题目
Problem
At a particular supermarket, the times taken to serve each customer in a queue at a
standard checkout may be modelled by a normal distribution with mean 240 seconds
and standard deviation 20 seconds.
There is a queue of 3 customers at a standard checkout.
Making a reasonable assumption about the times taken to serve these customers,
(a) find the probability that the total time taken to serve the 3 customers will be less
than 11 minutes.
(3)
(b) State the assumption you have made in part (a)
(1)
In the supermarket there is also an express checkout, which is reserved for customers
buying 10 or fewer items. The time taken to serve a customer at this express checkout
may be modelled by a normal distribution with mean 100 seconds and standard
deviation 8 seconds.
On a particular day Jiang has 8 items to pay for and has to choose whether to join a
queue of 3 customers waiting at a standard checkout or a queue of 7 customers waiting
at the express checkout.
Using a similar assumption to that made in part (a),
(c) find the probability that the total time taken to serve the 3 customers at the
standard checkout will exceed the total time taken to serve the 7 customers at
the express checkout.
(4)
(Total for Question 7 is 8 marks)
题目中文翻译
在某超市，标准收银台为每位顾客服务所需时间可由均值 240 秒、标准差 20 秒的正态分布建模。
标准收银台有 3 位顾客排队。
对这些顾客的服务时间作出合理假设后，
(a) 求为这 3 位顾客服务的总时间少于 11 分钟的概率。
(b) 写出你在 (a) 小题中所作的假设。
超市还有一个快速收银台，仅供购买不超过 10 件商品的顾客使用。
快速收银台为一位顾客服务所需时间可由均值 100 秒、标准差 8 秒的正态分布建模。
某天 Jiang 有 8 件商品要付款，需要在等待 3 位顾客的标准收银台队列和等待 7 位顾客的快速收银台队列之间作出选择。
采用与 (a) 小题类似的假设，
(c) 求标准收银台 3 位顾客的总服务时间超过快速收银台 7 位顾客的总服务时间的概率。
解答
(a)
解法一
思路
展开
假设三位顾客的服务时间相互独立。三个独立正态变量的总和仍服从正态分布，均值相加、方差相加。先建立总服务时间（秒）的分布，再把 11 分钟换成 660 秒后标准化。
答题过程
展开
Let P i P_i P i i i i
P i ∼ N ( 240 , 20 2 ) . P_i\sim\mathrm N(240,20^2). P i ∼ N ( 240 , 2 0 2 ) . Let
Q = P 1 + P 2 + P 3 . Q=P_1+P_2+P_3. Q = P 1 + P 2 + P 3 . Then
E ( Q ) = 3 ( 240 ) = 720 , Var ( Q ) = 3 ( 20 2 ) = 1200. \begin{align*}
\operatorname E(Q)
=&\,3(240)=720,\\
\operatorname{Var}(Q)
=&\,3(20^2)=1200.
\end{align*} E ( Q ) = Var ( Q ) = 3 ( 240 ) = 720 , 3 ( 2 0 2 ) = 1200. Therefore,
Q ∼ N ( 720 , 1200 ) . Q\sim\mathrm N(720,1200). Q ∼ N ( 720 , 1200 ) . Since 11 minutes is 660 seconds,
P ( Q < 660 ) = P ( Z < 660 − 720 1200 ) = P ( Z < − 1.732 … ) = 0.0416 . \begin{align*}
\operatorname P(Q<660)
=&\,\operatorname P\left(
Z<\frac{660-720}{\sqrt{1200}}
\right)\\
=&\,\operatorname P(Z<-1.732\ldots)\\
=&\,\boxed{0.0416}.
\end{align*} P ( Q < 660 ) = = = P ( Z < 1200 660 − 720 ) P ( Z < − 1.732 … ) 0.0416 .
解法二
思路
展开
官方评分资料也接受全程使用分钟。把三位顾客的总时间从秒换成分钟后，均值除以 60，方差除以 60 2 60^2 6 0 2
答题过程
展开
Let Q m = Q / 60 Q_m=Q/60 Q m = Q /60
E ( Q m ) = 720 60 = 12 , Var ( Q m ) = 1200 60 2 = 1 3 . \begin{align*}
\operatorname E(Q_m)
=&\,\frac{720}{60}=12,\\
\operatorname{Var}(Q_m)
=&\,\frac{1200}{60^2}\\
=&\,\frac13.
