题目
Problem
A company produces a certain type of mug. The masses of these mugs are normally distributed with mean μ \mu μ
229.1 229.6 230.9 231.2 231.7 229.1 \quad 229.6 \quad 230.9 \quad 231.2 \quad 231.7 229.1 229.6 230.9 231.2 231.7
(a) Find a 95% confidence interval for μ \mu μ
(4)
Sonia plans to take 20 random samples, each of 5 mugs. A 95% confidence interval for μ \mu μ
(b) Find the probability that more than 3 of these intervals will not contain μ \mu μ
(3)
(Total for Question 4 is 7 marks)
题目中文翻译
某公司生产一种特定的杯子。杯子的质量服从均值为 μ、标准差为 1.2 克的正态分布。抽取 5 个杯子的随机样本，测得其质量如下。
229.1 229.6 230.9 231.2 231.7
(a) 求 μ 的 95% 置信区间，区间端点保留到小数点后 1 位。
Sonia 计划取 20 个随机样本，每个样本包含 5 个杯子。对每个样本都要确定一个 μ 的 95% 置信区间。
(b) 求这 20 个区间中有超过 3 个不包含 μ 的概率。
解答
(a)
解法一
思路
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母体服从正态分布且标准差 σ = 1.2 \sigma=1.2 σ = 1.2 z 0.975 = 1.96 z_{0.975}=1.96 z 0.975 = 1.96 x ‾ ± 1.96 σ n \overline{x}\pm1.96\frac{\sigma}{\sqrt{n}} x ± 1.96 n σ 95 % 95\% 95%
答题过程
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The sample mean is
x ‾ = 229.1 + 229.6 + 230.9 + 231.2 + 231.7 5 = 230.5. \begin{align*}
\overline{x}
=&\,\frac{229.1+229.6+230.9+231.2+231.7}{5}\\
=&\,230.5.
\end{align*} x = = 5 229.1 + 229.6 + 230.9 + 231.2 + 231.7 230.5. Since the population standard deviation is known, the 95 % 95\% 95% μ \mu μ
x ‾ ± 1.96 σ n = 230.5 ± 1.96 ( 1.2 5 ) = ( 229.44815 … , 231.55185 … ) . \begin{align*}
&\,\overline{x}\pm1.96\frac{\sigma}{\sqrt{n}}\\
=&\,230.5\pm1.96\left(\frac{1.2}{\sqrt{5}}\right)\\
=&\,(229.44815\ldots,\ 231.55185\ldots).
\end{align*} = = x ± 1.96 n σ 230.5 ± 1.96 ( 5 1.2 ) ( 229.44815 … , 231.55185 … ) . Therefore, to 1 decimal place, the confidence interval is
( 229.4 , 231.6 ) grams . \boxed{(229.4,\ 231.6)\text{ grams}}. ( 229.4 , 231.6 ) grams .
(b)
解法一
思路
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每个 95 % 95\% 95% μ \mu μ 0.05 0.05 0.05 X X X μ \mu μ X ∼ B ( 20 , 0.05 ) X\sim B(20,0.05) X ∼ B ( 20 , 0.05 ) X ≥ 4 X\geq4 X ≥ 4
答题过程
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Let X X X μ \mu μ
X ∼ B ( 20 , 0.05 ) . X\sim\operatorname{B}(20,0.05). X ∼ B ( 20 , 0.05 ) . Therefore,
P ( X > 3 ) = 1 − P ( X ≤ 3 ) = 1 − 0.984098 … = 0.015901 … = 0.0159 (3 s.f.) . \begin{align*}
P(X>3)
=&\,1-P(X\leq3)\\
=&\,1-0.984098\ldots\\
=&\,0.015901\ldots\\
=&\,\boxed{0.0159}\quad\text{(3 s.f.)}.
\end{align*} P ( X > 3 ) = = = = 1 − P ( X ≤ 3 ) 1 − 0.984098 … 0.015901 … 0.0159 (3 s.f.) . Equivalently, if Y Y Y μ \mu μ Y ∼ B ( 20 , 0.95 ) Y\sim\operatorname{B}(20,0.95) Y ∼ B ( 20 , 0.95 ) P ( Y ≤ 16 ) P(Y\leq16) P ( Y ≤ 16 )
