题目
Problem
8.
A market stall sells cauliflowers classified by three different weights, small (S), medium (M) and large (L).
The weights, in grams, of these cauliflowers are modelled by independent normal distributions as shown below.
S ∼ N ( 265 , 78 2 ) S\sim N(265,78^2) S ∼ N ( 265 , 7 8 2 ) M ∼ N ( 625 , 42 2 ) M\sim N(625,42^2) M ∼ N ( 625 , 4 2 2 ) L ∼ N ( 886 , 45 2 ) L\sim N(886,45^2) L ∼ N ( 886 , 4 5 2 )
The random variable T T T T ∼ N ( a , b ) T\sim N(a,b) T ∼ N ( a , b )
(a) Find the value of a a a b b b
(3)
(b) Using standardisation, find the probability that a randomly selected large cauliflower weighs more than 3 times the weight of a randomly selected small cauliflower.
(5)
A random sample of 2 small cauliflowers is taken.
(c) Using standardisation, find the probability that the first cauliflower in this sample weighs at least 100 grams more than the mean weight of both cauliflowers.
(5)
(Total for Question 8 is 13 marks)
题目中文翻译
某市场摊位出售三种重量等级的花椰菜：小 (S)、中 (M) 和大 (L)。
这些花椰菜的重量（单位：克）分别由相互独立的正态分布建模，如下所示。
S ∼ N ( 265 , 78 2 ) S\sim N(265,78^2) S ∼ N ( 265 , 7 8 2 ) M ∼ N ( 625 , 42 2 ) M\sim N(625,42^2) M ∼ N ( 625 , 4 2 2 ) L ∼ N ( 886 , 45 2 ) L\sim N(886,45^2) L ∼ N ( 886 , 4 5 2 )
随机变量 T T T T ∼ N ( a , b ) T\sim N(a,b) T ∼ N ( a , b )
(a) 求 a a a b b b
(3)
(b) 使用标准化方法，求随机选取的一个大花椰菜重量大于随机选取的一个小花椰菜重量 3 倍的概率。
(5)
现在抽取 2 个小花椰菜的随机样本。
(c) 使用标准化方法，求样本中第一个花椰菜至少比两者平均重量重 100 克的概率。
(5)
解答
(a)
解法一
思路
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把两个中等花椰菜的重量记为 M 1 , M 2 M_1,M_2 M 1 , M 2 L L L T = M 1 + M 2 − L T=M_1+M_2-L T = M 1 + M 2 − L
答题过程
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Let
T = M 1 + M 2 − L , T=M_1+M_2-L, T = M 1 + M 2 − L , where M 1 M_1 M 1 M 2 M_2 M 2 L L L
a = E ( T ) = 625 + 625 − 886 = 364 , \begin{align*}
a=E(T)
=&\,625+625-886\\
=&\,\boxed{364},
\end{align*} a = E ( T ) = = 625 + 625 − 886 364 , and
b = Var ( T ) = 42 2 + 42 2 + ( − 1 ) 2 ( 45 2 ) = 5553 . \begin{align*}
b=\operatorname{Var}(T)
=&\,42^2+42^2+(-1)^2(45^2)\\
=&\,\boxed{5553}.
\end{align*} b = Var ( T ) = = 4 2 2 + 4 2 2 + ( − 1 ) 2 ( 4 5 2 ) 5553 .
(b)
解法一
思路
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所求事件为 L > 3 S L>3S L > 3 S X = L − 3 S X=L-3S X = L − 3 S X > 0 X>0 X > 0 L L L S S S
答题过程
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Let
X = L − 3 S . X=L-3S. X = L − 3 S . Since L L L S S S
E ( X ) = 886 − 3 ( 265 ) = 91 , \begin{align*}
E(X)
=&\,886-3(265)\\
=&\,91,
\end{align*} E ( X ) = = 886 − 3 ( 265 ) 91 , and
Var ( X ) = 45 2 + ( − 3 ) 2 ( 78 2 ) = 56781. \begin{align*}
\operatorname{Var}(X)
=&\,45^2+(-3)^2(78^2)\\
=&\,56781.
