Answer to Question #328963 in Statistics and Probability for Fayth Hippolyte

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 62kg and 66kg?

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,

$z=\cfrac{x-\mu} {\sigma}, \\ 1.282=\cfrac{70-66}{\sigma},\\ \sigma=\cfrac{4} {1.282} =3.12,\\ z_1=\cfrac{62-66} {3.12}=-1.28, \\ z_2=\cfrac{66-66}{3.12}=0,\\$

$P(62<X<66) =P(-1.28<Z<0)=\\ =P(Z<0)-P(Z<-1.28)=\\=0.5-0.1003=0.3997=39.97\%\\\text{ (from z-table). }$