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S1 2022 Jan Q5

A Level / Edexcel / S1

IAL 2022 Jan Paper · Question 5

题目

Problem

Jia writes a computer program that randomly generates values from a normal distribution. He sets the mean as 4040 and the standard deviation as 2.42.4

(a) Find the probability that a particular value generated by the computer program is less than 3737

(3)

Jia changes the mean to mm but leaves the standard deviation as 2.42.4 The computer program then randomly generates 22 independent values from this normal distribution. The probability that both of these values are greater than 3232 is 0.160.16

(b) Find the value of mm, giving your answer to 22 decimal places.

(4)

Jia now changes the mean to 44 and the standard deviation to 88 The computer program then randomly generates 55 independent values from this normal distribution.

(c) Find the probability that at least one of these values is negative.

(4)
题目中文翻译

Jia 编写了一个程序，用来从正态分布中随机生成数值。他把均值设为 4040，标准差设为 2.42.4。

(a) 求程序生成的某一个数值小于 3737 的概率。

Jia 把均值改为 mm，但标准差仍为 2.42.4。程序随后从该正态分布中独立随机生成 22 个数值。两个数值都大于 3232 的概率为 0.160.16。

(b) 求 mm，答案保留到小数点后 22 位。

Jia 现在把均值改为 44，标准差改为 88。程序随后从该正态分布中独立随机生成 55 个数值。

(c) 求至少有一个数值为负数的概率。

解答

(a)

解法一

思路

展开

标准化 3737，再查左尾概率。

答题过程

展开 P(X<37)= P(Z<37−402.4)= P(Z<−1.25)= 1−0.8944= 0.1056.\begin{align*} P(X<37) =&\,P\left(Z<\frac{37-40}{2.4}\right)\\ =&\,P(Z<-1.25)\\ =&\,1-0.8944\\ =&\,0.1056. \end{align*}

So the probability is

0.106\begin{align*} 0.106 \end{align*}

to 33 significant figures.

(b)

解法一

思路

展开

两次都大于 3232 的概率为 0.160.16，所以一次大于 3232 的概率是 0.16=0.4\sqrt{0.16}=0.4。因此 P(X<32)=0.6P(X<32)=0.6，对应 z=0.2533z=0.2533。

答题过程

展开

For one generated value,

P(X>32)=0.16=0.4.\begin{align*} P(X>32)=\sqrt{0.16}=0.4. \end{align*}

So

P(X<32)=0.6.\begin{align*} P(X<32)=0.6. \end{align*}

The corresponding standard normal value is

z=0.2533.\begin{align*} z=0.2533. \end{align*}

Hence

32−m2.4=0.2533.\begin{align*} \frac{32-m}{2.4}=0.2533. \end{align*}

Therefore

m= 32−0.2533(2.4)= 31.392…\begin{align*} m =&\,32-0.2533(2.4)\\ =&\,31.392\ldots \end{align*}

So

m=31.39.\begin{align*} m=31.39. \end{align*}

(c)

解法一

思路

展开

先求单个值为负的概率，再用补事件：至少一个为负 = 1−1- 全部不为负。

答题过程

展开

For one generated value,

P(X<0)= P(Z<0−48)= P(Z<−0.5)= 0.3085.\begin{align*} P(X<0) =&\,P\left(Z<\frac{0-4}{8}\right)\\ =&\,P(Z<-0.5)\\ =&\,0.3085. \end{align*}

So

P(X⩾0)=1−0.3085=0.6915.\begin{align*} P(X\geqslant0)=1-0.3085=0.6915. \end{align*}

Therefore

P(at least one negative)= 1−P(none negative)= 1−(0.6915)5= 0.8419…\begin{align*} P(\text{at least one negative}) =&\,1-P(\text{none negative})\\ =&\,1-(0.6915)^5\\ =&\,0.8419\ldots \end{align*}

So the probability is

0.842.\begin{align*} 0.842. \end{align*}