The army wants to give advanced training to 20% of those recruits with highest scores . What is the lowest scores acceptance for advanced training.

Given,

m=110,

$\sigma$=10

p=20% =0.2

$Z=\frac {x-m}{\sigma}$

$\therefore P(z \gt z_1) = \frac{20}{100} = 0.2$

Since 0.5-0.2 = 0.30 and corresponding to 0.30 the entry in the area table is 0.84

$\begin{align} Z&=\frac {x-110}{10}\\ 0.84&=\frac {x-110}{10}\\ 0.84+110& = x\\ x&=118.4\\ \end{align}$

The lowest score acceptance is 118.4