Question

Find P(-1.2 <2<-0.2)

Answer 1

The probability of P(–a < Z < –b) is illustrated below:

First separate the terms as the difference between z-scores:

$P(-a<Z<-b)=P(Z<-b)-P(Z<-a)$

Then express these as their respective probabilities under the standard normal distribution curve:

$P(Z<-b)-P(Z<-a)=\Phi(-b)-\Phi(-a)$

$=(1-\Phi(b))-(1-\Phi(a))$

$=1-\Phi(b)-1+\Phi(a)$

$=\Phi(a)-\Phi(b)$

The above calculations can also be seen clearly in the diagram below:

So,

$P(-1.2<Z<-0.2)=\Phi(1.2)-\Phi(0.2)=0.8849-0.5793=0.3056$ (using Z-score table)