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Merge pull request #51 from datawhalechina/修正矫正专用
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chp1 定理 18
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zhimin-z authored Jul 15, 2024
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Expand Up @@ -1491,7 +1491,7 @@ $$
$$
P[\frac{X}{m} \geq \frac{1}{2}] = P[Z \geq \frac{\frac{m}{2}-\mu}{\sigma}] = P[Z \geq \frac{\varepsilon\sqrt{m}}{\sqrt{1-\varepsilon^2}}]
$$
根据正态分布不等式(定理 20),有:
根据正态分布不等式(定理 21),有:
$$
P[Z \geq x] \geq \frac{1}{2}\left[1 - \sqrt{1-\exp\left(-\frac{2x^2}{\pi}\right)}\right] \geq \frac{1}{2}\left[1 - \sqrt{1-\exp\left(-x^2\right)}\right]
$$
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