Skill: Determine percentiles from normal distributions

A percentile tells you the value below which a certain percentage of observations fall. For example, the 90th percentile is the value that is greater than about $(90\%)$ of the observations.

For a normal distribution, use these steps:

1. Convert the percentile to a cumulative area.
2. Find the corresponding $z$-score using a standard-normal table or inverse-normal technology.
3. Convert the $z$-score to the original measurement using

$$x=\mu+z\sigma,$$

where $\mu$ is the mean and $\sigma$ is the standard deviation.

Example

Test scores are approximately normally distributed with mean $(70)$ and standard deviation $8$. Find the score at the 90th percentile.

Step 1: Identify the cumulative area

The 90th percentile means that $(90\%)$ of scores are below the desired score:

$$P(X\leq x)=0.90.$$

Step 2: Find the $z$-score

Use a standard-normal table to find the $z$-score with cumulative area $(0.9000)$. This gives approximately

$$z=1.28.$$

This means the 90th percentile is $(1.28)$ standard deviations above the mean.

Step 3: Convert to the original score

Substitute $\mu=70$, $(z=1.28)$, and $\sigma=8$ into the conversion formula:

$$x=\mu+z\sigma$$ $$x=70+(1.28)(8)$$ $$x=80.24.$$

Therefore, the score at the 90th percentile is approximately

$$\boxed{80.2}.$$

This means about $(90\%)$ of the scores are below $(80.2)$, not that $(80.2\%)$ of the score is being measured. The 50th percentile is the median, which equals the mean in a normal distribution.