The article Reliability of DomesticWaste Biofilm Reactors J

The article \"Reliability of Domestic-Waste Biofilm Reactors\" (J. of Envir. Engr., 1995: 785-790) suggests that substrate concentration (mg/cm^3) of influent to a reactor is normally distributed with mu = 0.30 and sigma = 0.06. What is the probability that the concentration exceeds .25? What is the probability that the concentration is at most .10? How would you characterize the largest 5% of all concentration values? What is the probability that the concentration differs from the mean value by at most 1.5 standard deviations? What value c is such that the interval (0.30-c, 0.30+c) includes 98% of all concentrations. If four reactors are randomly selected, what is the probability that at least one has the substrate concentration exceeding 0.33?

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.25      
u = mean =    0.3      
          
s = standard deviation =    0.06      
          
Thus,          
          
z = (x - u) / s =    -0.833333333      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -0.833333333   ) =    0.797671619 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.1      
u = mean =    0.3      
          
s = standard deviation =    0.06      
          
Thus,          
          
z = (x - u) / s =    -3.333333333      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -3.333333333   ) =    0.00042906 [ANSWER]

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.95      
          
Then, using table or technology,          
          
z =    1.644853627      
          
As x = u + z * s,          
          
where          
          
u = mean =    0.3      
z = the critical z score =    1.644853627      
s = standard deviation =    0.06      
          
Then          
          
x = critical value =    0.398691218  

Thus, x > 0.398691218 are the top 5%. [ANSWER]

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d)

z1 = lower z score =    -1.5      
z2 = upper z score =     1.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.066807201      
P(z < z2) =    0.933192799      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.866385597   [ANSWER]  

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