Question

The discharge of certain pollutants from a chemical plant is normally distributed, with a mean daily discharge of 25 mg/L. If the discharge exceeds 45 mg/L (which it does 0.5% of the time), then corrective measure must be taken. Calculate the standard deviation for the daily discharge of pollutants.

Answer 1

For normal distribution, P(X < A) = P(Z < (A - mean)/standard deviation)

Let denote standard devaition

Here, P(X > 45) = 0.005

P(X < 45) = 1 - 0.005 = 0.995

P(Z < (45 - 25)/) = 0.995

From standard normal distribution table, take the value of Z corresponding to probability of 0.995

(45 - 25)/ = 2.58

= 7.75 mg/L