题目

Problem

The random variable $X$ is normally distributed with mean $\mu$ and variance $36$ .

Given that

P(\mu-2k<X<\mu+2k)=0.6

(a) find the value of $k$ .

(4)

The random variable $Y$ is normally distributed with mean $\mu$ and standard deviation $\sigma$ .

Given that

2\mu=3\sigma^2 \quad\text{and}\quad P\left(Y>\frac32\mu\right)=0.0668

(b) find the value of $\mu$ and the value of $\sigma$ .

(5)

解答

(a)

解法一

思路

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$X$ 的标准差是 $6$ 。中间概率是 $0.6$ ，所以两边尾部总共是 $0.4$ ，每边是 $0.2$ 。因此

\begin{align*} P(X<\mu+2k)=0.8. \end{align*}

查表得 $z=0.8416$ 。

答题过程

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Since

\begin{align*} P(\mu-2k<X<\mu+2k)=0.6, \end{align*}

by symmetry,

\begin{align*} P(X<\mu+2k)=0.8. \end{align*}

The corresponding standard normal value is

\begin{align*} z=0.8416. \end{align*}

Since the variance is $36$ , the standard deviation is $6$ . So

\begin{align*} \frac{\mu+2k-\mu}{6}=0.8416. \end{align*}

Hence

\begin{align*} \frac{2k}{6}=&\,0.8416\\[3mm] k=&\,3(0.8416)\\[3mm] k=&\,2.5248. \end{align*}

Therefore

\begin{align*} k=2.52 \end{align*}

to $3$ significant figures.

(b)

解法一

思路

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由

\begin{align*} P\left(Y>\frac32\mu\right)=0.0668 \end{align*}

可知对应右尾概率 $0.0668$ ，所以 $z=1.5$ 。标准化后：

\begin{align*} \frac{\frac32\mu-\mu}{\sigma}=1.5. \end{align*}

再结合 $2\mu=3\sigma^2$ 解联立。

答题过程

展开

Since

\begin{align*} P\left(Y>\frac32\mu\right)=0.0668, \end{align*}

the corresponding standard normal value is

\begin{align*} z=1.5. \end{align*}

Therefore

\begin{align*} \frac{\frac32\mu-\mu}{\sigma}=1.5. \end{align*}

\begin{align*} \frac{\mu}{2\sigma}=1.5. \end{align*}

Hence

\begin{align*} \mu=3\sigma. \end{align*}

We are also given

\begin{align*} 2\mu=3\sigma^2. \end{align*}

Substitute $\mu=3\sigma$ :

\begin{align*} 2(3\sigma)=&\,3\sigma^2\\[3mm] 6\sigma=&\,3\sigma^2\\[3mm] \sigma=&\,2, \end{align*}

since $\sigma>0$ .

Then

\begin{align*} \mu=3(2)=6. \end{align*}

Therefore

\begin{align*} \mu=6,\qquad \sigma=2. \end{align*}

IAL 2024 Jan Q8

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程