CalcGospel 國際數學圖譜
返回

IAL 2021 Oct S3 Q5

A Level / Edexcel / S3

IAL 2021 Oct Paper · Question 5

题目

Problem

Assam produces bags of flour. The stated weight printed on the bags of flour is 3 kg.

The weights of the bags of flour are normally distributed with standard deviation 0.015 kg.

Assam weighs a random sample of 9 bags of flour and finds their mean weight is 2.977 kg.

(a) Calculate the 99% confidence interval for the mean weight of a bag of flour. Give your limits to 3 decimal places.

(3)

Assam decides to increase the amount of flour put into the bags.

(b) Explain why the confidence interval has led Assam to take this action.

(1)

After the increase a random sample of n bags of flour is taken. The sample mean weight of these n bags is 2.995 kg. A 95% confidence interval for μ\mu gave a lower limit of less than 2.991 kg.

(c) Find the maximum value of n.

(4)
(Total 8 marks)
题目中文翻译

Assam 生产面粉袋。袋上印的标称重量是 3 kg。

这些面粉袋的重量服从正态分布，标准差为 0.015 kg。

Assam 随机抽取 9 袋面粉，发现样本均值为 2.977 kg。

(a) 求面粉袋平均重量的 99% 置信区间。答案保留 3 位小数。

Assam 决定增加每袋面粉的装量。

(b) 解释为什么这个置信区间促使 Assam 做出这个决定。

增加后又随机抽取 n 袋面粉，样本均值为 2.995 kg。参数 μ\mu 的 95% 置信区间给出的下限小于 2.991 kg。

(c) 求 n 的最大值。

解答

(a)

解法一

思路

展开

总体标准差已知，且总体服从正态分布，所以使用标准正态分布构造均值的双侧置信区间。99%99\% 置信区间对应的临界值为 z0.995=2.5758z_{0.995}=2.5758。

答题过程

展开

For a 99%99\% confidence interval, the critical value is

z0.995=2.5758.z_{0.995}=2.5758.

Since σ=0.015\sigma=0.015 and n=9n=9, the confidence interval for μ\mu is

 xˉ±z0.995σn= 2.977±2.5758(0.0159)= 2.977±0.012879= (2.964121, 2.989879).\begin{align*} &\,\bar{x}\pm z_{0.995}\frac{\sigma}{\sqrt{n}}\\ =&\,2.977\pm2.5758\left(\frac{0.015}{\sqrt{9}}\right)\\ =&\,2.977\pm0.012879\\ =&\,(2.964121,\ 2.989879). \end{align*}

Therefore, to 33 decimal places, the 99%99\% confidence interval is

(2.964, 2.990) kg.\boxed{(2.964,\ 2.990)\text{ kg}}.

(b)

解法一

思路

展开

将袋上标示的 33 kg 与 (a) 的置信区间比较。整个区间都低于 33 kg，因此数据表明实际总体平均重量低于标称重量。

答题过程

展开

The stated weight of 33 kg does not lie within the 99%99\% confidence interval; in fact, the entire interval lies below 33 kg. This suggests that the mean weight is less than the stated weight, so Assam decides to increase the amount of flour in each bag.

(c)

解法一

思路

展开

95%95\% 置信区间的下限是 2.995−1.96(0.015/n)2.995-1.96(0.015/\sqrt{n})。把“下限小于 2.9912.991”直接写成不等式，注意移项时不等号方向，再求满足条件的最大正整数 nn。

答题过程

展开

The lower limit of the 95%95\% confidence interval is

2.995−1.96(0.015n).2.995-1.96\left(\frac{0.015}{\sqrt{n}}\right).

The given condition therefore gives

2.995−1.96(0.015n)<2.991.2.995-1.96\left(\frac{0.015}{\sqrt{n}}\right)<2.991.

Hence

1.96(0.015n)> 0.004n< 1.96(0.015)0.004< 7.35.\begin{align*} 1.96\left(\frac{0.015}{\sqrt{n}}\right) >&\,0.004\\ \sqrt{n} <&\,\frac{1.96(0.015)}{0.004}\\ <&\,7.35. \end{align*}

Since both sides are positive,

n<7.352=54.0225.n<7.35^2=54.0225.

Therefore, the maximum integer value of nn is

54.\boxed{54}.