To answer this question we will use the following formula for the z-score:
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma}, \\ where\text{ }x\text{ is the observed value, }\mu\text{ is the mean, and }\sigma\text{ is the standard deviation.} \end{gathered}[/tex]
Substituting = 2000h, = 40h and x=2100h we get:
[tex]z=\frac{2100h-2000h}{40h}.[/tex]
Simplifying the above result we get:
[tex]z=\frac{100h}{40h}=2.5.[/tex]
Using tables we get:
[tex]P(z<2.5)=0.99379[/tex]
Then:
[tex]P(z\geq2.5)=1-0.99379=0.00621.[/tex]
The above result as a percentage is:
[tex]0.621\%.[/tex]
Answer:
[tex]0.62\%.[/tex]
