题目
Problem
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
A farmer grows watermelons on a farm. The weights, in kilograms, of the watermelons are modelled by the random variable W W W
f ( w ) = { 1 9 ( 6 − w ) ( w − 2 ) , 2 ≤ w ≤ 5 , 0 , otherwise . f(w)=
\begin{cases}
\dfrac{1}{9}(6-w)(w-2), & 2\leq w\leq5,\\
0, & \text{otherwise}.
\end{cases} f ( w ) = ⎩ ⎨ ⎧ 9 1 ( 6 − w ) ( w − 2 ) , 0 , 2 ≤ w ≤ 5 , otherwise . (a) Sketch the graph of f ( w ) f(w) f ( w )
(2)
(b) State the mode of W W W
(1)
(c) Find the mean weight of these watermelons.
(4)
Given that E ( W 2 ) = 73 5 E(W^2)=\dfrac{73}{5} E ( W 2 ) = 5 73
(d) find the standard deviation of the weights of these watermelons, giving the answer to 3 significant figures.
(2)
(e) Show that the median lies between w = 3.7 w=3.7 w = 3.7 w = 3.8 w=3.8 w = 3.8
(5)
The farmer inspects the watermelons to check whether or not they are spoiled.
The number of spoiled watermelons found by the farmer in an area of 150 m 2 150\text{ m}^2 150 m 2
The farmer randomly selects an area of 900 m 2 900\text{ m}^2 900 m 2
(f) Use a normal approximation to find the probability that the farmer finds fewer than 20 spoiled watermelons.
(5)
(Total for Question 6 is 19 marks)
题目中文翻译
本题中，你必须展示计算过程的所有步骤。
完全依赖计算器技术的解答不可接受。
一名农民在农场种植西瓜。西瓜的重量（单位：千克）由随机变量 W W W
f ( w ) = { 1 9 ( 6 − w ) ( w − 2 ) , 2 ≤ w ≤ 5 , 0 , 其他情况 . f(w)=
\begin{cases}
\dfrac{1}{9}(6-w)(w-2), & 2\leq w\leq5,\\
0, & \text{其他情况}.
\end{cases} f ( w ) = ⎩ ⎨ ⎧ 9 1 ( 6 − w ) ( w − 2 ) , 0 , 2 ≤ w ≤ 5 , 其他情况 . (a) 画出 f ( w ) f(w) f ( w )
(b) 写出 W W W
(c) 求这些西瓜的平均重量。
已知 E ( W 2 ) = 73 5 E(W^2)=\dfrac{73}{5} E ( W 2 ) = 5 73
(d) 求这些西瓜重量的标准差，答案保留 3 位有效数字。
(e) 证明中位数位于 w = 3.7 w=3.7 w = 3.7 w = 3.8 w=3.8 w = 3.8
农民检查西瓜是否变质。在农场面积为 150 m 2 150\text{ m}^2 150 m 2
农民从农场随机选取面积为 900 m 2 900\text{ m}^2 900 m 2
(f) 使用正态近似，求农民发现少于 20 个变质西瓜的概率。
（第 6 题共 19 分）
解答
(a)
解法一
思路
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先由两个因式确定零点，再看二次项系数判断抛物线开口方向。密度只在给定区间内存在，区间外为零。
答题过程
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The graph is a downward-opening quadratic with roots w = 2 w=2 w = 2 w = 6 w=6 w = 6
On the required interval, it starts at ( 2 , 0 ) (2,0) ( 2 , 0 ) w = 4 w=4 w = 4
f ( 5 ) = 1 9 . f(5)=\frac19. f ( 5 ) = 9 1 . Also, f ( w ) = 0 f(w)=0 f ( w ) = 0 w < 2 w<2 w < 2 w > 5 w>5 w > 5
(b)
解法一
思路
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密度曲线最高点的横坐标就是众数。由 (a) 的对称轴可直接读出。
答题过程
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The maximum of the quadratic occurs at w = 4 w=4 w = 4
Therefore,
Mode of W = 4 kg . \boxed{\text{Mode of }W=4\text{ kg}}. Mode of W = 4 kg .
(c)
解法一
思路
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连续型随机变量的均值为 ∫ w f ( w ) d w \int wf(w)\,\mathrm{d}w ∫ w f ( w ) d w [ 2 , 5 ] [2,5] [ 2 , 5 ]
答题过程
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E ( W ) = 1 9 ∫ 2 5 w ( 6 − w ) ( w − 2 ) d w = 1 9 ∫ 2 5 ( − w 3 + 8 w 2 − 12 w ) d w = 1 9 [ − w 4 4 + 8 w 3 3 − 6 w 2 ] 2 5 = 3.75 kg . \begin{align*}
E(W)
=&\,\frac19\int_2^5w(6-w)(w-2)\,\mathrm{d}w\\[4mm]
=&\,\frac19\int_2^5\big(-w^3+8w^2-12w\big)\,\mathrm{d}w\\[4mm]
=&\,\frac19\left[-\frac{w^4}{4}+\frac{8w^3}{3}-6w^2\right]_2^5\\[4mm]
=&\,\boxed{3.75\text{ kg}}.
