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CIE 9231 2025 June Paper 41 Q3

A Level / CIE / FS

CIE 9231 2025 June Paper 41 Paper · Question 3

题目

Problem

Eggs in a supermarket are sold in boxes of six. A supermarket manager wishes to model the number of broken eggs in the boxes sold in the store. A random sample of 2000 boxes is taken and the number of broken eggs recorded. The observed frequencies are shown in the table below.

Number of broken eggs0123456
Observed frequency1844143110101

(a) Use the data to estimate the probability that an egg is broken. Give your answer correct to 4 significant figures.

[1]

It is decided to carry out a goodness of fit test at the 0.5% significance level to determine whether a binomial distribution fits the data.

The observed frequencies and the expected frequencies are given in the following table.

Number of broken eggs0123456
Observed frequency1844143110101
Expected frequency1831.3aa6.0160.1190.0010.0000.000

(b) Show that a=162.6a = 162.6 correct to 1 decimal place.

[1]

(c) Carry out a goodness of fit test at the 0.5% level of significance to test whether a binomial distribution is a satisfactory model for the data.

[5]

(d) Give a reason why a binomial distribution may not be a suitable model in this situation.

[1]
题目中文翻译

超市中的鸡蛋按每盒六个出售。一位超市经理希望建立模型描述店内售出盒装鸡蛋中的破损鸡蛋数量。随机抽取 2000 盒,并记录破损鸡蛋的数量。观测频数如下表所示。

破损鸡蛋数0123456
观测频数1844143110101

(a) 使用数据估计一个鸡蛋破损的概率。答案精确到 4 位有效数字。

决定在 0.5% 显著性水平下进行拟合优度检验,以判断二项分布是否适合这些数据。

观测频数和期望频数如下表所示。

破损鸡蛋数0123456
观测频数1844143110101
期望频数1831.3aa6.0160.1190.0010.0000.000

(b) 证明 a=162.6a = 162.6,精确到 1 位小数。

(c) 在 0.5% 显著性水平下进行拟合优度检验,检验二项分布是否是这些数据的合适模型。

(d) 给出一个理由说明为什么在这种情境下二项分布可能不是合适的模型。

解答