题目
Problem
A task was designed to take a person 25 minutes to complete. The random variable T T T
[ 25 − k , 25 + 3 k ] [25-k,25+3k] [ 25 − k , 25 + 3 k ] where k > 0 k>0 k > 0
(a) Find
P ( 25 − 0.5 k < T < 25 + 2.5 k ) P\Bigg(
\begin{aligned}
&\,25-0.5k<T\\
&\,\quad<25+2.5k
\end{aligned}
\Bigg) P ( 25 − 0.5 k < T < 25 + 2.5 k ) (1)
Given that E ( T 2 ) = 918.76 E(T^2)=918.76 E ( T 2 ) = 918.76
(b) find the value of k k k
(6)
The task is to be completed by 50 people. Assuming that people complete the task independently,
(c) calculate the probability that at least 20 of these people complete the task in less than 25 minutes.
(3)
(Total for Question 3 is 10 marks)
题目中文翻译
一项任务设计为由一个人在 25 分钟内完成。随机变量 T T T T T T
[ 25 − k , 25 + 3 k ] [25-k,25+3k] [ 25 − k , 25 + 3 k ] 上服从均匀分布，其中 k > 0 k>0 k > 0
(a) 求
P ( 25 − 0.5 k < T < 25 + 2.5 k ) P\Bigg(
\begin{aligned}
&\,25-0.5k<T\\
&\,\quad<25+2.5k
\end{aligned}
\Bigg) P ( 25 − 0.5 k < T < 25 + 2.5 k ) (1)
已知 E ( T 2 ) = 918.76 E(T^2)=918.76 E ( T 2 ) = 918.76
(b) 求 k k k
(6)
该任务将由 50 人完成。假设每个人完成任务的情况相互独立，
(c) 计算其中至少 20 人在少于 25 分钟内完成任务的概率。
(3)
（第 3 题共 10 分）
解答
(a)
解法一
思路
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均匀分布在任一子区间内的概率等于“子区间长度除以整个分布区间的长度”。
答题过程
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The full interval has length
( 25 + 3 k ) − ( 25 − k ) = 4 k . \begin{align*}
&\,(25+3k)-(25-k)\\[4mm]
=&\,4k.
\end{align*} = ( 25 + 3 k ) − ( 25 − k ) 4 k . and the required interval has length
( 25 + 2.5 k ) − ( 25 − 0.5 k ) = 3 k . \begin{align*}
&\,(25+2.5k)\\[4mm]
&\,\hspace{2pt}-(25-0.5k)\\[4mm]
=&\,3k.
\end{align*} = ( 25 + 2.5 k ) − ( 25 − 0.5 k ) 3 k . Write the endpoints of the required interval as
c = 25 − 0.5 k , d = 25 + 2.5 k . \begin{align*}
c=&\,25-0.5k,\\[4mm]
d=&\,25+2.5k.
\end{align*} c = d = 25 − 0.5 k , 25 + 2.5 k . Therefore,
P ( c < T < d ) = 3 k 4 k = 3 4 . P(c<T<d)=\frac{3k}{4k}=\boxed{\frac34}. P ( c < T < d ) = 4 k 3 k = 4 3 .
(b)
解法一
思路
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先用均匀分布的期望与方差公式求出均值和方差，再利用“二阶矩等于方差加上均值平方”建立关于 k k k k > 0 k>0 k > 0
答题过程
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For a uniform distribution on [ 25 − k , 25 + 3 k ] [25-k,25+3k] [ 25 − k , 25 + 3 k ]
E ( T ) = ( 25 − k ) + ( 25 + 3 k ) 2 = 25 + k . \begin{align*}
&\,E(T)\\[4mm]
=&\,\frac{(25-k)+(25+3k)}2\\[4mm]
=&\,25+k.
\end{align*} = = E ( T ) 2 ( 25 − k ) + ( 25 + 3 k ) 25 + k . and
Var ( T ) = ( 4 k ) 2 12 = 4 k 2 3 . \begin{align*}
&\,\operatorname{Var}(T)\\[4mm]
=&\,\frac{(4k)^2}{12}\\[4mm]
=&\,\frac{4k^2}{3}.
\end{align*} = = Var ( T ) 12 ( 4 k ) 2 3 4 k 2 . Using the second-moment identity,
E ( T 2 ) = Var ( T ) + [ E ( T ) ] 2 . \begin{align*}
&\,E(T^2)\\[4mm]
=&\,\operatorname{Var}(T)+[E(T)]^2.
\end{align*} = E ( T 2 ) Var ( T ) + [ E ( T ) ] 2 . 918.76 = 4 k 2 3 + ( 25 + k ) 2 = 7 k 2 3 + 50 k + 625. \begin{align*}
918.76
=&\,\frac{4k^2}{3}+(25+k)^2 \\[4mm]
=&\,\frac{7k^2}{3}+50k+625.
