题目
Problem
The random variables X X X Y Y Y X ∼ P o ( 3 ) X\sim\mathrm{Po}(3) X ∼ Po ( 3 ) Y ∼ P o ( 2 ) Y\sim\mathrm{Po}(2) Y ∼ Po ( 2 )
(a) Find P ( 2 < X < 5 ) \mathrm{P}(2<X<5) P ( 2 < X < 5 )
(2)
(b) Find P ( X + Y > 2 ) \mathrm{P}(X+Y>2) P ( X + Y > 2 )
(3)
(c) The total of 100 100 100 X X X 150 150 150 Y Y Y T T T
Use a suitable approximating distribution to find P ( T < 560 ) \mathrm{P}(T<560) P ( T < 560 )
(4)
题目中文翻译
随机变量 X X X Y Y Y X ∼ P o ( 3 ) X\sim\mathrm{Po}(3) X ∼ Po ( 3 ) Y ∼ P o ( 2 ) Y\sim\mathrm{Po}(2) Y ∼ Po ( 2 )
(a) 求 P ( 2 < X < 5 ) \mathrm{P}(2<X<5) P ( 2 < X < 5 )
(2)
(b) 求 P ( X + Y > 2 ) \mathrm{P}(X+Y>2) P ( X + Y > 2 )
(3)
(c) 用 T T T X X X 100 100 100 Y Y Y 150 150 150
用一个合适的近似分布求 P ( T < 560 ) \mathrm{P}(T<560) P ( T < 560 )
(4)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有 题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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严格不等号 2 < X < 5 2<X<5 2 < X < 5 3 3 3 4 4 4
答题过程
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Since X X X
P ( 2 < X < 5 ) = P ( X = 3 ) + P ( X = 4 ) = e − 3 ( 3 3 3 ! + 3 4 4 ! ) ≈ 0.392 . \begin{align*}
\mathrm P(2<X<5)
=&\,\mathrm P(X=3)+\mathrm P(X=4)\\[4mm]
=&\,\mathrm e^{-3}\bigg(\frac{3^3}{3!}+\frac{3^4}{4!}\bigg)\\[4mm]
\approx&\,\boxed{0.392}.
\end{align*} P ( 2 < X < 5 ) = = ≈ P ( X = 3 ) + P ( X = 4 ) e − 3 ( 3 ! 3 3 + 4 ! 3 4 ) 0.392 .
(b)
解法一
思路
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独立 Poisson 变量相加后参数相加，再用补事件求总数大于二的概率。
答题过程
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Independence gives X + Y ∼ P o ( 3 + 2 ) = P o ( 5 ) X+Y\sim\mathrm{Po}(3+2)=\mathrm{Po}(5) X + Y ∼ Po ( 3 + 2 ) = Po ( 5 )
P ( X + Y > 2 ) = 1 − P ( X + Y ≤ 2 ) = 1 − e − 5 ( 1 + 5 + 5 2 2 ! ) ≈ 0.875 . \begin{align*}
\mathrm P(X+Y>2)
=&\,1-\mathrm P(X+Y\le2)\\[4mm]
=&\,1-\mathrm e^{-5}\bigg(1+5+\frac{5^2}{2!}\bigg)\\[4mm]
\approx&\,\boxed{0.875}.
\end{align*} P ( X + Y > 2 ) = = ≈ 1 − P ( X + Y ≤ 2 ) 1 − e − 5 ( 1 + 5 + 2 ! 5 2 ) 0.875 .
解法二
思路
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也可直接列出总数为 0 , 1 , 2 0,1,2 0 , 1 , 2
答题过程
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The pairs ( X , Y ) (X,Y) ( X , Y ) X + Y ≤ 2 X+Y\le2 X + Y ≤ 2
( 0 , 0 ) , ( 1 , 0 ) , ( 0 , 1 ) , ( 2 , 0 ) , ( 1 , 1 ) , ( 0 , 2 ) . (0,0),\ (1,0),\ (0,1),\ (2,0),\ (1,1),\ (0,2). ( 0 , 0 ) , ( 1 , 0 ) , ( 0 , 1 ) , ( 2 , 0 ) , ( 1 , 1 ) , ( 0 , 2 ) . Using independence, their probabilities sum to
P ( X + Y ≤ 2 ) = e − 5 ( 1 + 3 + 2 ) + e − 5 ( 3 2 2 ! + 3 ( 2 ) + 2 2 2 ! ) = e − 5 ( 1 + 5 + 12.5 ) . \begin{align*}
\mathrm P(X+Y\le2)
=&\,\mathrm e^{-5}(1+3+2)\\[4mm]
&\,+\mathrm e^{-5}\bigg(\frac{3^2}{2!}+3(2)+\frac{2^2}{2!}\bigg)\\[4mm]
=&\,\mathrm e^{-5}(1+5+12.5).
\end{align*} P ( X + Y ≤ 2 ) = = e − 5 ( 1 + 3 + 2 ) + e − 5 ( 2 ! 3 2 + 3 ( 2 ) + 2 ! 2 2 ) e − 5 ( 1 + 5 + 12.5 ) . Therefore
P ( X + Y > 2 ) = 1 − 18.5 e − 5 ≈ 0.875 . \begin{align*}
\mathrm P(X+Y>2)=&\,1-18.5\mathrm e^{-5}\\[4mm]
\approx&\,\boxed{0.875}.
\end{align*} P ( X + Y > 2 ) = ≈ 1 − 18.5 e − 5 0.875 .
(c)
解法一
思路
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这里是许多个独立观测值的总和，参数分别乘以观测次数；不能把单个随机变量的倍数当作总和。
答题过程
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The sum of the independent observations has distribution
T ∼ P o ( 100 ( 3 ) + 150 ( 2 ) ) = P o ( 600 ) . T\sim\mathrm{Po}(100(3)+150(2))=\mathrm{Po}(600). T ∼ Po ( 100 ( 3 ) + 150 ( 2 )) = Po ( 600 ) . Since 600 > 15 600>15 600 > 15 N ∼ N ( 600 , 600 ) N\sim\mathrm N(600,600) N ∼ N ( 600 , 600 )
P ( T < 560 ) = P ( T ≤ 559 ) ≈ P ( N < 559.5 ) = Φ ( 559.5 − 600 600 ) = Φ ( − 1.65341 … ) ≈ 0.0491 . \begin{align*}
\mathrm P(T<560)
=&\,\mathrm P(T\le559)\\[4mm]
\approx&\,\mathrm P(N<559.5)\\[4mm]
=&\,\Phi\bigg(\frac{559.5-600}{\sqrt{600}}\bigg)\\[4mm]
=&\,\Phi(-1.65341\ldots)\\[4mm]
\approx&\,\boxed{0.0491}.
\end{align*} P ( T < 560 ) = ≈ = = ≈ P ( T ≤ 559 ) P ( N < 559.5 ) Φ ( 600 559.5 − 600 ) Φ ( − 1.65341 … ) 0.0491 . 