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 LESS than 8.40?
0.9971
O 0.3264
• 0.0029
• 0.6736

Answer 1

Using the normal distribution and the central limit theorem, it is found that there is a 0.0029 = 0.29% probability that the sample average is LESS than 8.40.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

In this problem:

• The mean is of $\mu = 8.5$.

• The standard deviation is of $\sigma = 0.22$.

• A sample of 37 is taken, hence $n = 37, s = \frac{0.22}{\sqrt{37}} = 0.0362$.

The probability that the sample average is LESS than 8.40 is the p-value of Z when X = 8.4, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{8.4 - 8.5}{0.0362}$

$Z = -2.76$

$Z = -2.76$ has a p-value of 0.0029.

0.0029 = 0.29% probability that the sample average is LESS than 8.40.

To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213