一位生物学家认为，某种青蛙的质量服从正态分布。他随机选取这种青蛙 $120$ 只，记录每只青蛙的质量 $x$，单位为克。结果汇总如下。

这位生物学家利用这 $120$ 只青蛙的质量，计算相应总体均值和方差的无偏估计，结果分别为 $236.5\text{ g}$ 和 $134.56\text{ g}^2$。

大部分相应的期望频数如下表所示，均精确到小数点后 $2$ 位。这些期望频数是根据生物学家的模型计算的。

(a) 求 $c$ 和 $d$ 的值，答案精确到小数点后两位。

(b) 在 $5\%$ 的显著性水平下进行拟合优度检验，考察正态分布是否是这些数据的合适模型。

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。

用给定均值和方差将边界标准化。计算各区间的正态概率，再乘以样本量。

Under the fitted model $X\sim N(236.5,134.56)$, the standard deviation is $11.6$. Thus

The upper-tail expectation is

To two decimal places, $c=12.10$ and $d=2.57$.

先用正态区间概率求 $c$，再利用全部期望频数的和为 $120$ 求 $d$。使用表中已舍入的数据会改变最后一位。

Using $X\sim N(236.5,11.6^2)$,

The expected frequencies sum to $120$. Using the rounded table entries and $c=12.10$ gives

The difference from the direct upper-tail value $2.57$ is due to using the rounded entries in the table.

两端分别合并期望频数小于 $5$ 的格。均值与方差是用样本估计的，所以再扣除两个自由度。

Test $H_0$ that a normal distribution fits the frog masses against $H_1$ that it does not. Combine the first two classes and combine the last two classes. The resulting observed counts are $11,20,39,40,10$. Recomputing the expected counts from the fitted normal distribution without intermediate rounding gives approximately $9.2944,25.2203,39.7136,31.1013,14.6705$. All five exceed $5$.

The statistic is

There are $5-1-2=2$ degrees of freedom. The upper $5\%$ critical value is $5.991$. Since the statistic is smaller, we fail to reject $H_0$. There is insufficient evidence at the $5\%$ level that a normal distribution is unsuitable for these frog masses.