Question

g The pH measurements of water specimens from various locations along a given river basin are Normally distributed, with mean 8 and standard deviation 0.3. You take water specimens from four randomly selected locations on this river basin. What is the probability that the mean pH measurement of these four specimens is greater than 8.2

Answer 1

0.0918 = 9.18% probability that the mean pH measurement of these four specimens is greater than 8.2

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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

`Z = (X - \mu)/(\sigma)`

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean 8 and standard deviation 0.3.

This means that
`\mu = 8, \sigma = 0.3`

Sample of 4:

This means that
`n = 4, s = (0.3)/(√(4)) = 0.15`

What is the probability that the mean pH measurement of these four specimens is greater than 8.2?

This is 1 subtracted by the pvalue of Z when X = 8.2. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (8.2 - 8)/(0.15)`

`Z = 1.33`

`Z = 1.33` has a pvalue of 0.9082

1 - 0.9082 = 0.0918

0.0918 = 9.18% probability that the mean pH measurement of these four specimens is greater than 8.2