题目
Problem
A machine produces packets of cereal with weights, X kg, such that X ~ N(μ, σ^2).
A random sample of these packets is taken and a 95% confidence interval for μ is found to be (509.1572, 510.8428).
(a) Find a 90% confidence interval for μ based on the same sample of packets.
You must show your working and give your confidence limits to one decimal place.
(6)
Using three different random samples, three 90% confidence intervals are to be found.
(b) Calculate the probability that at least 1 of these intervals will contain μ.
(2)
(Total for Question 3 is 8 marks)
题目中文翻译
某机器生产一批批谷物包装，重量 X kg，且 X ~ N(μ, σ^2)。
从这些包装中抽取一个随机样本，得到 μ 的 95% 置信区间为 (509.1572, 510.8428)。
(a) 根据同一样本，求 μ 的 90% 置信区间。你必须写出计算过程，并将置信区间端点保留到小数点后 1 位。
使用三个不同的随机样本，将分别求出三个 90% 置信区间。
(b) 计算有至少 1 个这样的区间包含 μ 的概率。
解答
(a)
解法一
思路
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原 95 % 95\% 95% x ‾ \overline{x} x 2 × 1.96 × σ n 2\times1.96\times\frac{\sigma}{\sqrt n} 2 × 1.96 × n σ 90 % 90\% 90% 1.6449 1.6449 1.6449
答题过程
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The sample mean is the midpoint of the given confidence interval:
x ‾ = 509.1572 + 510.8428 2 = 510. \begin{align*}
\overline{x}
=&\,\frac{509.1572+510.8428}{2}\\
=&\,510.
\end{align*} x = = 2 509.1572 + 510.8428 510. Let the standard error be σ n \frac{\sigma}{\sqrt n} n σ 95 % 95\% 95%
2 ( 1.96 ) σ n = 510.8428 − 509.1572 = 1.6856. \begin{align*}
2(1.96)\frac{\sigma}{\sqrt n}
=&\,510.8428-509.1572\\
=&\,1.6856.
\end{align*} 2 ( 1.96 ) n σ = = 510.8428 − 509.1572 1.6856. Therefore,
σ n = 1.6856 2 ( 1.96 ) = 0.43. \begin{align*}
\frac{\sigma}{\sqrt n}
=&\,\frac{1.6856}{2(1.96)}\\
=&\,0.43.
\end{align*} n σ = = 2 ( 1.96 ) 1.6856 0.43. For a 90 % 90\% 90% z = 1.6449 z=1.6449 z = 1.6449
510 ± 1.6449 ( 0.43 ) = 510 ± 0.707307 … = ( 509.29269 … , 510.70731 … ) . \begin{align*}
&\,510\pm1.6449(0.43)\\
=&\,510\pm0.707307\ldots\\
=&\,(509.29269\ldots,\ 510.70731\ldots).
\end{align*} = = 510 ± 1.6449 ( 0.43 ) 510 ± 0.707307 … ( 509.29269 … , 510.70731 … ) . Thus, to 1 decimal place, the 90 % 90\% 90%
( 509.3 , 510.7 ) . \boxed{(509.3,\ 510.7)}. ( 509.3 , 510.7 ) .
(b)
解法一
思路
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三个区间来自三个不同的随机样本，所以可把它们视为独立。每个 90 % 90\% 90% μ \mu μ 0.9 0.9 0.9 Y Y Y μ \mu μ Y ∼ B ( 3 , 0.9 ) Y\sim B(3,0.9) Y ∼ B ( 3 , 0.9 )
答题过程
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Let Y Y Y μ \mu μ
Y ∼ B ( 3 , 0.9 ) . Y\sim\operatorname{B}(3,0.9). Y ∼ B ( 3 , 0.9 ) . Therefore,
P ( Y ≥ 1 ) = 1 − P ( Y = 0 ) = 1 − ( 0.1 ) 3 = 0.999 . \begin{align*}
P(Y\geq1)
=&\,1-P(Y=0)\\
=&\,1-(0.1)^3\\
=&\,\boxed{0.999}.
\end{align*} P ( Y ≥ 1 ) = = = 1 − P ( Y = 0 ) 1 − ( 0.1 ) 3 0.999 . 