# Finding Percentiles with the Normal Distribution

A percentile is the value in a normal distribution that has a specified percentage of observations below it. Percentiles are often used in standardized tests like the GRE and in comparing height and weight of children to gauge their development relative to their peers. The table below shows a portion of the percentile ranks for verbal and quantitative scores on the GRE exam. For example, if you scored 166 on the quantitative reasoning portion of the GRE, then 91% of those who took the test scored lower than you. If your score was 153, then 51% of those taking the exam scored lower than you.

Scaled
Score |
Quantitative Reasoning Percentile Rank |

170 |
97 |

169 |
97 |

168 |
95 |

167 |
93 |

166 |
91 |

165 |
89 |

164 |
87 |

163 |
85 |

162 |
82 |

161 |
79 |

160 |
76 |

159 |
73 |

158 |
70 |

157 |
67 |

156 |
63 |

155 |
59 |

154 |
55 |

153 |
51 |

We might ask, "In a population of 60 year old men with a mean BMI = 29 and s=6, what is the 90^{th} percentile for BMI?"

Previously, we started with a value, and asked what was the probability of values less (or greater) than that. However, we are now asking a problem that runs in the other direction. We are given the proportion or probability (90^{th} percentile) and asked what value of BMI that corresponds to.

Consequently, we use the previous concept and equation, but we work backwards.

First, we go the Z table and find the probability closest to 0.90 and determine what the corresponding Z score is.

For any normal distribution a probability of 90% corresponds to a Z score of about 1.28.

We also could have computed this using R by using the qnorm() function to find the Z score corresponding to a 90 percent probability.

> qnorm(0.90)

[1] 1.281552

So, given a normal distribution with μ =29 and σ =6, what value of BMI corresponds to a Z score of 1.28?

We know that Z=(x-μ)/σ.

Previously, we knew x, μ, and σ and computed Z. Now, we know Z, μ, and σ, and we need to compute X, the value corresponding to the 90^{th} percentile for this distribution. We can do this by rearranging the equation to solve for "x".

x = μ + Z σ

In this case, x = 29 + 1.28(6) = 36.7

** Conclusion:** 90% of 60 year old men have BMI values less than 36.7.

**Test Yourself**

The height of 12 month old boys is normally distributed with μ=76.4, σ=2.9 cm. What is the *10 ^{th} percentile* for height?