The discharge of suspended solids from a phosphate mine is normally distributed with mean daily discharge 27 milligrams per liter (mg/l) and standard deviation 14 mg/l. in what proportion of the days will the daily discharge be less than 13 mg/l?

To solve this problem we use the z test. The formula for z score is:

z = (x - u) / s

where,

x = sample value = 13 and below

u = mean value = 27

s = standard deviation = 14

z = (13 - 27) / 14 = - 1

using the z table to get the standard normal probabilities, the p value at z = - 1 is:

P = 0.1587

This means that 15.87% that the discharge is 13 mg/L and below.