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相扑选手

Sumo Wrestlers

专题
Statistics / 统计
难度
L2

题目详情

两名相扑选手（A 和 B）在连续 NN 天中每天进行一场比赛。设 XiX_i 表示第 ii 天 A 的比赛结果（A 输记为 0，A 赢记为 1），并设 Xˉn\bar{X}_n 表示 A 获胜场次所占的比例。

假设每场比赛相互独立，且设 pp 表示相扑选手 A 获胜的概率。如果 p=0.63p = 0.63，并且 σ=0.025\sigma = 0.025，那么为了使 P(0.57≤Xˉn≤0.69)>0.88P(0.57 \leq \bar{X}_n \leq 0.69) \gt 0.88，两名选手至少需要进行多少场比赛？

Two Sumo wrestlers (A and B) have one wrestling match each day for NN days. Let XiX_i denote the outcome of a wrestling match on day ii for A (0 if A lost and 1 if A won) and let Xˉn\bar{X}_n denote the fraction of matches that A won.

Assume that each match is independent of the other matches and let pp denote the probability that sumo A wins the match. If p=0.63p = 0.63, and σ=0.025\sigma = 0.025 how many matches must the sumo wrestlers have in order for P(0.57≤Xˉn≤0.69)>0.88P(0.57 \leq \bar{X}_n \leq 0.69) \gt 0.88.

解析

第 1 步：定义问题

设 Xi∼Bernoulli(p)X_i \sim \text{Bernoulli}(p)，相互独立，且 Xˉn=1n∑i=1nXi\bar{X}_n = \frac{1}{n}\sum_{i=1}^n X_i。

  • 已知 p=0.63p = 0.63，目标标准差为 SD(Xˉn)=σ=0.025\text{SD}(\bar{X}_n) = \sigma = 0.025
  • 对于 Bernoulli 变量：Var(Xˉn)=p(1−p)n\text{Var}(\bar{X}_n) = \frac{p(1-p)}{n}
  • 我们希望：P(0.57≤Xn^≤0.69)>0.88P(0.57 \leq \hat{X_n} \leq 0.69) > 0.88

第 2 步：计算 Xˉn\bar{X}_n 的标准差

对于 Bernoulli 随机变量： Var(Xi)=p(1−p)=0.63∗0.37=0.2331\text{Var}(X_i) = p(1-p) = 0.63 * 0.37 = 0.2331

因此 X^n\hat{X}_n 的标准差为： SD(Xn^)=p(1−p)n=0.2331n\text{SD}(\hat{X_n}) = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.2331}{n}}

题目还给了 σ=0.025\sigma = 0.025。令 SD(Xˉn)=σ\text{SD}(\bar{X}_n) = \sigma，得到：

n=p(1−p)σ2=0.63⋅0.370.0252≈373n = \frac{p(1-p)}{\sigma^2} = \frac{0.63 \cdot 0.37}{0.025^2} \approx 373

第 3 步：使用正态近似（CLT）

根据中心极限定理：

Xˉn≈N(p,p(1−p)n)=N(0.63,0.0252)\bar{X}_n \approx N(p, \frac{p(1-p)}{n}) = N(0.63, 0.025^2)

我们要求 P(0.57≤Xˉn≤0.69)=P(0.57−0.630.025≤Z≤0.69−0.630.025)=P(−2.4≤Z≤2.4)≈0.983P(0.57 \le \bar{X}_n \le 0.69) = P\left(\frac{0.57-0.63}{0.025} \le Z \le \frac{0.69-0.63}{0.025}\right) = P(-2.4 \le Z \le 2.4) \approx 0.983

这个概率大于 0.88，所以 n≈373n \approx 373 场比赛已经足够。

n≈373 场比赛\boxed{n \approx 373 \text{ 场比赛}}

Original Explanation

Step #1: Define the Problem

Let Xi∼Bernoulli(p)X_i \sim \text{Bernoulli}(p), independent, and Xˉn=1n∑i=1nXi\bar{X}_n = \frac{1}{n}\sum_{i=1}^n X_i.

  • We are given p=0.63p = 0.63 and desired SD(Xˉn)=σ=0.025\text{SD}(\bar{X}_n) = \sigma = 0.025
  • For Bernoulli: Var(Xˉn)=p(1−p)n\text{Var}(\bar{X}_n) = \frac{p(1-p)}{n}
  • We want: P(0.57≤Xn^≤0.69)>0.88P(0.57 \leq \hat{X_n} \leq 0.69) > 0.88

Step #2: Standard Deviation of Xn

For a Bernoulli random variable: Var(Xi)=p(1−p)=0.63∗0.37=0.2331\text{Var}(X_i) = p(1-p) = 0.63 * 0.37 = 0.2331

Standard deviation of X^n\hat{X}_n: SD(Xn^)=p(1−p)n=0.2331n\text{SD}(\hat{X_n}) = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.2331}{n}}

We are also told σ=0.025\sigma = 0.025. Set SD(Xˉn)=σ\text{SD}(\bar{X}_n) = \sigma:

n=p(1−p)σ2=0.63⋅0.370.0252≈373n = \frac{p(1-p)}{\sigma^2} = \frac{0.63 \cdot 0.37}{0.025^2} \approx 373

Step #3: Using Normal Approximation (CLT)

Using central limit theorem:

Xˉn≈N(p,p(1−p)n)=N(0.63,0.0252)\bar{X}_n \approx N(p, \frac{p(1-p)}{n}) = N(0.63, 0.025^2)

We want P(0.57≤Xˉn≤0.69)=P(0.57−0.630.025≤Z≤0.69−0.630.025)=P(−2.4≤Z≤2.4)≈0.983P(0.57 \le \bar{X}_n \le 0.69) = P\left(\frac{0.57-0.63}{0.025} \le Z \le \frac{0.69-0.63}{0.025}\right) = P(-2.4 \le Z \le 2.4) \approx 0.983

This exceeds 0.88, so n≈373n \approx 373 matches is sufficient.

n≈373 matches\boxed{n \approx 373 \text{ matches}}