Question

A company's manufacturing process uses 500 gallons of water at a time. A scrubbing machine then removes most of a chemical pollutant, but the machine isn't perfect and it is costly to operate. Since there's a fine if the discharged water exceeds the legal maximum, the company sets the machine to attain an average 75ppm for the batches of water treated. They believe the machine's output can be described by a normal model with standard deviation 4.2.
a) what percent of the batches of water discharged exceed the 80ppm standard?

Answer 1

I also need help in determining what equations to use for the following two (still pertaining to the original post:
1) The company's lawyers insist that they have not more than 2% of the water over the limit. To what mean value should the company set the scrubbing bubble machine?
2) Because achieving a mean that low would raise the costs too much, they decide to leave the mean set at 75ppm and try to reduce the standard deviation to achieve the only 2% over goal. find the new standard deviation needed

Answer 2

Let \[z=\frac{ x-\mu }{ \sigma }\] where x=ppm of pollutant, μ=75 and σ=4.2
1)\[P(x>80)=P(z>\frac{ 80-75 }{ 4.2 })=P(z>1.19)=1-P(z \le 1.19)=1-0.8831=0.1169\] That means a 11.69%
2) We need to determine the value Zx that makes:
\[P(x >80)=0.02\rightarrow P(x \le 80)=1-0.02=0.98=P(z \le Z_x)=0.98\] and this value is Zx=2.05 Now we do\[Z_x=2.05=\frac{ 80-\mu }{ 4.2 }\rightarrow \mu=80-4.2 \times 2.05=71.37\] We need to adjust the scrubbing machine to 71.37 ppm
3)Same as before but now we keep the original μ=75 and find the required σ\[\frac{ 80-75 }{ \sigma }=2.05\rightarrow \sigma =5/2.05=2.43\] Which means we need to reduce the standard deviation of the scrubbing machine to 2.43 ppm