\end{align*} E ( W ) = = = = 9 1 ∫ 2 5 w ( 6 − w ) ( w − 2 ) d w 9 1 ∫ 2 5 ( − w 3 + 8 w 2 − 12 w ) d w 9 1 [ − 4 w 4 + 3 8 w 3 − 6 w 2 ] 2 5 3.75 kg .
(d)
解法一
思路
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使用 Var ( W ) = E ( W 2 ) − [ E ( W ) ] 2 \operatorname{Var}(W)=E(W^2)-[E(W)]^2 Var ( W ) = E ( W 2 ) − [ E ( W ) ] 2
答题过程
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Var ( W ) = 73 5 − ( 15 4 ) 2 = 43 80 . \begin{align*}
\operatorname{Var}(W)
=&\,\frac{73}{5}-\left(\frac{15}{4}\right)^2\\[4mm]
=&\,\frac{43}{80}.
\end{align*} Var ( W ) = = 5 73 − ( 4 15 ) 2 80 43 . Hence
sd ( W ) = 43 80 = 0.73314 … = 0.733 kg (3 s.f.). \begin{align*}
\operatorname{sd}(W)
=&\,\sqrt{\frac{43}{80}}\\[4mm]
=&\,0.73314\ldots\\[4mm]
=&\,\boxed{0.733\text{ kg}}\text{ (3 s.f.).}
\end{align*} sd ( W ) = = = 80 43 0.73314 … 0.733 kg (3 s.f.).
(e)
解法一
思路
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中位数 m m m P ( W ≤ m ) = 0.5 P(W\leq m)=0.5 P ( W ≤ m ) = 0.5
答题过程
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For 2 ≤ w ≤ 5 2\leq w\leq5 2 ≤ w ≤ 5
F ( w ) = 1 9 ∫ 2 w ( − t 2 + 8 t − 12 ) d t . F(w)=\frac19\int_2^w\big(-t^2+8t-12\big)\,\mathrm{d}t. F ( w ) = 9 1 ∫ 2 w ( − t 2 + 8 t − 12 ) d t . Therefore,
F ( 3.7 ) = 0.460 … < 0.5 , F ( 3.8 ) = 0.504 … > 0.5. \begin{align*}
F(3.7)&=0.460\ldots<0.5,\\[4mm]
F(3.8)&=0.504\ldots>0.5.
\end{align*} F ( 3.7 ) F ( 3.8 ) = 0.460 … < 0.5 , = 0.504 … > 0.5. Since the cumulative probability passes through 0.5 0.5 0.5
3.7 < median < 3.8 . \boxed{3.7<\text{median}<3.8}. 3.7 < median < 3.8 .
解法二
思路
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官方评分标准也接受直接令累计概率等于 0.5 0.5 0.5
答题过程
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Let the median be m m m
1 9 ∫ 2 m ( − w 2 + 8 w − 12 ) d w = 1 2 . \frac19\int_2^m\big(-w^2+8w-12\big)\,\mathrm{d}w=\frac12. 9 1 ∫ 2 m ( − w 2 + 8 w − 12 ) d w = 2 1 . This gives
2 m 3 − 24 m 2 + 72 m − 37 = 0. 2m^3-24m^2+72m-37=0. 2 m 3 − 24 m 2 + 72 m − 37 = 0. The roots are approximately 0.65 0.65 0.65 3.79 3.79 3.79 7.6 7.6 7.6 3.79 … 3.79\ldots 3.79 … [ 2 , 5 ] [2,5] [ 2 , 5 ]
m ≈ 3.79 , \boxed{m\approx3.79}, m ≈ 3.79 , which again shows that 3.7 < m < 3.8 3.7<m<3.8 3.7 < m < 3.8
(f)
解法一
思路
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面积从 150 m 2 150\text{ m}^2 150 m 2 900 m 2 900\text{ m}^2 900 m 2 19.5 19.5 19.5
答题过程
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Let X X X
X ∼ Po ( 30 ) . X\sim\operatorname{Po}(30). X ∼ Po ( 30 ) . Using the normal approximation,
X ≈ N ( 30 , 30 ) . X\approx N(30,30). X ≈ N ( 30 , 30 ) . Hence
P ( X < 20 ) ≈ P ( Z < 19.5 − 30 30 ) = P ( Z < − 1.917 … ) = 0.0276 (approximately). \begin{align*}
P(X<20)
&\approx P\left(Z<\frac{19.5-30}{\sqrt{30}}\right)\\[4mm]
&=P(Z<-1.917\ldots)\\[4mm]
&=\boxed{0.0276}\text{ (approximately).}
\end{align*} P ( X < 20 ) ≈ P ( Z < 30 19.5 − 30 ) = P ( Z < − 1.917 … ) = 0.0276 (approximately). 
