题目

Problem

The random variable X is defined as

X=4A-3B

where A and B are independent and

A \sim N(15,5^2) \qquad B \sim N(10,4^2)

(a) Find $P(X<40)$

(4)

The random variable C is such that $C \sim N(20,\sigma^2)$ The random variables $C_1, C_2$ and $C_3$ are independent and each has the same distribution as C The random variable D is defined as

D=\sum_{i=1}^3 C_i

Given that $P(A+B+D<76)=0.2420$ and that A, B and D are independent,

(b) showing your working clearly, find the standard deviation of C

(6)

(Total for Question 7 is 10 marks)

题目中文翻译

随机变量 X 定义为

X=4A-3B

其中 A 和 B 相互独立，并且

A \sim N(15,5^2) \qquad B \sim N(10,4^2)

(a) 求 $P(X<40)$ 。

随机变量 C 满足 $C \sim N(20,\sigma^2)$ 。随机变量 $C_1, C_2, C_3$ 相互独立，并且都与 C 服从同样的分布。随机变量 D 定义为

D=\sum_{i=1}^3 C_i

已知 $P(A+B+D<76)=0.2420$ ，且 A、B、D 相互独立，

(b) 请清楚写出过程，求 C 的标准差。

解答

(a)

解法一

思路

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$X=4A-3B$ 是两个独立正态变量的线性组合，因此仍服从正态分布。计算均值时保留系数符号，计算方差时则使用系数的平方；确定 $X$ 的分布后再标准化求概率。

答题过程

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Since $A$ and $B$ are independent,

\begin{align*} \operatorname E(X) =&\,4\operatorname E(A)-3\operatorname E(B)\\ =&\,4(15)-3(10)\\ =&\,30 \end{align*}

and

\begin{align*} \operatorname{Var}(X) =&\,4^2\operatorname{Var}(A)\\ &\,\hspace{2pt}+(-3)^2\operatorname{Var}(B)\\ =&\,4^2(5^2)+3^2(4^2)\\ =&\,544. \end{align*}

Therefore,

X\sim\mathrm N(30,544).

Hence

\begin{align*} \operatorname P(X<40) =&\,\operatorname P\left( Z<\frac{40-30}{\sqrt{544}} \right)\\ =&\,\operatorname P(Z<0.4287\ldots)\\ =&\,\boxed{0.666}. \end{align*}

(b)

解法一

思路

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先由三个独立同分布的 $C_i$ 求出 $D$ 的均值和方差，再与独立的 $A$ 、 $B$ 合并。已知左尾概率为 $0.2420$ ，查标准正态分布得到对应的 $z$ 值约为 $-0.700$ ，由标准化方程反解 $\sigma$ 。

答题过程

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Since

D=C_1+C_2+C_3,

where $C_1,C_2,C_3$ are independent and each has distribution $\mathrm N(20,\sigma^2)$ ,

\begin{aligned} \operatorname E(D)=&\,3(20)=60,\\ \operatorname{Var}(D)=&\,3\sigma^2. \end{aligned}

Let

S=A+B+D.

Since $A$ , $B$ and $D$ are independent,

\begin{align*} \operatorname E(S) =&\,15+10+60\\ =&\,85 \end{align*}

and

\begin{align*} \operatorname{Var}(S) =&\,5^2+4^2+3\sigma^2\\ =&\,41+3\sigma^2. \end{align*}

Thus

S\sim\mathrm N(85,41+3\sigma^2).

Given that $\operatorname P(S<76)=0.2420$ , the corresponding standard normal value is

\operatorname P(Z<-0.700)\approx0.2420.

Therefore,

\frac{76-85}{\sqrt{41+3\sigma^2}}=-0.700.

Rearranging,

\begin{align*} \frac{-9}{\sqrt{41+3\sigma^2}} =&\,-0.700\\ \sqrt{41+3\sigma^2} =&\,\frac{9}{0.700}\\ 41+3\sigma^2 =&\,\left(\frac{9}{0.700}\right)^2\\ 3\sigma^2 =&\,\left(\frac{9}{0.700}\right)^2-41\\ \sigma =&\,6.437\ldots. \end{align*}

Therefore, the standard deviation of $C$ is

\boxed{6.44}.

IAL 2023 June S3 Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程