Question

Rolling blackouts have become a way of life in many major cities around the world. With electrical demand far outstripping production, grid operators have come to rely on this systematic means of keeping the distribution system from experiencing complete failure. In one city, the blackout schedule has each region without power on average 2.5 hours per day. The standard deviation of blackout times in this city is 0.31 hours. A local hospital in the area purchases a generator that will provide power to nonessential systems for 3.5 hours. If the blackout times follow a normal distribution, what percentage of time will the hospital find themselves without power to these nonessential systems

Answer 1

= 94.99%

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:

`P ((x -u)/(\sigma ) )`

`P =((3.5-2.5)/(0.31) )\\\\=P(Z=3.2258)`

`[NORMSDIST(3.2258)]\\\\= 0.9994\\\\= 94.99%`

= 94.99%