Statistics / Unit 3: The Normal Distribution
3.5 Finding a Value from a Percentage
Notes
Sometimes you know the percentile and need the value. Find the matching z-score, then use $x = \mu + z\sigma$.
- Useful z-scores: the 90th percentile is about $z = 1.28$ and the 75th percentile is about $z = 0.67$.
- From the 68-95-99.7 rule: the top $2.5\%$ starts at $z = 2$, the bottom $16\%$ ends at $z = -1$.
- Example: With $\mu = 70$ and $\sigma = 10$, the 75th percentile is $70 + 0.67(10) = 76.7$.
Practice
Test scores are normal with $\mu = 70$ and $\sigma = 10$. The 90th percentile has $z \approx 1.28$. What score is at the 90th percentile?
$70 + 1.28(10) = 82.8$.
Light bulb lifetimes are normal with mean 1000 hours and standard deviation 50 hours. Using the 68-95-99.7 rule, the shortest-lasting $2.5\%$ of bulbs last less than about:
The bottom $2.5\%$ is below $\mu - 2\sigma = 1000 - 100 = 900$.
The scores on a test are normal with $\mu = 200$ and $\sigma = 20$. The 75th percentile has $z \approx 0.67$. About what score is the 75th percentile?
$200 + 0.67(20) = 213.4$.
Ready for more? Take the Unit 3 test.