb{P}}(X \in \left\lbrack x_{1},x_{2} \right\rbrack) = {\mathbb{P}}(X \in \left\lbrack x_{3},x_{4} \right\rbrack),{\mathbb{P}}(X \in \left\lbrack x_{2},x_{3} \right\rbrack)\)

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