Est 5.20.docx
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5.20. Suponga una distribución de Poisson con ¿Cuál es la probabilidad de que a) x=1? b) x<1 c) x>1 d) x ≤1 P (x=1)= P (x<1)= P (x>1)= P (x≤1)=
𝑒 −1 (1)1 1!
=
0.367879 (1) 1
𝑒 −5.0 (5.0)0 0! 𝑒 −2 (2)2 2! 𝑒 −1(1)1 1!
+ +
=
= 0.367879
0.00673795 (1) 1
𝑒 −3 (3)3 3!
+
𝑒 −4(4)4
𝑒 −5.0 (5.0)0 0!
4!
= 0.00673795 +
𝑒 −5 (5)5 5!
= 0.37461695
= 6.103038
𝑒 −1 (1)1 1!
=
0.367879 (1) 1
𝑒 −5.0 (5.0)0 0! 𝑒 −2 (2)2 2! 𝑒 −1(1)1 1!
+ +
=
= 0.367879
0.00673795 (1) 1
𝑒 −3 (3)3 3!
+
𝑒 −4(4)4
𝑒 −5.0 (5.0)0 0!
4!
= 0.00673795 +
𝑒 −5 (5)5 5!
= 0.37461695
= 6.103038
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