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?

Question

Quizplus · Accepted Answer

To ensure no more than 2% of the treated water exceeds the limit, we use the Z-score formula associated with the 98th percentile (since 100% - 2% = 98%) of a normal distribution. The Z-score for 98% is approximately 2.05. The formula to calculate the target mean ($\mu$) is $\mu = X - Z\sigma$, where $X$ is the maximum allowable concentration (80 ppm), $Z$ is the Z-score (2.05), and $\sigma$ is the standard deviation (4.2 ppm). Plugging in the values: $\mu = 80 - (2.05)(4.2) = 80 - 8.61 = 71.39$ ppm, which is approximately 71.374 ppm.

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