Quantiles of a normal distribution
The cumulative distribution function translates an x-value into a probability, \(P(X \le x)\). However sometimes it is important to work in the opposite direction. Given a cumulative probability, we want to find the corresponding value from the distribution. If we are provided with a probability, \(p\), then the value \(x\) such that
\[ P(X \le x) = p \]is the p'th quantile of the distribution of X.
We now give an example to illustrate the use of quantiles for a normally distributed random variable.
Example
If the weight of a Fuji apple has the following normal distribution
\[ X \;\; \sim \; \; \NormalDistn(\mu=180, \sigma=10) \]then we might want to find the apple weight that will be exceeded with 95% probability. In other words, we want to find the apple weight \(x\) such that
\[ P(X \lt x) \;\;= \;\; 0.05 \]In terms of z-scores, this is equivalent to
\[ P(X \lt x) \;= \; P\left(\frac {X-180} {10} \lt \frac {x-180} {10}\right) \;= \; P\left(Z \lt \frac {x-180} {10}\right) \;= \; 0.05 \]The relevant value for the standard normal distribution can be evaluated in Excel by typing in a spreadsheet cell
=NORM.S.INV(0.05)
to get the z-score −1.645.
\[ P(Z \lt -1.645) \;\;=\;\; 0.05 \]This means that there is a 5% probability of getting a value from a normal distribution that is more than 1.645 standard deviations below the mean. In the context of this example, this is the same as getting an apple weight below
\[ x \;=\; 180 - 1.645 \times 10 \;=\; 163.55 \text{ grams} \]Important normal quantiles
The following quantiles of the standard normal distribution are particularly important.
In particular, \(P(-1.96 \le Z \le 1.96) = 0.95\) so there is a 95% probability that any normally distributed random variable \(X\) is within 1.96 standard deviations of the distribution's mean.