题目
Problem
6.
A particular lift has a maximum load capacity of 700 kg.
The weights of men are normally distributed with mean 80 kg and standard
deviation 10 kg.
The weights of women are normally distributed with mean 69 kg and standard
deviation 5 kg.
You may assume that weights of people are independent.
(a) Find the probability that when 6 men and 3 women are in the lift, the load
exceeds 700 kg.
(4)
A sign in the lift states: “Maximum number of people in the lift is c c c
(b) Find the value of c c c
(6)
(Total for Question 6 is 10 marks)
题目中文翻译
一部电梯的最大载重为 700 千克。
男性体重服从均值 80 千克、标准差 10 千克的正态分布。
女性体重服从均值 69 千克、标准差 5 千克的正态分布。
你可以假设人的体重相互独立。
(a) 求当电梯中有 6 名男性和 3 名女性时，载重超过 700 千克的概率。
(4)
电梯内的标牌写着：“电梯中的最大人数为 c c c
(b) 求 c c c
(6)
解答
(a)
解法一
思路
展开
独立正态随机变量之和仍服从正态分布。总重量的均值等于各人均值之和，方差等于各人方差之和；注意相加的是方差，而不是标准差。
答题过程
展开
Let X X X
E ( X ) = 6 ( 80 ) + 3 ( 69 ) = 687 , \begin{align*}
E(X)=&\,6(80)+3(69)\\
=&\,687,
\end{align*} E ( X ) = = 6 ( 80 ) + 3 ( 69 ) 687 , and
Var ( X ) = 6 ( 10 2 ) + 3 ( 5 2 ) = 675. \begin{align*}
\operatorname{Var}(X)
=&\,6(10^2)+3(5^2)\\
=&\,675.
\end{align*} Var ( X ) = = 6 ( 1 0 2 ) + 3 ( 5 2 ) 675. Therefore,
X ∼ N ( 687 , 675 ) . X\sim N(687,675). X ∼ N ( 687 , 675 ) . Hence
P ( X > 700 ) = P ( Z > 700 − 687 675 ) = P ( Z > 0.5003 … ) = 0.3084 … ≈ 0.308 . \begin{align*}
P(X>700)
=&\,P\bigg(
Z>\frac{700-687}{\sqrt{675}}
\bigg)\\
=&\,P(Z>0.5003\ldots)\\
=&\,0.3084\ldots\\
\approx&\,\boxed{0.308}.
\end{align*} P ( X > 700 ) = = = ≈ P ( Z > 675 700 − 687 ) P ( Z > 0.5003 … ) 0.3084 … 0.308 .
(b)
解法一
思路
展开
要保证无论乘客性别如何都满足条件，应采用全是男性的最不利情形，因为男性体重的均值和标准差都更大。设人数为 n n n 0.025 0.025 0.025 1.96 1.96 1.96 n n n
答题过程
展开
The greatest risk of exceeding the load occurs when all n n n Y Y Y
Y ∼ N ( 80 n , 100 n ) . Y\sim N(80n,100n). Y ∼ N ( 80 n , 100 n ) . The requirement is
P ( Y > 700 ) < 0.025. P(Y>700)<0.025. P ( Y > 700 ) < 0.025. Since P ( Z > 1.96 ) = 0.025 P(Z>1.96)=0.025 P ( Z > 1.96 ) = 0.025
700 − 80 n 10 n > 1.96. \frac{700-80n}{10\sqrt n}>1.96. 10 n 700 − 80 n > 1.96. Since n > 0 n>0 n > 0 10 n 10\sqrt n 10 n t = n t=\sqrt n t = n t > 0 t>0 t > 0
700 − 80 t 2 > 19.6 t 80 t 2 + 19.6 t − 700 < 0. \begin{align*}
700-80t^2>&\,19.6t\\
80t^2+19.6t-700<&\,0.
\end{align*} 700 − 80 t 2 > 80 t 2 + 19.6 t − 700 < 19.6 t 0. The positive root of the corresponding quadratic equation is
t = − 19.6 + 19.6 2 + 4 ( 80 ) ( 700 ) 160 = 2.8380 … \begin{align*}
t
=&\,\frac{-19.6+
\sqrt{19.6^2+4(80)(700)}}{160}\\
=&\,2.8380\ldots
\end{align*} t = = 160 − 19.6 + 19. 6 2 + 4 ( 80 ) ( 700 ) 2.8380 … Therefore,
n < 2.8380 … n < 8.0546 … \begin{align*}
\sqrt n<&\,2.8380\ldots\\
n<&\,8.0546\ldots
\end{align*} n < n < 2.8380 … 8.0546 … The greatest whole number satisfying this inequality is 8 8 8
c = 8 . \boxed{c=8}. c = 8 .
解法二
思路
展开
也可以先将含 n \sqrt n n n n n n < 8.75 n<8.75 n < 8.75
答题过程
展开
For n n n Y Y Y Y ∼ N ( 80 n , 100 n ) Y\sim N(80n,100n) Y ∼ N ( 80 n , 100 n )
P ( Y > 700 ) < 0.025 ⟹ 700 − 80 n 10 n > 1.96. \begin{align*}
P(Y>700)<0.025
\quad\Longrightarrow\quad
\frac{700-80n}{10\sqrt n}>&\,1.96.
\end{align*} P ( Y > 700 ) < 0.025 ⟹ 10 n 700 − 80 n > 1.96. Since n > 0 n>0 n > 0
700 − 80 n > 19.6 n > 0 , 700-80n>19.6\sqrt n>0, 700 − 80 n > 19.6 n > 0 , so we must have n < 700 / 80 = 8.75 n<700/80=8.75 n < 700/80 = 8.75
( 700 − 80 n ) 2 > ( 19.6 n ) 2 490000 − 112000 n + 6400 n 2 > 384.16 n 6400 n 2 − 112384.16 n + 490000 > 0. \begin{align*}
(700-80n)^2>&\,(19.6\sqrt n)^2\\
490000-112000n+6400n^2>&\,384.16n\\
6400n^2-112384.16n+490000>&\,0.
\end{align*} ( 700 − 80 n ) 2 > 490000 − 112000 n + 6400 n 2 > 6400 n 2 − 112384.16 n + 490000 > ( 19.6 n ) 2 384.16 n 0. The two roots of the corresponding quadratic equation are
n 1 = 112384.16 − 112384.16 2 − 4 ( 6400 ) ( 490000 ) 12800 = 8.054671 … , n 2 = 112384.16 + 112384.16 2 − 4 ( 6400 ) ( 490000 ) 12800 = 9.505353 … . \begin{align*}
n_1=&\,\frac{112384.16-\sqrt{112384.16^2-4(6400)(490000)}}{12800}\\
=&\,8.054671\ldots,\\
n_2=&\,\frac{112384.16+\sqrt{112384.16^2-4(6400)(490000)}}{12800}\\
=&\,9.505353\ldots.
\end{align*} n 1 = = n 2 = = 12800 112384.16 − 112384.1 6 2 − 4 ( 6400 ) ( 490000 ) 8.054671 … , 12800 112384.16 + 112384.1 6 2 − 4 ( 6400 ) ( 490000 ) 9.505353 … . The quadratic is positive outside these roots. Combining this with 0 < n < 8.75 0<n<8.75 0 < n < 8.75
0 < n < 8.054671 … . 0<n<8.054671\ldots. 0 < n < 8.054671 … . Hence the greatest possible integer number of occupants is
c = 8 . \boxed{c=8}. c = 8 . 