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

(a) 证明当 $0 \le a \le 4$ 时，

随机抽取了 150 个此类航班，并将延误情况整理如下表。

(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。

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

For $0\leqslant a\leqslant4$,

as required.

也可以用梯形面积求区间概率。密度函数在 $[0,5)$ 上是一条直线，因此 $a$ 到 $a+1$ 下方的面积就是宽为 1、两条平行边分别为 $f(a)$ 与 $f(a+1)$ 的梯形面积。

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

Hence the required probability is its area:

as required.

先求累计分布函数，再用区间两端的累计概率之差。将 $a+1$ 与 $a$ 代入后，平方差自然化为题目要求的一次式；端点 $a=4$ 时使用 $F(5)=1$，同样适用。

For $0\leqslant t\leqslant5$,

Thus the cumulative distribution function is

For $0\leqslant a\leqslant4$, both endpoints lie in $[0,5]$. Since $T$ is continuous,

as required.

使用卡方拟合优度检验。由 (a) 令 $a=0,1,2,3,4$，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。

The hypotheses are:

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

Therefore, the expected frequencies are

The test statistic is

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

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

Since $10.1>9.488$, the test statistic lies in the critical region. Therefore, $H_0$ is rejected. There is sufficient evidence at the $5\%$ significance level that the given probability density function is not a suitable model for these flight delays.

由 (a) 求出的五格期望频数仍为 $6,18,30,42,54$。展开卡方公式，利用观察频数与期望频数总和均为 150，便可直接计算 $\sum O^2/E-150$。

The hypotheses are

From part (a), the expected frequencies are

Since $\sum O=\sum E=150$,

There are 5 classes and no estimated parameters, so the number of degrees of freedom is $5-1=4$. The critical value at the $5\%$ significance level is $9.488$.

Since $10.1>9.488$, reject $H_0$. There is sufficient evidence at the $5\%$ significance level that the given p.d.f. is not a suitable model for these flight delays.