Answer:
= 94.99%
Step-by-step explanation:
From the given information ,
it is obvious that the black out schedule has each region without power on average μ = 2.5 hours per day.
The standard deviation of blackout times in this city is б = 0.31 hours
A local hospital in the area purchases a generator that will provide power to nonessential systems for 3.5 hours
The percentage of time will the hospital find themselves without power to these nonessential systems is as follows:
[tex]P (\frac{x -u}{\sigma } )[/tex]
[tex]P =(\frac{3.5-2.5}{0.31} )\\\\=P(Z=3.2258)[/tex]
[tex][NORMSDIST(3.2258)]\\\\= 0.9994\\\\= 94.99%[/tex]
= 94.99%