\end{align*} Var ( X ) = = 4 5 2 + ( − 3 ) 2 ( 7 8 2 ) 56781. Hence
X ∼ N ( 91 , 56781 ) . X\sim\operatorname{N}(91,56781). X ∼ N ( 91 , 56781 ) . Therefore,
P ( L > 3 S ) = P ( X > 0 ) = P ( Z > 0 − 91 56781 ) = P ( Z > − 0.381891 … ) = 0.649 (3 s.f.) . \begin{align*}
P(L>3S)
=&\,P(X>0)\\
=&\,P\big(
Z>\frac{0-91}{\sqrt{56781}}
\big)\\
=&\,P(Z>-0.381891\ldots)\\
=&\,\boxed{0.649}\quad\text{(3 s.f.)}.
\end{align*} P ( L > 3 S ) = = = = P ( X > 0 ) P ( Z > 56781 0 − 91 ) P ( Z > − 0.381891 … ) 0.649 (3 s.f.) .
(c)
解法一
思路
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设两棵小花椰菜的重量为 S 1 , S 2 S_1,S_2 S 1 , S 2 ( S 1 − S 2 ) / 2 ⩾ 100 (S_1-S_2)/2\geqslant100 ( S 1 − S 2 ) /2 ⩾ 100
答题过程
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The required event is
S 1 ⩾ S 1 + S 2 2 + 100. S_1\geqslant\frac{S_1+S_2}{2}+100. S 1 ⩾ 2 S 1 + S 2 + 100. This is equivalent to
S 1 − S 2 2 ⩾ 100. \frac{S_1-S_2}{2}\geqslant100. 2 S 1 − S 2 ⩾ 100. Let
Y = S 1 − S 2 2 . Y=\frac{S_1-S_2}{2}. Y = 2 S 1 − S 2 . Since S 1 S_1 S 1 S 2 S_2 S 2
E ( Y ) = 265 − 265 2 = 0 , \begin{align*}
E(Y)
=&\,\frac{265-265}{2}\\
=&\,0,
\end{align*} E ( Y ) = = 2 265 − 265 0 , and
Var ( Y ) = 78 2 + 78 2 2 2 = 3042. \begin{align*}
\operatorname{Var}(Y)
=&\,\frac{78^2+78^2}{2^2}\\
=&\,3042.
\end{align*} Var ( Y ) = = 2 2 7 8 2 + 7 8 2 3042. Therefore,
P ( Y ⩾ 100 ) = P ( Z ⩾ 100 3042 ) = P ( Z ⩾ 1.813 … ) = 0.0349 (3 s.f.) . \begin{align*}
P(Y\geqslant100)
=&\,P\big(
Z\geqslant\frac{100}{\sqrt{3042}}
\big)\\
=&\,P(Z\geqslant1.813\ldots)\\
=&\,\boxed{0.0349}\quad\text{(3 s.f.)}.
\end{align*} P ( Y ⩾ 100 ) = = = P ( Z ⩾ 3042 100 ) P ( Z ⩾ 1.813 … ) 0.0349 (3 s.f.) .
解法二
思路
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也可以先把原不等式两边乘以 2，直接研究重量差 R = S 1 − S 2 R=S_1-S_2 R = S 1 − S 2 R ⩾ 200 R\geqslant200 R ⩾ 200 2 ( 78 2 ) 2(78^2) 2 ( 7 8 2 )
答题过程
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The required event can also be written as
S 1 − S 2 ⩾ 200. S_1-S_2\geqslant200. S 1 − S 2 ⩾ 200. Let R = S 1 − S 2 R=S_1-S_2 R = S 1 − S 2
E ( R ) = 265 − 265 = 0 , Var ( R ) = 78 2 + 78 2 = 12168. \begin{align*}
E(R)=&\,265-265=0,\\
\operatorname{Var}(R)
=&\,78^2+78^2\\
=&\,12168.
\end{align*} E ( R ) = Var ( R ) = = 265 − 265 = 0 , 7 8 2 + 7 8 2 12168. Therefore,
P ( R ⩾ 200 ) = P ( Z ⩾ 200 12168 ) = P ( Z ⩾ 1.813 … ) = 0.0349 (3 s.f.) . \begin{align*}
P(R\geqslant200)
=&\,P\big(
Z\geqslant\frac{200}{\sqrt{12168}}
\big)\\
=&\,P(Z\geqslant1.813\ldots)\\
=&\,\boxed{0.0349}\quad\text{(3 s.f.)}.
\end{align*} P ( R ⩾ 200 ) = = = P ( Z ⩾ 12168 200 ) P ( Z ⩾ 1.813 … ) 0.0349 (3 s.f.) . 