The article \"Reliability of Domestic-Waste Biofilm Reactors\" suggests that sub

Question

The article "Reliability of Domestic-Waste Biofilm Reactors" suggests that substrate concentration (mg/cm^3) of influent to a reactor is normally distrubuted with u=.30 and var=.06

d. what is the probability that the concentration differs from the mean value by at most 1.5 std?

e. what value c is such that the interval (.30-c, .30+c) includes 98% of all concentrations.

f. if four reactors are randomly selected, what is the probability that at least one has the substrate concentraion exceeding .33?

Explanation / Answer

D)

That means a z score of -1.5 to 1.5.

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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e)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.98      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.01      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -2.326347874      
By symmetry,          
z2 =    2.326347874      
          
As          
          
u = mean =    0.3      
s = standard deviation = sqrt(0.06) =   0.244948974      
          
Then          
          
c = z*s = 2.326347874*0.244948974 = 0.569836525 [ANSWER]

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f)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0.33      
u = mean =    0.3      
          
s = standard deviation =    0.244948974      
          
Thus,          
          
z = (x - u) / s =    0.122474487      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.122474487   ) =    0.451261625

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    4      
p = the probability of a success =    0.451261625      
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.090669523

Thus, P(at least one) = 1 - P(0) =   0.909330477 [ANSWER]      

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