The output of a generator is normally distributed with mean

The output of a generator is normally distributed with mean of 3500 kW and a standard deviation of 30 kW. What would the lower specification limit be, such that only 0.45% of the generators would exceed the limit?

Solution

U = 3500
SD = 30

For P =0.45% = 0.0045 exceed

So, P = 1 - 0.0045 = 0.9955 less

At P = 0.9955 , z = 2.61

2.61 = (x - 3500) / 30

78.3 = x- 3500

x = 3578.3 ---> ANSWER