题目
Problem
A potter makes decorative tiles in two colours, red and yellow. The length, R R R Y Y Y N ( 12 , 0.8 2 ) N(12, 0.8^2) N ( 12 , 0. 8 2 ) R R R Y Y Y
A red tile and a yellow tile are chosen at random.
(a) Find the probability that the yellow tile is longer than the red tile.
(4)
Taruni buys 3 red tiles and 1 yellow tile.
(b) Find the probability that the total length of the 3 red tiles is less than 4 times the length of the yellow tile.
(7)
Stefan defines the random variable X = a R + b Y X = aR + bY X = a R + bY a a a b b b a a a b b b X X X
(c) Find the value of a a a b b b
(7)
(Total 18 marks)
题目中文翻译
一位陶工制作红色和黄色两种装饰瓷砖。红色瓷砖的长度 R R R Y Y Y N ( 12 , 0.8 2 ) N(12, 0.8^2) N ( 12 , 0. 8 2 ) R R R Y Y Y
随机选取一块红砖和一块黄砖。
(a) 求黄色瓷砖比红色瓷砖更长的概率。
Taruni 购买 3 块红砖和 1 块黄砖。
(b) 求 3 块红砖的总长度小于黄色瓷砖长度 4 倍的概率。
Stefan 定义随机变量 X = a R + b Y X=aR+bY X = a R + bY a a a b b b a a a b b b X X X
(c) 求 Stefan 应该选取的 a a a b b b
解答
(a)
解法一
思路
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“黄色瓷砖比红色瓷砖长”可写成 Y − R > 0 Y-R>0 Y − R > 0 D = Y − R D=Y-R D = Y − R D D D
答题过程
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Let
D = Y − R . D=Y-R. D = Y − R . Since R R R Y Y Y
E ( D ) = 12 − 15 = − 3 , Var ( D ) = 0.8 2 + 1.5 2 = 2.89. \begin{align*}
E(D)=&\,12-15=-3,\\
\operatorname{Var}(D)=&\,0.8^2+1.5^2=2.89.
\end{align*} E ( D ) = Var ( D ) = 12 − 15 = − 3 , 0. 8 2 + 1. 5 2 = 2.89. Therefore,
D ∼ N ( − 3 , 2.89 ) . D\sim N(-3,2.89). D ∼ N ( − 3 , 2.89 ) . Hence
P ( Y > R ) = P ( D > 0 ) = P ( Z > 0 − ( − 3 ) 2.89 ) = P ( Z > 1.7647 … ) = 0.03881 … \begin{align*}
P(Y>R)=&\,P(D>0)\\
=&\,P\bigg(Z>\frac{0-(-3)}{\sqrt{2.89}}\bigg)\\
=&\,P(Z>1.7647\ldots)\\
=&\,0.03881\ldots
\end{align*} P ( Y > R ) = = = = P ( D > 0 ) P ( Z > 2.89 0 − ( − 3 ) ) P ( Z > 1.7647 … ) 0.03881 … Thus, to 3 significant figures,
P ( Y > R ) = 0.0388 . \boxed{P(Y>R)=0.0388}. P ( Y > R ) = 0.0388 .
(b)
解法一
思路
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先分别建立三块红色瓷砖总长度与黄色瓷砖长度四倍的分布。所求事件是前者小于后者，因此令两者之差 L = 4 Y − ( R 1 + R 2 + R 3 ) L=4Y-(R_1+R_2+R_3) L = 4 Y − ( R 1 + R 2 + R 3 ) P ( L > 0 ) P(L>0) P ( L > 0 )
答题过程
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Let
T = R 1 + R 2 + R 3 , T=R_1+R_2+R_3, T = R 1 + R 2 + R 3 , where the three red tile lengths are independent. Then
E ( T ) = 3 ( 15 ) = 45 , Var ( T ) = 3 ( 1.5 2 ) = 6.75. \begin{align*}
E(T)=&\,3(15)=45,\\
\operatorname{Var}(T)=&\,3(1.5^2)=6.75.
\end{align*} E ( T ) = Var ( T ) = 3 ( 15 ) = 45 , 3 ( 1. 5 2 ) = 6.75. Thus,
T ∼ N ( 45 , 6.75 ) . T\sim N(45,6.75). T ∼ N ( 45 , 6.75 ) . Also,
E ( 4 Y ) = 4 ( 12 ) = 48 , Var ( 4 Y ) = 4 2 ( 0.8 2 ) = 10.24 , \begin{align*}
E(4Y)=&\,4(12)=48,\\
\operatorname{Var}(4Y)=&\,4^2(0.8^2)=10.24,
\end{align*} E ( 4 Y ) = Var ( 4 Y ) = 4 ( 12 ) = 48 , 4 2 ( 0. 8 2 ) = 10.24 , so
4 Y ∼ N ( 48 , 10.24 ) . 4Y\sim N(48,10.24). 4 Y ∼ N ( 48 , 10.24 ) . Now let
L = 4 Y − T . L=4Y-T. L = 4 Y − T . Since T T T Y Y Y
E ( L ) = 48 − 45 = 3 , Var ( L ) = 10.24 + 6.75 = 16.99. \begin{align*}
E(L)=&\,48-45=3,\\
\operatorname{Var}(L)=&\,10.24+6.75=16.99.
