Question

A company's manufacturing process results in wastewater that
contains an amount X (measured in parts per million) of a chemical
pollutant. A "scrubbing" machine removes most of the chemical
pollutant before pumping the wastewater into a nearby lake. The
company must pay a large fine if the treated water contains more
than 80 parts per million of the chemical pollutant, so the company
sets the scrubbing machine to attain an expected chemical pollutant
value E(X) = μ = 75 in the treated water. The distribution of
chemical pollutant X in the treated water can be described by a
normal model with expected value E(X) = 75 and standard deviation
SD(X) = 4.2.

Right-click this link standard normal table to open a standard
normal table in a separate window or tab.

**Question 1**. What is the probability that the
treated water exceeds 80 parts per million of chemical
pollutant?

a .6170 b .8830 c .3830 d .2340 e.1170

**Question 2**. The company's lawyers want the
probability to be .025 that the treated water exceeds the 80 parts
per million limit. To what mean value μ should the company set the
scrubbing machine? (the standard deviation does not change).

a.78.325 b. 88.232 c. 79.895 d. 71.768 e. 80.105

Answer #1

Question 1) P(x > 80.0)=?

The z-score at x = 80.0 is,

z = 1.1905

This implies that

P(x > 80.0) = P(z > 1.1905) = 1 - 0.883

P(x > 80.0) = **0.117 Option e**

Question 2) Given in the question

P(X < 80) = 0.975

This implies that

P(Z < 1.96) = 0.975

With the help of formula for z, we can say that

Option d

PS: you have to refer z score table to find the final
probabilities.

Please hit thumps up if the answer helped you

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