一位交通专家声称，在一段繁忙高速公路上每天发生的故障次数服从均值为 $0.7$ 的 Poisson 分布。记录了 200 天内每天的故障次数。下表给出了观察频数，以及使用该专家的分布得到的一些期望频数。

(a) 求 $m$ 和 $n$ 的值，精确到 3 位小数。

(b) 在 5% 显著性水平下进行拟合优度检验，以调查该专家的声称。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

把泊松概率乘以 $200$ 得到期望频数。至少五次的概率用补集计算，最后才取三位小数。

For $X\sim\operatorname{Poisson}(0.7)$,

For the last category,

Thus $m=69.522$ and $n=0.157$, to three decimal places.

将尾部小期望频数的格与相邻格合并。用合并后的格数计算自由度，泊松均值已给定。

The null hypothesis is that $\operatorname{Poisson}(0.7)$ fits the daily breakdown counts. The alternative is that it does not. Combine counts of $3$, $4$ and $5$ or more. The observed count of this combined category is $7+3+3=13$, and its expected count is $200\Pr(X\ge3)=6.828316\ldots$. The remaining expected counts are all above $5$.

Using unrounded expected frequencies,

With four categories and no estimated parameter, there are $3$ degrees of freedom. The upper $5\%$ critical value is $7.815$. Since $7.15600\ldots<7.815$, we fail to reject $H_0$. There is insufficient evidence at the $5\%$ level that the expert’s Poisson model fails to fit these breakdown counts.