CalcGospel 國際數學圖譜
返回

IAL 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)

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)\\[3mm] =&\,P(Z<-1.25)\\[3mm] =&\,1-0.8944\\[3mm] =&\,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)\\[3mm] =&\,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)\\[3mm] =&\,P(Z<-0.5)\\[3mm] =&\,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})\\[3mm] =&\,1-(0.6915)^5\\[3mm] =&\,0.8419\ldots \end{align*}

So the probability is

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