Question

A water treatment plant needs to maintain the pH of the water in the reservoir at a certain level. To monitor this, they take 2 oz. of water at 37 locations every hour, measure the pH at each of those locations, and find their average. If the pH level of the reservoir is ok, the results at each location will have varying results, with an average pH of 8.5 and a standard deviation of 0.22. If the pH level of the reservoir is ok, what is the probability that the sample average is MORE than 8.47?A.-0.833 B.0.2033 C.0.7967 D.-0.14 E.0.4443 F.0.5557

Answer 1

Answer:

C.) 0.7967

Explanation:

Hello,

Normal distribution is used here as:

$z=\frac{pH_{sample}-pH_{av}}{stand_{dev}*\sqrt{n} }$

In this case, the sample's pH is 8.47, the average pH 8.5, standard deviation 0.22 and sample size 37, thus:

$z= \frac{8.47-8.5}{0.22/\sqrt{37} }\\z=-0.8295$

Now, based on the table for standard normal distribution which comes from a symmetric bell-like graph, the probability is 0.797 which matched with the C answer.

Best regards.