SOLUTIONS MANUAL for Stochastic Modeling: Analysis and ...

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Pr{Y 2,6 > 30} = 1− Pr{Y 2,6 ≤ 30}
= 1−
30 ∑
j=0
e −6(6) (6(6)) j
j!
≈ 0.819
4. We want the smallest m such that
Pr{Y 30 − Y 6 >m}≤0.05
m∑ e −24 24 j
Pr{Y 30 − Y 6 >m} =1−
j!
m =32doesit.
5. If we assume that the accident rate is still λ = 1/week, then
Pr{Y t+24 − Y t ≤ 20} = Pr{Y 24 ≤ 20}
j=0
=
20 ∑
j=0
e −24 24 j
j!
≈ 0.24
Therefore, there is nearly a 1 in 4 chance of seeing 20 or fewer accidents even if the
rate is still 1/week. This is not overwhelming evidence in favor of a lower rate.
6. t corresponds to square meters of metal and λ = 1/50 defect/meter 2
(a)
Pr{Y 200 ≥ 7} = 1− Pr{Y 200 ≤ 6}
= 1−
≈ 0.111
6∑
j=0
e − 1 55 (200) ( 200
50 )j
j!
(b) For an out-of-control process λ ≥ 4/50. For λ =4/50 we want
Pr{Y 200 >c}≥0.95