Question

Consider that in a population of Filipino adults ages 18 to 65, Body Mass Index (BMI) is normally distributed with a mean of 28 and a standard deviation of 6. What is the BMI mark of the bottom 20% of the distribution of this population? *

Answer 1

Let $X=$ BMI, $X\sim N(\mu, \sigma^2 )$

Given $\mu=28, \sigma=6$

$$P(X< x)=P(Z<\dfrac{x-\mu}{\sigma})$$

$$=P(Z<\dfrac{x-28}{6})=0.2$$

$$\dfrac{x-28}{6}\approx-0.8416$$

$$x\approx22.95$$