To transform any normal distribution to the standard normal distribution $Z$ is to standardize the random variable. Scale by the factor $Z = \frac{X - \mu}{\sigma}$ Once standardized, use [[R]] or a lookup table to solve the relevant probabilities. You may need to transform back to the original scaling depending on the question asked (the probability from the standard table will be accurate, however if you are asked to calculate a bound in the original scale you must transform back). Let $X\sim N(\mu, \sigma^2)$. We can show that $E \Big [\frac{X - \mu}{\sigma} \Big ] = \frac{1}{\sigma}E[X - \mu] = \frac{1}{\sigma}(E[X] - \mu) = \frac{1}{\sigma}(\mu - \mu) = 0$ and $V \Big [\frac{X - \mu}{\sigma} \Big ] = \frac{1}{\sigma^2}V[X - \mu] = \frac{1}{\sigma^2}V[X] = \frac{1}{\sigma^2} \cdot \sigma^2 = 0$ Thus we have shown that $X \sim N(\mu, \sigma^2) \Rightarrow \frac{X - \mu}{\sigma} \sim N(0,1)$ > [!Example] Example > Suppose $X$ is a random variable $X \sim N(12,4)$. > > Find $k$ such that $P(X>k) = 0.10$. > > $Z$-score tables often present the area under the standard normal curve to the left of the $Z$-score. To simplify, we can consider the $P(X>k) = 1 - P(X<=k) = 1 - 0.10 = 0.90$ > > Formally, we can state this as > $P(Z <= \frac{k-12}{\sqrt4}) = 0.90$ > > Using a lookup table, we find that $\Phi(1.285) \approx 0.90$ > > Finally we can solve for $k$. > > $\frac{k-12}{\sqrt4} = 1.285$ > $k = 1.285 * \sqrt4 + 12$ > $k \approx 14.56$ > > We can also use R to identify the Z-score with the `qnorm` function. > ```R > qnorm(1 - 0.1, 12, sqrt(4)) > > 14.56 > ``` If you know the area under the standard normal curve (e.g., the probability) but not the critical value, use `qnorm`. If you know the critical value and want to find the probability, use `pnorm`. Note that these both take standard deviation, not variance, as parameters.