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IAL 2023 June S3 Q4

A Level / Edexcel / S3

IAL 2023 June Paper · Question 4

题目

Problem

It is suggested that the delay, in hours, of certain flights from a particular country may be modelled by the continuous random variable, T, with probability density function

f(t)={2t25,0≤t<50,otherwisef(t)= \begin{cases} \frac{2t}{25}, & 0 \le t < 5 \\ 0, & \text{otherwise} \end{cases}

(a) Show that for 0≤a≤40 \le a \le 4

P(a≤T<a+1)=125(2a+1)P(a \le T < a+1)=\frac{1}{25}(2a+1)
(3)

A random sample of 150 of these flights is taken. The delays are summarised in the table below.

Delay (t hours)Frequency
0≤t<10 \le t < 110
1≤t<21 \le t < 213
2≤t<32 \le t < 324
3≤t<43 \le t < 435
4≤t<54 \le t < 568

(b) Test, at the 5% significance level, whether the given probability density function is a suitable model for these delays. You should state your hypotheses, expected frequencies, test statistic and the critical value used.

(8)
(Total for Question 4 is 11 marks)
题目中文翻译

据推测，某国某些航班的延误时间（小时）可由连续随机变量 TT 建模，其概率密度函数为

f(t)={2t25,0≤t<50,otherwisef(t)= \begin{cases} \frac{2t}{25}, & 0 \le t < 5 \\ 0, & \text{otherwise} \end{cases}

(a) 证明当 0≤a≤40 \le a \le 4 时，

P(a≤T<a+1)=125(2a+1)P(a \le T < a+1)=\frac{1}{25}(2a+1)

随机抽取了 150 个此类航班，并将延误情况整理如下表。

延误时间（t 小时）频数
0≤t<10 \le t < 110
1≤t<21 \le t < 213
2≤t<32 \le t < 324
3≤t<43 \le t < 435
4≤t<54 \le t < 568

(b) 在 5% 显著性水平下，检验给定的概率密度函数是否适合作为这些延误时间的模型。 你应写出原假设、期望频数、检验统计量和所用临界值。

解答

(a)

解法一

思路

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区间概率等于概率密度函数在该区间下的面积。对 2t25\frac{2t}{25} 从 aa 积分到 a+1a+1，再展开平方差，即可自然推出题目给定的形式。

答题过程

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For 0⩽a⩽40\leqslant a\leqslant4,

P⁡(a⩽T<a+1)= ∫aa+12t25 dt= [t225]aa+1= (a+1)2−a225= a2+2a+1−a225= 125(2a+1),\begin{align*} \operatorname P(a\leqslant T<a+1) =&\,\int_a^{a+1}\frac{2t}{25}\,\mathrm{d}t\\ =&\,\left[\frac{t^2}{25}\right]_a^{a+1}\\ =&\,\frac{(a+1)^2-a^2}{25}\\ =&\,\frac{a^2+2a+1-a^2}{25}\\ =&\,\frac{1}{25}(2a+1), \end{align*}

as required.

解法二

思路

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官方评分资料也接受用梯形面积求区间概率。密度函数在 [0,5)[0,5) 上是一条直线，因此 aa 到 a+1a+1 下方的面积就是宽为 1、两条平行边分别为 f(a)f(a) 与 f(a+1)f(a+1) 的梯形面积。

答题过程

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Over the interval [a,a+1)[a,a+1), the graph of f(t)=2t25f(t)=\dfrac{2t}{25} is a straight line. The sum of the two parallel sides of the trapezium is

f(a)+f(a+1)= 2a25+2(a+1)25= 4a+225.\begin{align*} f(a)+f(a+1) =&\,\frac{2a}{25}+\frac{2(a+1)}{25}\\ =&\,\frac{4a+2}{25}. \end{align*}

Hence the required probability is its area:

P⁡(a⩽T<a+1)= 12(4a+225)(1)= 2a+125,\begin{align*} \operatorname P(a\leqslant T<a+1) =&\,\frac12\left(\frac{4a+2}{25}\right)(1)\\ =&\,\frac{2a+1}{25}, \end{align*}

as required.

(b)

解法一

思路

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使用卡方拟合优度检验。由 (a) 令 a=0,1,2,3,4a=0,1,2,3,4，先得到五个区间的理论概率，再乘样本量 150 求期望频数。题目没有用样本估计任何参数，因此自由度是类别数减 1。

答题过程

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The hypotheses are:

H0: The given p.d.f. is a suitable model for the delays,H1: The given p.d.f. is not a suitable model for the delays.\begin{aligned} H_0:&\ \text{The given p.d.f. is a suitable model for the delays},\\ H_1:&\ \text{The given p.d.f. is not a suitable model for the delays}. \end{aligned}

From part (a), for a=0,1,2,3,4a=0,1,2,3,4, the class probabilities are

125,325,525,725,925.\frac1{25},\quad\frac3{25},\quad\frac5{25}, \quad\frac7{25},\quad\frac9{25}.

Therefore, the expected frequencies are

150(125)= 6,150(325)= 18,150(525)= 30,150(725)= 42,150(925)= 54.\begin{align*} 150\left(\frac1{25}\right)=&\,6,\\ 150\left(\frac3{25}\right)=&\,18,\\ 150\left(\frac5{25}\right)=&\,30,\\ 150\left(\frac7{25}\right)=&\,42,\\ 150\left(\frac9{25}\right)=&\,54. \end{align*}

The test statistic is

χ2= ∑(O−E)2E= (10−6)26 +(13−18)218 +(24−30)230 +(35−42)242 +(68−54)254= 10.0518…= 10.1.\begin{align*} \chi^2 =&\,\sum\frac{(O-E)^2}{E}\\ =&\,\frac{(10-6)^2}{6}\\ &\,\hspace{2pt}+\frac{(13-18)^2}{18}\\ &\,\hspace{4pt}+\frac{(24-30)^2}{30}\\ &\,\hspace{6pt}+\frac{(35-42)^2}{42}\\ &\,\hspace{8pt}+\frac{(68-54)^2}{54}\\ =&\,10.0518\ldots\\ =&\,10.1. \end{align*}

There are five classes and no parameters have been estimated from the sample, so

ν=5−1=4.\nu=5-1=4.

At the 5%5\% significance level, the critical value is

χ42(0.05)=9.488.\chi^2_4(0.05)=9.488.

Since 10.1>9.48810.1>9.488, the test statistic lies in the critical region. Therefore, H0H_0 is rejected. There is sufficient evidence at the 5%5\% significance level that the given probability density function is not a suitable model for these flight delays.