Answer to Question #327857 in Statistics and Probability for faye

Question

The weights of students in a certain school are normally distributed with a mean weight of 66 kg. 10% have a weight greater than 70kg. What percentage of students weighs between 52kg and 66kg?  

How did we get our answer for this problem above? 

Answer 1

$P(X>70)=0.1,\\ P(X<70)=1-P(X>70)=\\=1-0.1=0.9.$

From z-table the 90^th percentile

is z = 1.282.

So,

$z=\cfrac{x-\mu} {\sigma}, \\ 1.282=\cfrac{70-66}{\sigma},\\ \sigma=\cfrac{4} {1.282} =3.12,\\ z_1=\cfrac{52-66} {3.12}=-4.49, \\ z_2=\cfrac{66-66}{3.12}=0,\\$

$P(52<X<66) =P(-4.49<Z<0)=\\ =P(Z<0)-P(Z<-4.49)=\\=0.5-0=0.5=50\%\\\text{ (from z-table). }$

We found the standard deviation and then the searched probability.