A water treatment plant needs to maintain the pH of the water in the reservoir a

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 8.47?

Explanation / Answer

Solution

Answer: First Option as explained below

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ……………………..(1)

P(X t) = P[{(X - µ)/ } {(t - µ)/ }] = P[Z {(t - µ)/ }] ……………………(2)

X bar ~ N(µ, 2/n),………………………………………………………………….(3),

where X bar is average of a sample size n from population of X.

So, P(X bar t) = P[Z {(n)(t - µ)/ }] ……………………………………………(4)

Now, to work out the solution,

Let X = pH value of the sample from any selected location.

Given n = 37,

Given, ‘the results at each location will have varying results, with an average pH of 8.5 and a standard deviation of 0.22.’ => µ = 8.5 and = 0.22

Also given, ‘If the pH level of the reservoir is ok, what is the probability that the sample average is LESS than 8.47?” => we want P(X bar 8.47).

Now, by (4), P(X bar 8.47) = P[Z {(37)(8.47 – 8.5)/0.22}]

= P[Z - 0.83] = 0.2033 ANSWER

[probability value is directly read off from Standard Normal Distribution Tables.]

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