An electronic product takes an average of 9 hours to move through an assembly line. If the standard deviation of 0.9 hours, what is the probability that an item will take between 7.5 and 7.9 hours to move through the assembly line?

Answer :

Given information :

The population mean $(\mu )=9$

The population standard deviation $(\sigma )=0.9$

Solution :

→ The probability that an item will take between 7.5 and 7.9 hours to move through the assembly line :

$$\therefore P(7.5<x<7.9)=P(\frac{7.5–9}{0.9}<\frac{x–\mu }{\sigma }<\frac{7.9–9}{0.9})$$ $$=P(-1.67<z<-1.22)$$ $$=P(z<-1.22)–P(z<-1.67)$$ $$=0.111–0.047$$ $$=0.064$$