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How confident is the algorithm that I am at least as good as mu - sigma? #133

Closed Answered by vivekjoshy
jauggy asked this question in Q&A
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$$\Pr(\mu-1\sigma \le X \le \mu+1\sigma) \approx 68.27%$$

$$\Pr(\mu-2\sigma \le X \le \mu+2\sigma) \approx 95.45%$$

$$\Pr(\mu-3\sigma \le X \le \mu+3\sigma) \approx 99.73%$$

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