Preventive examination

2.8% of the population suffers from disease XY. A preventive examination identifies 5.1% of healthy individuals as sick, and 91.4% of those examined who are actually sick are also identified as sick.
What percentage of examined individuals is identified as healthy?
What percentage of examined individuals identified as healthy are actually healthy?

Final Answer:

a =  92.4836 %
b =  105.0997 %

Step-by-step explanation:

$$\begin{gathered} r=2.8\%=\frac{2.8}{100}=0.028 \\ {p}_0=5.1\%=\frac{5.1}{100}=0.051 \\ {p}_1=91.4\%=\frac{91.4}{100}=0.914 \\ n=p_0\cdot (1-r)+p_1\cdot r=0.051\cdot (1-0.028)+0.914\cdot 0.028\doteq 0.0752 \\ a=100\cdot (1-n)=100\cdot (1-0.0752)=92.4836\% \end{gathered}$$

$b=100\cdot \frac{1-r}{1-n}=100\cdot \frac{1-0.028}{1-0.0752}=105.0997\%$

Try another example