2, 0.0025)$, we know that the mean $\mu = 0.2$ and the variance $\sigma^2 = 0.0025$. Therefore, the standard deviation $\sigma = \sqrt{0.0025} = 0.05$. To standardize $X$, we use the transformation $Z = \frac{X - \mu}{\sigma}$, where $Z \sim N(0,1)$.
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