\end{align*} E ( L ) = Var ( L ) = 48 − 45 = 3 , 10.24 + 6.75 = 16.99. Therefore,
L ∼ N ( 3 , 16.99 ) . L\sim N(3,16.99). L ∼ N ( 3 , 16.99 ) . The required probability is
P ( T < 4 Y ) = P ( L > 0 ) = P ( Z > 0 − 3 16.99 ) = P ( Z > − 0.7278 … ) = 0.7666 … \begin{align*}
P(T<4Y)=&\,P(L>0)\\
=&\,P\bigg(Z>\frac{0-3}{\sqrt{16.99}}\bigg)\\
=&\,P(Z>-0.7278\ldots)\\
=&\,0.7666\ldots
\end{align*} P ( T < 4 Y ) = = = = P ( L > 0 ) P ( Z > 16.99 0 − 3 ) P ( Z > − 0.7278 … ) 0.7666 … Hence, to 3 significant figures,
P ( T < 4 Y ) = 0.767 . \boxed{P(T<4Y)=0.767}. P ( T < 4 Y ) = 0.767 .
(c)
解法一
思路
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先用 E ( X ) = 780 E(X)=780 E ( X ) = 780 a a a b b b a a a b b b R R R Y Y Y X = a R + b Y X=aR+bY X = a R + bY a 2 Var ( R ) + b 2 Var ( Y ) a^2\operatorname{Var}(R)+b^2\operatorname{Var}(Y) a 2 Var ( R ) + b 2 Var ( Y ) b b b
答题过程
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The condition on the mean gives
E ( X ) = 15 a + 12 b = 780 , a + 0.8 b = 52. \begin{align*}
E(X)=&\,15a+12b=780,\\
a+0.8b=&\,52.
\end{align*} E ( X ) = a + 0.8 b = 15 a + 12 b = 780 , 52. Hence
a = 52 − 0.8 b . a=52-0.8b. a = 52 − 0.8 b . Since R R R Y Y Y
Var ( X ) = 2.25 a 2 + 0.64 b 2 . \operatorname{Var}(X)=2.25a^2+0.64b^2. Var ( X ) = 2.25 a 2 + 0.64 b 2 . Substituting a = 52 − 0.8 b a=52-0.8b a = 52 − 0.8 b
Var ( X ) = 2.25 ( 52 − 0.8 b ) 2 + 0.64 b 2 = 2.08 b 2 − 187.2 b + 6084. \begin{align*}
\operatorname{Var}(X)=&\,2.25(52-0.8b)^2+0.64b^2\\
=&\,2.08b^2-187.2b+6084.
\end{align*} Var ( X ) = = 2.25 ( 52 − 0.8 b ) 2 + 0.64 b 2 2.08 b 2 − 187.2 b + 6084. To minimise the variance, differentiate with respect to b b b
d d b Var ( X ) = 4.16 b − 187.2 , 4.16 b − 187.2 = 0 , b = 45. \begin{align*}
\frac{\mathrm{d}}{\mathrm{d}b}\operatorname{Var}(X)=&\,4.16b-187.2,\\
4.16b-187.2=&\,0,\\
b=&\,45.
\end{align*} d b d Var ( X ) = 4.16 b − 187.2 = b = 4.16 b − 187.2 , 0 , 45. Since
d 2 d b 2 Var ( X ) = 4.16 > 0 , \frac{\mathrm{d}^2}{\mathrm{d}b^2}\operatorname{Var}(X)=4.16>0, d b 2 d 2 Var ( X ) = 4.16 > 0 , this stationary point gives the minimum variance. Therefore,
a = 52 − 0.8 ( 45 ) = 16. \begin{align*}
a=&\,52-0.8(45)\\
=&\,16.
\end{align*} a = = 52 − 0.8 ( 45 ) 16. Thus Stefan should use
a = 16 , b = 45 . \boxed{a=16,\qquad b=45}. a = 16 , b = 45 .
解法二
思路
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也可以不求導，而把约束代入后的方差二次式配方。平方项的系数为正，因此当平方项等于零时，方差取得最小值；由此直接读出 b b b a a a
答题过程
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The mean condition gives
a = 52 − 0.8 b . a=52-0.8b. a = 52 − 0.8 b . Therefore,
Var ( X ) = 2.08 b 2 − 187.2 b + 6084 = 2.08 ( b − 45 ) 2 + 1872. \begin{align*}
\operatorname{Var}(X)=&\,2.08b^2-187.2b+6084\\
=&\,2.08(b-45)^2+1872.
\end{align*} Var ( X ) = = 2.08 b 2 − 187.2 b + 6084 2.08 ( b − 45 ) 2 + 1872. Since 2.08 ( b − 45 ) 2 ⩾ 0 2.08(b-45)^2\geqslant0 2.08 ( b − 45 ) 2 ⩾ 0
b = 45. b=45. b = 45. It follows that
a = 52 − 0.8 ( 45 ) = 16. \begin{align*}
a=&\,52-0.8(45)\\
=&\,16.
\end{align*} a = = 52 − 0.8 ( 45 ) 16. Hence
a = 16 , b = 45 . \boxed{a=16,\qquad b=45}. a = 16 , b = 45 . 