A company's manufacturing process uses 500 gallons of water at a time. A "scrubbing" machine then removes most of a chemical pollutant before pumping the
Water into a nearby lake. To meet federal regulations the treated water must not contain
More than 80 parts per million (ppm.) of the chemical. The machine's output can be
Described by a normal model with standard deviation 4.2 ppm. The company's lawyers
Insist that not more than 2% of the treated water should be over the limit. To achieve
This, to what mean should the company set the scrubbing machine?
A) 80 ppm.
B) 75 ppm.
C) 71.374 ppm.
D) 88.626 ppm.
E) 69.459 ppm.

Question

Quizplus · Accepted Answer

To find the mean that will ensure that at most 2% of the treated water exceeds 80 ppm, we need to use the normal distribution and the z-score formula.

Let X be the chemical level in the treated water, and let μ be the mean of the normal distribution. We want to find μ such that P(X > 80) ≤ 0.02.

Using the z-score formula, we can standardize X as follows:

z = (X - μ) / σ

where σ is the standard deviation of the normal distribution, which is given as 4.2 ppm.

Now we can rewrite the inequality as:

P(X > 80) = P(z > (80 - μ) / 4.2) ≤ 0.02

Looking at the z-score table, we find that the z-score corresponding to the right tail area of 0.02 is approximately 2.05.

Therefore, we have:

(80 - μ) / 4.2 = 2.05

Solving for μ, we get:

μ = 80 - 2.05 × 4.2 ≈ 71.374 ppm

Thus, the best choice is C, where the mean of the scrubbing machine's output is set at 71.374 ppm, which will ensure that no more than 2% of the treated water exceeds the federal limit of 80 ppm.

A Company's Manufacturing Process Uses 500 Gallons of Water at a Time