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A company's manufacturing process uses 500 litres 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. Because there is a fine charged if regulations are not met, the company sets the machine to attain an average of 75 ppm in the treated water. The machine's output can be described by a Normal model with a standard deviation 4.2 ppm.
-What percent of the batches of water discharged exceed the 80 ppm standard?
A) 88.3%
B) 11.7%
C) 61.7%
D) 38.3%
E) -11.7%

Question

Quizplus · Accepted Answer

To find the percentage of batches exceeding 80 ppm, we calculate the z-score for 80 ppm given the mean of 75 ppm and the standard deviation of 4.2 ppm. The z-score formula is $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value of interest (80 ppm), $\mu$ is the mean (75 ppm), and $\sigma$ is the standard deviation (4.2 ppm). Plugging in the values, we get $z = \frac{(80 - 75)}{4.2} = \frac{5}{4.2} \approx 1.19$. Looking up 1.19 in a standard normal distribution table or using a calculator, we find that the area to the right of this z-score (which represents the percentage of batches exceeding 80 ppm) is approximately 11.7%.

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