Question

A company's manufacturing process results in wastewater that contains an amount X (measured in parts per million) of a chemical pollutant. A "scrubbing" machine removes most of the chemical pollutant before pumping the wastewater into a nearby lake. The company must pay a large fine if the treated water contains more than 80 parts per million of the chemical pollutant, so the company sets the scrubbing machine to attain an expected chemical pollutant value E(X) = μ = 75 in the treated water. The distribution of chemical pollutant X in the treated water can be described by a normal model with expected value E(X) = 75 and standard deviation SD(X) = 4.2.

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Question 1. What is the probability that the treated water exceeds 80 parts per million of chemical pollutant?

a .6170 b .8830 c .3830 d .2340 e.1170

Question 2. The company's lawyers want the probability to be .025 that the treated water exceeds the 80 parts per million limit. To what mean value μ should the company set the scrubbing machine? (the standard deviation does not change).

a.78.325 b. 88.232 c. 79.895 d. 71.768 e. 80.105

Answer 1

This is a normal distribution question with

Question 1) P(x > 80.0)=?
The z-score at x = 80.0 is,

z = 1.1905
This implies that
P(x > 80.0) = P(z > 1.1905) = 1 - 0.883
P(x > 80.0) = 0.117 Option e
Question 2) Given in the question
P(X < 80) = 0.975
This implies that
P(Z < 1.96) = 0.975
With the help of formula for z, we can say that
Option d
PS: you have to refer z score table to find the final probabilities.
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