题目

Problem

The distance an athlete can throw a discus is normally distributed with mean $40$ m and standard deviation $4$ m.

(a) Using standardisation, show that the probability that this athlete throws the discus less than $38.8$ m is $0.3821$ .

(2)

This athlete enters a discus competition. To qualify for the final, they have $3$ attempts to throw the discus a distance of more than $38.8$ m. Once they qualify, they do not use any of their remaining attempts.

Given that they qualified for the final and that throws are independent,

(b) find the probability that this athlete qualified for the final on their second throw with a distance of more than $44$ m.

(5)

题目中文翻译

一名运动员掷铁饼的距离服从均值为 $40$ m、标准差为 $4$ m 的正态分布。

(a) 利用标准化证明该运动员掷出少于 $38.8$ m 的概率为 $0.3821$ 。

该运动员参加铁饼比赛。为了晋级决赛，他有 $3$ 次机会掷出超过 $38.8$ m；一旦晋级，就不再使用余下的试掷机会。

已知该运动员已经晋级决赛，且各次投掷相互独立，

(b) 求他在第二投以超过 $44$ m 的成绩晋级决赛的概率。

解答

(a)

解法一

思路

展开

设 $X$ 是投掷距离，则 $X\sim N(40,4^2)$ 。标准化 $38.8$ 得到 $z=-0.3$ ，再用标准正态表。

答题过程

展开

Let $X\sim N(40,4^2)$ .

Then

\begin{align*} P(X<38.8) =&\,P\left(Z<\frac{38.8-40}{4}\right)\\ =&\,P(Z<-0.3)\\ =&\,1-P(Z<0.3)\\ =&\,1-0.6179\\ =&\,0.3821. \end{align*}

(b)

解法一

思路

展开

这是条件概率。已知已经 qualified，要求“第二投才 qualified 且第二投超过 $44$ ”。第二投才 qualified 表示第一投没有超过 $38.8$ ，第二投超过 $44$ 。分母是三次内至少一次超过 $38.8$ 。

答题过程

展开

From part (a),

\begin{align*} P(X<38.8)=0.3821. \end{align*}

So the probability of qualifying in at most $3$ attempts is

\begin{align*} P(\text{qualify}) =&\,1-P(\text{all 3 throws are less than }38.8)\\ =&\,1-(0.3821)^3\\ =&\,0.9442\ldots. \end{align*}

Also,

\begin{align*} P(X>44) =&\,P\left(Z>\frac{44-40}{4}\right)\\ =&\,P(Z>1)\\ =&\,1-0.8413\\ =&\,0.1587. \end{align*}

The probability that the first throw does not qualify and the second throw is more than $44$ m is

\begin{align*} 0.3821(0.1587). \end{align*}

Therefore

\begin{align*} P(\text{second throw is over }44\mid \text{qualify}) =&\,\frac{0.3821(0.1587)}{0.9442\ldots}\\ =&\,0.06422\ldots. \end{align*}

So the probability is

\begin{align*} 0.0642. \end{align*}

解法二

思路

展开

把第一投、第二投和第三投晋级视为三个互斥事件，分别计算后相加，得到条件概率的分母。

答题过程

展开

Let $p_F=0.3821$ and $p_Q=1-p_F=0.6179$ . The three mutually exclusive qualifying paths give

\begin{align*} P(\text{qualify}) =&\,p_Q+p_Fp_Q+p_F^2p_Q\\ =&\,0.9442\ldots. \end{align*}

The required conditional probability is therefore

\begin{align*} P(\text{2nd over }44\mid\text{qualify}) =&\,\frac{p_F(0.1587)}{0.9442\ldots}\\ =&\,0.06422\ldots. \end{align*}

So the probability is $0.0642$ .

S1 2024 Jan Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程