题目
Problem
The random variables D D D E E E D ∼ P o ( 1.7 ) D\sim\mathrm{Po}(1.7) D ∼ Po ( 1.7 ) E ∼ P o ( 2.1 ) E\sim\mathrm{Po}(2.1) E ∼ Po ( 2.1 )
(a) Find P ( D + E < 4 ) \mathrm{P}(D+E<4) P ( D + E < 4 )
(3)
(b) Given that D + E < 4 D+E<4 D + E < 4 P ( D = 1 ) \mathrm{P}(D=1) P ( D = 1 )
(3)
(c) The random variable T T T 100 100 100 D D D
Use the normal approximation to the Poisson distribution to find P ( T > 190 ) \mathrm{P}(T>190) P ( T > 190 )
(4)
题目中文翻译
随机变量 D D D E E E D ∼ P o ( 1.7 ) D\sim\mathrm{Po}(1.7) D ∼ Po ( 1.7 ) E ∼ P o ( 2.1 ) E\sim\mathrm{Po}(2.1) E ∼ Po ( 2.1 )
(a) 求 P ( D + E < 4 ) \mathrm{P}(D+E<4) P ( D + E < 4 )
(3)
(b) 已知 D + E < 4 D+E<4 D + E < 4 P ( D = 1 ) \mathrm{P}(D=1) P ( D = 1 )
(3)
(c) 随机变量 T T T 100 100 100 D D D
用正态分布近似泊松分布，求 P ( T > 190 ) \mathrm{P}(T>190) P ( T > 190 )
(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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独立 Poisson 变量相加仍为 Poisson，参数相加。小于 4 4 4 0 , 1 , 2 , 3 0,1,2,3 0 , 1 , 2 , 3
答题过程
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Since D D D E E E
D + E ∼ P o ( 1.7 + 2.1 ) = P o ( 3.8 ) . D+E\sim\mathrm{Po}(1.7+2.1)=\mathrm{Po}(3.8). D + E ∼ Po ( 1.7 + 2.1 ) = Po ( 3.8 ) . Therefore
P ( D + E < 4 ) = e − 3.8 ( 1 + 3.8 + 3.8 2 2 ! + 3.8 3 3 ! ) ≈ 0.4734848 ≈ 0.473 . \begin{align*}
\mathrm P(D+E<4)
=&\,\mathrm e^{-3.8}\bigg(1+3.8+\frac{3.8^2}{2!}+\frac{3.8^3}{3!}\bigg)\\[4mm]
\approx&\,0.4734848\\[4mm]
\approx&\,\boxed{0.473}.
\end{align*} P ( D + E < 4 ) = ≈ ≈ e − 3.8 ( 1 + 3.8 + 2 ! 3. 8 2 + 3 ! 3. 8 3 ) 0.4734848 0.473 .
(b)
解法一
思路
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条件概率的分子要同时满足两件事。当 D = 1 D=1 D = 1 4 4 4 E ≤ 2 E\le2 E ≤ 2
答题过程
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The required conditional probability is
P ( D = 1 ∣ D + E < 4 ) = P ( D = 1 , D + E < 4 ) P ( D + E < 4 ) . \begin{align*}
&\,\mathrm P(D=1\mid D+E<4)\\[4mm]
=&\,\frac{\mathrm P(D=1,\ D+E<4)}{\mathrm P(D+E<4)}.
\end{align*} = P ( D = 1 ∣ D + E < 4 ) P ( D + E < 4 ) P ( D = 1 , D + E < 4 ) . If D = 1 D=1 D = 1 D + E < 4 D+E<4 D + E < 4 E ≤ 2 E\le2 E ≤ 2
P ( D = 1 ) P ( E ≤ 2 ) = 1.7 e − 3.8 ( 1 + 2.1 + 2.1 2 2 ! ) ≈ 0.2017508. \begin{align*}
&\,\mathrm P(D=1)\mathrm P(E\le2)\\[4mm]
=&\,1.7\mathrm e^{-3.8}\bigg(1+2.1+\frac{2.1^2}{2!}\bigg)\\[4mm]
\approx&\,0.2017508.
\end{align*} = ≈ P ( D = 1 ) P ( E ≤ 2 ) 1.7 e − 3.8 ( 1 + 2.1 + 2 ! 2. 1 2 ) 0.2017508. Using the unrounded probability from (a),
P ( D = 1 ∣ D + E < 4 ) ≈ 0.2017508 0.4734848 ≈ 0.426 . \begin{align*}
\mathrm P(D=1\mid D+E<4)\approx&\,\frac{0.2017508}{0.4734848}\\[4mm]
\approx&\,\boxed{0.426}.
\end{align*} P ( D = 1 ∣ D + E < 4 ) ≈ ≈ 0.4734848 0.2017508 0.426 .
(c)
解法一
思路
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总和的 Poisson 参数乘以 100 100 100 170 170 170 190 190 190 190.5 190.5 190.5
答题过程
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The total of 100 100 100 D D D T ∼ P o ( 170 ) T\sim\mathrm{Po}(170) T ∼ Po ( 170 ) 170 > 15 170>15 170 > 15 Y ∼ N ( 170 , 170 ) Y\sim\mathrm N(170,170) Y ∼ N ( 170 , 170 )
With a continuity correction,
P ( T > 190 ) = P ( T ≥ 191 ) ≈ P ( Y > 190.5 ) = 1 − Φ ( 190.5 − 170 170 ) = 1 − Φ ( 1.572 … ) ≈ 0.0579 . \begin{align*}
\mathrm P(T>190)
=&\,\mathrm P(T\ge191)\\[4mm]
\approx&\,\mathrm P(Y>190.5)\\[4mm]
=&\,1-\Phi\bigg(\frac{190.5-170}{\sqrt{170}}\bigg)\\[4mm]
=&\,1-\Phi(1.572\ldots)\\[4mm]
\approx&\,\boxed{0.0579}.
\end{align*} P ( T > 190 ) = ≈ = = ≈ P ( T ≥ 191 ) P ( Y > 190.5 ) 1 − Φ ( 170 190.5 − 170 ) 1 − Φ ( 1.572 … ) 0.0579 . 