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author | Julian T <julian@jtle.dk> | 2021-02-17 12:35:27 +0100 |
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committer | Julian T <julian@jtle.dk> | 2021-02-17 12:35:53 +0100 |
commit | cfcd3eefcbc547a8b69d9f5830b7a27c0bdb60ce (patch) | |
tree | 8b038689dc88f2d129bc1804973ce4f53b5ebb57 /sem6/prob/m2 | |
parent | 1e6f7a495318addcbb6bc5ae7465a2eb1d889acd (diff) |
Changes to prob
Diffstat (limited to 'sem6/prob/m2')
-rw-r--r-- | sem6/prob/m2/opgaver.md | 20 |
1 files changed, 20 insertions, 0 deletions
diff --git a/sem6/prob/m2/opgaver.md b/sem6/prob/m2/opgaver.md index a5d0cdb..0ce9c77 100644 --- a/sem6/prob/m2/opgaver.md +++ b/sem6/prob/m2/opgaver.md @@ -50,4 +50,24 @@ $$ ## Opgave 3 +Først skal man finde $\lambda$. + +$$ + \int_{0}^{\infty} \lambda e^{- \frac x {100}} \mathrm{dx} = 1 \\ + \left[ - \lambda 100 \cdot e^{- \frac x {100}}\right]_{0}^{\infty} = 1 \\ + \lambda \cdot 100 = 1 \\ + \lambda = \frac 1 {100} +$$ + +Nu kan man sætte 50 til 150 ind. + +$$ + P(50 < x \leq 150) = \int_{50}^{150} f(x) \mathrm{dx} = - e^{- \frac {150} {100}} + e^{ - \frac {50} {100}} = 0.3834 +$$ + +Derefter kan vi tage fra 0 til 100. + +$$ + P(x < 100) = \int_{0}^{100} f(x) \mathrm{dx} = - e^{- \frac {100} {100}} = - \frac 1 e +$$ |