\end{align*} 918.76 = = 3 4 k 2 + ( 25 + k ) 2 3 7 k 2 + 50 k + 625. Hence
7 k 2 + 150 k − 881.28 = 0. 7k^2+150k-881.28=0. 7 k 2 + 150 k − 881.28 = 0. Solving,
D = 150 2 + 4 ( 7 ) ( 881.28 ) = 47175.84. \begin{align*}
D=&\,150^2+4(7)(881.28)\\[4mm]
=&\,47175.84.
\end{align*} D = = 15 0 2 + 4 ( 7 ) ( 881.28 ) 47175.84. k = − 150 ± D 14 = 4.8 or − 26.228 … \begin{align*}
k=&\,\frac{-150\pm\sqrt D}{14}\\[4mm]
=&\,4.8\quad\text{or}\\[4mm]
&\,-26.228\ldots
\end{align*} k = = 14 − 150 ± D 4.8 or − 26.228 … Since k > 0 k>0 k > 0
k = 4.8 . \boxed{k=4.8}. k = 4.8 .
解法二
思路
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也可以直接积分求 E ( T 2 ) E(T^2) E ( T 2 ) 4 k 4k 4 k 1 / ( 4 k ) 1/(4k) 1/ ( 4 k ) k k k
答题过程
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Write the endpoints of the full interval as
a = 25 − k , b = 25 + 3 k . \begin{align*}
a=&\,25-k,\\[4mm]
b=&\,25+3k.
\end{align*} a = b = 25 − k , 25 + 3 k . The probability density is 1 / ( 4 k ) 1/(4k) 1/ ( 4 k ) b − a = 4 k b-a=4k b − a = 4 k
E ( T 2 ) = 1 4 k ∫ a b t 2 d t = [ t 3 12 k ] a b = b 3 − a 3 12 k = ( b − a ) ( b 2 + a b + a 2 ) 12 k = b 2 + a b + a 2 3 . \begin{align*}
&\,E(T^2)\\[4mm]
=&\,\frac1{4k}\int_a^b t^2\,\mathrm{d}t\\[4mm]
=&\,\Big[\frac{t^3}{12k}\Big]_a^b\\[4mm]
=&\,\frac{b^3-a^3}{12k}\\[4mm]
=&\,\frac{(b-a)(b^2+ab+a^2)}{12k}\\[4mm]
=&\,\frac{b^2+ab+a^2}{3}.
\end{align*} = = = = = E ( T 2 ) 4 k 1 ∫ a b t 2 d t [ 12 k t 3 ] a b 12 k b 3 − a 3 12 k ( b − a ) ( b 2 + ab + a 2 ) 3 b 2 + ab + a 2 . Expanding the three terms,
b 2 = 625 + 150 k + 9 k 2 , a b = 625 + 50 k − 3 k 2 , a 2 = 625 − 50 k + k 2 . \begin{align*}
b^2=&\,625+150k+9k^2,\\[4mm]
ab=&\,625+50k-3k^2,\\[4mm]
a^2=&\,625-50k+k^2.
\end{align*} b 2 = ab = a 2 = 625 + 150 k + 9 k 2 , 625 + 50 k − 3 k 2 , 625 − 50 k + k 2 . Hence
E ( T 2 ) = 1875 + 150 k + 7 k 2 3 = 625 + 50 k + 7 k 2 3 . \begin{align*}
&\,E(T^2)\\[4mm]
=&\,\frac{1875+150k+7k^2}{3}\\[4mm]
=&\,625+50k+\frac{7k^2}{3}.
\end{align*} = = E ( T 2 ) 3 1875 + 150 k + 7 k 2 625 + 50 k + 3 7 k 2 . Since E ( T 2 ) = 918.76 E(T^2)=918.76 E ( T 2 ) = 918.76
7 k 2 + 150 k − 881.28 = 0. 7k^2+150k-881.28=0. 7 k 2 + 150 k − 881.28 = 0. Solving this equation and using k > 0 k>0 k > 0
k = 4.8 . \boxed{k=4.8}. k = 4.8 .
(c)
解法一
思路
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对一名参与者而言，25 位于均匀分布区间中距左端点 k k k 1 / 4 1/4 1/4
答题过程
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For one person,
P ( T < 25 ) = 25 − ( 25 − k ) 4 k = 1 4 . \begin{align*}
&\,P(T<25)\\[4mm]
=&\,\frac{25-(25-k)}{4k}\\[4mm]
=&\,\frac14.
\end{align*} = = P ( T < 25 ) 4 k 25 − ( 25 − k ) 4 1 . Let X X X
X ∼ B ( 50 , 1 4 ) . X\sim\operatorname{B}\big(50,\frac14\big). X ∼ B ( 50 , 4 1 ) . Therefore,
P ( X ⩾ 20 ) = 1 − P ( X ⩽ 19 ) = 1 − 0.9861 = 0.0139 . \begin{align*}
&\,P(X\geqslant20)\\[4mm]
=&\,1-P(X\leqslant19) \\[4mm]
=&\,1-0.9861 \\[4mm]
=&\,\boxed{0.0139}.
\end{align*} = = = P ( X ⩾ 20 ) 1 − P ( X ⩽ 19 ) 1 − 0.9861 0.0139 . 