题目
Problem
Navtej travels to work by train. A train leaves the station every 7 minutes and Navtej’s
arrival at the station is independent of when the train is due to leave.
(a) Write down a suitable model for the distribution of the time, T minutes, that he has
to wait for a train to leave.
(1)
(b) Find the mean and standard deviation of T
(3)
During a 10-week period, Navtej travels to work by train on 46 occasions.
(c) Estimate the probability that the mean length of time that he has to wait for a train
to leave is between 3.4 and 3.6 minutes.
(5)
(d) State a necessary assumption for the calculation in part (c).
(1)
(Total for Question 4 is 10 marks)
题目中文翻译
Navtej 乘火车上班。火车每 7 分钟发出一班，而 Navtej 到站的时间与火车预定出发时间无关。
(a) 写出等待火车出发的时间 T（分钟）的一个合适分布模型。
(b) 求 T 的均值和标准差。
在 10 周期间，Navtej 共乘火车上班 46 次。
(c) 估计他等待火车出发的平均时间介于 3.4 到 3.6 分钟之间的概率。
(d) 写出 (c) 小题计算所需的一个必要假设。
解答
(a)
解法一
思路
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火车每隔 7 7 7 7 7 7 [ 0 , 7 ] [0,7] [ 0 , 7 ]
答题过程
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T ∼ U ( 0 , 7 ) . \boxed{T\sim U(0,7)}. T ∼ U ( 0 , 7 ) .
(b)
解法一
思路
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对连续均匀分布 U ( a , b ) U(a,b) U ( a , b ) a + b 2 \frac{a+b}{2} 2 a + b ( b − a ) 2 12 \frac{(b-a)^2}{12} 12 ( b − a ) 2
答题过程
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For T ∼ U ( 0 , 7 ) T\sim U(0,7) T ∼ U ( 0 , 7 )
E ( T ) = 0 + 7 2 = 3.5 . E(T)=\frac{0+7}{2}=\boxed{3.5}. E ( T ) = 2 0 + 7 = 3.5 . Also,
Var ( T ) = ( 7 − 0 ) 2 12 = 49 12 . \begin{align*}
\operatorname{Var}(T)
=&\,\frac{(7-0)^2}{12}\\
=&\,\frac{49}{12}.
\end{align*} Var ( T ) = = 12 ( 7 − 0 ) 2 12 49 . Therefore, the standard deviation is
sd ( T ) = 49 12 = 7 12 = 2.0207 … ≈ 2.02 minutes . \begin{align*}
\operatorname{sd}(T)
=&\,\sqrt{\frac{49}{12}}\\
=&\,\frac{7}{\sqrt{12}}\\
=&\,2.0207\ldots\\
\approx&\,\boxed{2.02}\text{ minutes}.
\end{align*} sd ( T ) = = = ≈ 12 49 12 7 2.0207 … 2.02 minutes .
(c)
解法一
思路
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这里研究的是 46 46 46 T ˉ \bar T T ˉ 3.5 3.5 3.5 1 46 \frac{1}{46} 46 1
答题过程
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By the central limit theorem,
T ˉ ∼ ˙ N ( 3.5 , 49 12 ( 46 ) ) = N ( 3.5 , 49 552 ) . \bar{T}\mathrel{\dot\sim}
N\bigg(3.5,\frac{49}{12(46)}\bigg)
=N\bigg(3.5,\frac{49}{552}\bigg). T ˉ ∼ ˙ N ( 3.5 , 12 ( 46 ) 49 ) = N ( 3.5 , 552 49 ) . The standard deviation of T ˉ \bar T T ˉ
49 552 = 0.2979 … \sqrt{\frac{49}{552}}=0.2979\ldots 552 49 = 0.2979 … Therefore,
P ( 3.4 < T ˉ < 3.6 ) = P ( 3.4 − 3.5 49 / 552 < Z < 3.6 − 3.5 49 / 552 ) = P ( − 0.3356 … < Z < 0.3356 … ) = 0.2628 … ≈ 0.263 . \begin{align*}
P(3.4<\bar T<3.6)
=&\,P\bigg(
\frac{3.4-3.5}{\sqrt{49/552}}<Z\\
&\,\hspace{18pt}<
\frac{3.6-3.5}{\sqrt{49/552}}
\bigg)\\
=&\,P(-0.3356\ldots<Z<0.3356\ldots)\\
=&\,0.2628\ldots\\
\approx&\,\boxed{0.263}.
\end{align*} P ( 3.4 < T ˉ < 3.6 ) = = = ≈ P ( 49/552 3.4 − 3.5 < Z < 49/552 3.6 − 3.5 ) P ( − 0.3356 … < Z < 0.3356 … ) 0.2628 … 0.263 .
(d)
解法一
思路
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中心极限定理用于样本均值时，需要各次观测可以看作彼此独立、来自同一分布。写出其中一个必要假设即可；这里采用最直接的独立性假设。
答题过程
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The waiting times on the 46 occasions are assumed to be independent.
