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IAL 2022 Jan S3 Q1

A Level / Edexcel / S3

IAL 2022 Jan Paper · Question 1

题目

Problem

The weights, x kg, of each of 10 watermelons selected at random from Priya’s shop were recorded. The results are summarised as follows

∑x=114.2∑x2=1310.464\sum x=114.2 \qquad \sum x^2=1310.464

(a) Calculate unbiased estimates of the mean and the variance of the weights of the watermelons in Priya’s shop.

(3)

Priya researches the weight of watermelons, for the variety she has in her shop, and discovers that the weights of these watermelons are normally distributed with a standard deviation of 0.8 kg

(b) Calculate a 95% confidence interval for the mean weight of watermelons in Priya’s shop. Give the limits of your confidence interval to 2 decimal places.

(4)

Priya claims that the confidence interval in part (b) suggests that nearly all of the watermelons in her shop weigh more than 10.5 kg

(c) Use your answer to part (b) to estimate the smallest proportion of watermelons in her shop that weigh less than 10.5 kg

(3)
(Total for Question 1 is 10 marks)
题目中文翻译

Priya 商店里随机抽取了 10 个西瓜并称重，重量为 x 千克。结果汇总如下：

∑x=114.2∑x2=1310.464\sum x=114.2 \qquad \sum x^2=1310.464

(a) 求 Priya 商店中这些西瓜重量的均值和方差的无偏估计。

Priya 调查了她店里这种品种西瓜的重量，并发现这些西瓜重量服从标准差为 0.8 千克的正态分布。

(b) 求 Priya 商店中西瓜平均重量的 95% 置信区间。将置信区间的上下限保留 2 位小数。

Priya 认为 (b) 小题中的置信区间表明，她店里几乎所有西瓜都重于 10.5 千克。

(c) 利用 (b) 小题的答案，估计她店里重量小于 10.5 千克的西瓜所占的最小比例。

解答

(a)

解法一

思路

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样本均值是总体均值的无偏估计。总体方差的无偏估计要用分母 n−1n-1；题目给出 ∑x\sum x 和 ∑x2\sum x^2，所以先求 xˉ\bar x，再用 ∑x2−nxˉ2n−1\frac{\sum x^2-n\bar{x}^2}{n-1}。

答题过程

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The unbiased estimate of the population mean is

xˉ=∑xn=114.210=11.42 kg.\bar{x}=\frac{\sum x}{n}=\frac{114.2}{10}=\boxed{11.42}\text{ kg}.

The unbiased estimate of the population variance is

s2= ∑x2−nxˉ2n−1= 1310.464−10(11.42)29= 0.7 kg2.\begin{align*} s^2 =&\,\frac{\sum x^2-n\bar{x}^2}{n-1}\\ =&\,\frac{1310.464-10(11.42)^2}{9}\\ =&\,\boxed{0.7}\text{ kg}^2. \end{align*}

(b)

解法一

思路

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总体标准差已知为 0.80.8 kg，因此使用标准正态临界值 1.961.96，而不是用样本方差和 tt 分布。把 xˉ=11.42\bar{x}=11.42、σ=0.8\sigma=0.8、n=10n=10 代入均值的 95%95\% 置信区间公式。

答题过程

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For a 95%95\% confidence interval, the critical value is z=1.96z=1.96. Therefore,

xˉ±zσn= 11.42±1.960.810= 11.42±0.495845…\begin{align*} \bar{x}\pm z\frac{\sigma}{\sqrt n} =&\,11.42\pm1.96\frac{0.8}{\sqrt{10}}\\ =&\,11.42\pm0.495845\ldots \end{align*}

This gives

(10.9241…,11.9158…).(10.9241\ldots,11.9158\ldots).

Hence, to 2 decimal places, the 95%95\% confidence interval is

(10.92,11.92) kg.\boxed{(10.92,11.92)}\text{ kg}.

(c)

解法一

思路

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总体标准差固定时，总体均值越大，低于 10.510.5 kg 的比例越小。因此应从 (b) 的置信区间中选取最大的可能均值，即上限约 11.9211.92 kg。以此建立单个西瓜重量的正态模型，再求左尾概率。

答题过程

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To estimate the smallest proportion below 10.510.5 kg, use the upper endpoint of the confidence interval as the population mean. Let YY be the weight of a watermelon. Then

Y∼N(11.9158…,0.82).Y\sim N(11.9158\ldots,0.8^2).

Therefore,

P(Y<10.5)= P(Z<10.5−11.9158…0.8)= P(Z<−1.7698…)= 0.03838…≈ 0.038.\begin{align*} P(Y<10.5) =&\,P\bigg( Z<\frac{10.5-11.9158\ldots}{0.8} \bigg)\\ =&\,P(Z<-1.7698\ldots)\\ =&\,0.03838\ldots\\ \approx&\,\boxed{0.038}. \end{align*}

Thus the estimated smallest proportion is approximately 0.0380.038, or 3.8%3.8\%.