题目

Problem

A company produces small bags of flour and large bags of flour. The weight, $X$ grams, of flour in a small bag is normally distributed with mean $502$ and standard deviation $3$ . One of these small bags of flour is selected at random.

(a) (i) Using standardisation find $P(X>508)$

(2)

(ii) Hence, find $P(496<X<508)$

(2)

A random sample of $4$ small bags of flour is taken.

(b) Find the probability that exactly $2$ of these small bags each contain more than $508$ grams of flour.

(3)

The weight, $Y$ grams, of flour in a large bag is normally distributed with mean $1024$ and standard deviation $\sigma$ . The $85$ th percentile for the weight of flour in a large bag is $1038.51$ to $2$ decimal places.

(2)

Given that $P(X>k)=P(Y<2k)=p$ , where $k$ and $p$ are constants,

(d) (i) find the value of $k$

(ii) find the value of $p$

(5)

解答

(a)(i)

解法一

思路

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已知 $X\sim N(502,3^2)$ 。标准化时用

\begin{align*} Z=\frac{X-\mu}{\sigma}. \end{align*}

答题过程

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Since

\begin{align*} X\sim N(502,3^2), \end{align*}

we standardise:

\begin{align*} P(X>508) =&\,P\left(Z>\frac{508-502}{3}\right)\\[3mm] =&\,P(Z>2). \end{align*}

From the standard normal distribution,

\begin{align*} P(Z>2)=0.0228. \end{align*}

Therefore

\begin{align*} P(X>508)=0.0228. \end{align*}

(a)(ii)

解法一

思路

展开

$496$ 和 $508$ 都离均值 $502$ 相差 $6$ ，也就是 $2$ 个标准差。因此中间概率是 $1$ 减去两边尾部概率。上一小题已经求出一边尾部概率是 $0.0228$ 。

答题过程

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Since

\begin{align*} \frac{508-502}{3}=2 \end{align*}

and

\begin{align*} \frac{496-502}{3}=-2, \end{align*}

we have

\begin{align*} P(496<X<508) =&\,P(-2<Z<2)\\[3mm] =&\,1-2P(Z>2)\\[3mm] =&\,1-2(0.0228)\\[3mm] =&\,0.9544. \end{align*}

\begin{align*} P(496<X<508)=0.954. \end{align*}

(b)

解法一

思路

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每一袋超过 $508$ g 的概率是上一小题的 $0.0228$ 。抽 $4$ 袋，要求刚好 $2$ 袋超过 $508$ g，用二项分布：

\begin{align*} \binom42p^2(1-p)^2. \end{align*}

答题过程

展开

Let $p=P(X>508)=0.0228$ .

For exactly $2$ bags out of $4$ ,

\begin{align*} P(\text{exactly }2) =&\,\binom42p^2(1-p)^2\\[3mm] =&\,6(0.0228)^2(1-0.0228)^2\\[3mm] =&\,0.002978\ldots. \end{align*}

Therefore the probability is

\begin{align*} 0.00298 \end{align*}

to $3$ significant figures.

(c)

解法一

思路

展开

$1038.51$ 是第 $85$ 百分位，所以

\begin{align*} P(Y<1038.51)=0.85. \end{align*}

查标准正态表，左侧概率 $0.85$ 对应 $z\approx1.0364$ 。建立标准化方程即可。

答题过程

展开

Since $1038.51$ is the $85$ th percentile,

\begin{align*} P(Y<1038.51)=0.85. \end{align*}

The corresponding standard normal value is

\begin{align*} z=1.0364\ldots. \end{align*}

Hence

\begin{align*} \frac{1038.51-1024}{\sigma} =&\,1.0364\ldots\\[3mm] \sigma =&\,\frac{14.51}{1.0364\ldots}\\[3mm] =&\,14.000\ldots. \end{align*}

Therefore

\begin{align*} \sigma=14 \end{align*}

to the nearest gram.

(d)(i)

解法一

思路

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两个概率相等：

\begin{align*} P(X>k)=P(Y<2k). \end{align*}

由于一个是右尾、一个是左尾，要让面积相等，标准化后的边界应互为相反数。也就是

\begin{align*} \frac{k-502}{3}=\frac{1024-2k}{14}. \end{align*}

答题过程

展开

For $X$ ,

\begin{align*} P(X>k)=P\left(Z>\frac{k-502}{3}\right). \end{align*}

For $Y$ ,

\begin{align*} P(Y<2k)=P\left(Z<\frac{2k-1024}{14}\right). \end{align*}

For these probabilities to be equal, the two cut-off values are opposite in sign:

\begin{align*} \frac{k-502}{3}=-\frac{2k-1024}{14}. \end{align*}

\begin{align*} \frac{k-502}{3} =&\,\frac{1024-2k}{14}\\[3mm] 14(k-502)=&\,3(1024-2k)\\[3mm] 14k-7028=&\,3072-6k\\[3mm] 20k=&\,10100\\[3mm] k=&\,505. \end{align*}

(d)(ii)

解法一

思路

展开

用 $k=505$ 回到 $P(X>k)$ 。因为 $505$ 比 $502$ 大 $3$ ，刚好是 $1$ 个标准差。

答题过程

展开

Using $k=505$ ,

\begin{align*} p =&\,P(X>505)\\[3mm] =&\,P\left(Z>\frac{505-502}{3}\right)\\[3mm] =&\,P(Z>1)\\[3mm] =&\,0.1587\ldots. \end{align*}

Therefore

\begin{align*} p=0.159 \end{align*}

to $3$ significant figures.

IAL 2025 May Q6

题目

解答

(a)(i)

解法一

思路

答题过程

(a)(ii)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

(d)(i)

解法一

思路

答题过程

(d)(ii)

解法一

思路

答题过程