\end{align*} E ( Q m ) = Var ( Q m ) = = 60 720 = 12 , 6 0 2 1200 3 1 . Thus
Q m ∼ N ( 12 , 1 3 ) . Q_m\sim\mathrm N\left(12,\frac13\right). Q m ∼ N ( 12 , 3 1 ) . Its standard deviation is 1 / 3 1/\sqrt3 1/ 3
P ( Q m < 11 ) = P ( Z < − 3 ) = P ( Z < − 1.732 … ) = 0.0416 . \begin{align*}
\operatorname P(Q_m<11)
=&\,\operatorname P(Z<-\sqrt3)\\
=&\,\operatorname P(Z<-1.732\ldots)\\
=&\,\boxed{0.0416}.
\end{align*} P ( Q m < 11 ) = = = P ( Z < − 3 ) P ( Z < − 1.732 … ) 0.0416 .
(b)
解法一
思路
展开
在求总服务时间的方差时直接相加了三位顾客各自的方差，这要求各位顾客的服务时间相互独立。
答题过程
展开
It is assumed that the times taken to serve the customers are independent.
(c)
解法一
思路
展开
分别建立标准收银台三位顾客与快速收银台七位顾客的总时间，再令 S S S S > 0 S>0 S > 0
答题过程
展开
Let Q Q Q
Q ∼ N ( 720 , 1200 ) . Q\sim\mathrm N(720,1200). Q ∼ N ( 720 , 1200 ) . Let R R R
E ( R ) = 7 ( 100 ) = 700 , Var ( R ) = 7 ( 8 2 ) = 448. \begin{align*}
\operatorname E(R)
=&\,7(100)=700,\\
\operatorname{Var}(R)
=&\,7(8^2)=448.
\end{align*} E ( R ) = Var ( R ) = 7 ( 100 ) = 700 , 7 ( 8 2 ) = 448. Thus
R ∼ N ( 700 , 448 ) . R\sim\mathrm N(700,448). R ∼ N ( 700 , 448 ) . Define S = Q − R S=Q-R S = Q − R
E ( S ) = 720 − 700 = 20 , Var ( S ) = 1200 + 448 = 1648. \begin{align*}
\operatorname E(S)
=&\,720-700=20,\\
\operatorname{Var}(S)
=&\,1200+448=1648.
\end{align*} E ( S ) = Var ( S ) = 720 − 700 = 20 , 1200 + 448 = 1648. Therefore,
S ∼ N ( 20 , 1648 ) . S\sim\mathrm N(20,1648). S ∼ N ( 20 , 1648 ) . The required probability is
P ( Q > R ) = P ( S > 0 ) = P ( Z > 0 − 20 1648 ) = P ( Z > − 0.4926 … ) = 0.689 . \begin{align*}
\operatorname P(Q>R)
=&\,\operatorname P(S>0)\\
=&\,\operatorname P\left(
Z>\frac{0-20}{\sqrt{1648}}
\right)\\
=&\,\operatorname P(Z>-0.4926\ldots)\\
=&\,\boxed{0.689}.
\end{align*} P ( Q > R ) = = = = P ( S > 0 ) P ( Z > 1648 0 − 20 ) P ( Z > − 0.4926 … ) 0.689 .
解法二
思路
展开
也可以把两队总时间之差直接换算为分钟。秒数差除以 60 后，均值为 1 / 3 1/3 1/3 103 / 225 103/225 103/225
答题过程
展开
Let S m = S / 60 S_m=S/60 S m = S /60
E ( S m ) = 20 60 = 1 3 , Var ( S m ) = 1648 60 2 = 103 225 . \begin{align*}
\operatorname E(S_m)
=&\,\frac{20}{60}\\
=&\,\frac13,\\
\operatorname{Var}(S_m)
=&\,\frac{1648}{60^2}\\
=&\,\frac{103}{225}.
\end{align*} E ( S m ) = = Var ( S m ) = = 60 20 3 1 , 6 0 2 1648 225 103 . Hence
S m ∼ N ( 1 3 , 103 225 ) . S_m\sim\mathrm N\left(\frac13,\frac{103}{225}\right). S m ∼ N ( 3 1 , 225 103 ) . Its standard deviation is
103 15 . \frac{\sqrt{103}}{15}. 15 103 . Therefore,
P ( S m > 0 ) = P ( Z > − 5 103 ) = P ( Z > − 0.4926 … ) = 0.689 . \begin{align*}
\operatorname P(S_m>0)
=&\,\operatorname P\left(
Z>-\frac5{\sqrt{103}}
\right)\\
=&\,\operatorname P(Z>-0.4926\ldots)\\
=&\,\boxed{0.689}.
\end{align*} P ( S m > 0 ) = = = P ( Z > − 103 5 ) P ( Z > − 0.4926 … ) 0.689 . 