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S1 2022 Oct Q5

A Level / Edexcel / S1

IAL 2022 Oct Paper · Question 5

题目

Problem

The weights, WW grams, of kiwi fruit grown on a farm are normally distributed with mean 8080 grams and standard deviation 88 grams. The table shows the classifications of the kiwi fruit by their weight, where kk is a positive constant.

\multicolumn2∣c∣Small\multicolumn3c∣LargeTinyPetiteExtraJumboMegaw<6666⩽w<7070⩽w<8080⩽w<kw⩾k\begin{array}{|c|c|c|c|c|} \hline \multicolumn{2}{|c|}{\text{Small}}&\multicolumn{3}{c|}{\text{Large}}\\ \hline \text{Tiny}&\text{Petite}&\text{Extra}&\text{Jumbo}&\text{Mega}\\ \hline w<66&66\leqslant w<70&70\leqslant w<80&80\leqslant w<k&w\geqslant k\\ \hline \end{array}

One kiwi fruit is selected at random from those grown on the farm.

(a) Find the probability that this kiwi fruit is Large.

(3)

35%35\% of the kiwi fruit are Jumbo.

(b) Find the value of kk to one decimal place.

(4)

75%75\% of Tiny kiwi fruit weigh more than yy grams.

(c) Find the value of yy giving your answer to one decimal place.

(5)
题目中文翻译

某农场种植的猕猴桃重量 WW g 服从平均数为 8080 g、标准差为 88 g 的正态分布。题目表格按重量给出了猕猴桃的分类，其中 kk 为正常数。

从该农场种植的猕猴桃中随机选取一个。

(a) 求这个猕猴桃属于 Large 类的概率。

35%35\% 的猕猴桃属于 Jumbo 类。

(b) 求 kk 的值，答案保留到小数点后 11 位。

Tiny 类猕猴桃中有 75%75\% 的重量大于 yy g。

(c) 求 yy 的值，答案保留到小数点后 11 位。

解答

(a)

解法一

思路

展开

Large 包含 Extra、Jumbo、Mega，也就是 w⩾70w\geqslant70。

答题过程

展开 P(W⩾70)= P(Z⩾70−808)= P(Z⩾−1.25)= P(Z⩽1.25)= 0.8944.\begin{align*} P(W\geqslant70) =&\,P\left(Z\geqslant\frac{70-80}{8}\right)\\ =&\,P(Z\geqslant-1.25)\\ =&\,P(Z\leqslant1.25)\\ =&\,0.8944. \end{align*}

(b)

解法一

思路

展开

Jumbo 是 80⩽W<k80\leqslant W<k，概率为 0.350.35。由于 P(W<80)=0.5P(W<80)=0.5，所以 P(W<k)=0.85P(W<k)=0.85。

答题过程

展开

Since 35%35\% are Jumbo,

P(80⩽W<k)=0.35.\begin{align*} P(80\leqslant W<k)=0.35. \end{align*}

As P(W<80)=0.5P(W<80)=0.5,

P(W<k)=0.85.\begin{align*} P(W<k)=0.85. \end{align*}

The corresponding standard normal value is

z=1.0364.\begin{align*} z=1.0364. \end{align*}

So

k−808=1.0364.\begin{align*} \frac{k-80}{8}=1.0364. \end{align*}

Hence

k=80+1.0364(8)=88.2912.\begin{align*} k=80+1.0364(8)=88.2912. \end{align*}

Therefore

k=88.3.\begin{align*} k=88.3. \end{align*}

(c)

解法一

思路

展开

Tiny 是 W<66W<66。其中 75%75\% 大于 yy，所以 25%25\% 小于 yy。先算 P(W<66)P(W<66)，再取它的 25%25\% 作为 P(W<y)P(W<y)。

答题过程

展开 P(W<66)= P(Z<66−808)= P(Z<−1.75)= 0.0401.\begin{align*} P(W<66) =&\,P\left(Z<\frac{66-80}{8}\right)\\ =&\,P(Z<-1.75)\\ =&\,0.0401. \end{align*}

Since 75%75\% of Tiny kiwi fruit weigh more than yy grams,

P(W<y)=0.25(0.0401)=0.010025.\begin{align*} P(W<y)=0.25(0.0401)=0.010025. \end{align*}

This corresponds to approximately

z=−2.326.\begin{align*} z=-2.326. \end{align*}

So

y−808=−2.326.\begin{align*} \frac{y-80}{8}=-2.326. \end{align*}

Hence

y=80−2.326(8)=61.392.\begin{align*} y=80-2.326(8)=61.392. \end{align*}

Therefore

y=61.4.\begin{align*} y=61.4. \end{align*}

解法二

思路

展开

也可以直接计算 P(W>y)P(W>y)：先取所有非 Tiny 猕猴桃，再加上 Tiny 中位于 yy 上方的 75%75\%。

答题过程

展开

As

P(W<66)=P(Z<−1.75)=0.0401,P(W<66)=P(Z<-1.75)=0.0401,

we have

P(W>y)= P(W⩾66)+0.75P(W<66)= 0.9599+0.75(0.0401)= 0.989975.\begin{align*} P(W>y) =&\,P(W\geqslant66)+0.75P(W<66)\\ =&\,0.9599+0.75(0.0401)\\ =&\,0.989975. \end{align*}

Thus P(W<y)=0.010025P(W<y)=0.010025, giving z=−2.326z=-2.326. Hence

y= 80+8(−2.326)= 61.4\begin{align*} y =&\,80+8(-2.326)\\ =&\,61.4 \end{align*}

to one decimal place.