A remediation company has identified stream Z as one which is at risk of contami
ID: 3254622 • Letter: A
Question
A remediation company has identified stream Z as one which is at risk of contamination, which occurs when the pollutant exceeds 24 p.u.v (parts per unit volume). Unfortunately their permits do not allow them to test the water in stream Z, so they must rely on data from connected streams X and Y. Suppose that stream X and Y meet to form stream Z and that the total contaminant concentration of a stream Z is given by
Z= X +.4 Y
The variable X represents the contribution of contaminants (puv) of stream 1; and is normally distributed as N(20,8).
The variable Y represents the contribution of contaminants (puv) of stream 2; and is normally distributed as N(10,7).
The company in charge of remediation of the stream has found that the contaminant contributions of streams 1 and 2 are correlated with a correlation coefficient of rho=.4
What is the probability that stream Z will be contaminated?
Explanation / Answer
Answer:
A remediation company has identified stream Z as one which is at risk of contamination, which occurs when the pollutant exceeds 24 p.u.v (parts per unit volume). Unfortunately their permits do not allow them to test the water in stream Z, so they must rely on data from connected streams X and Y. Suppose that stream X and Y meet to form stream Z and that the total contaminant concentration of a stream Z is given by
Z= X +.4 Y
The variable X represents the contribution of contaminants (puv) of stream 1; and is normally distributed as N(20,8).
The variable Y represents the contribution of contaminants (puv) of stream 2; and is normally distributed as N(10,7).
The company in charge of remediation of the stream has found that the contaminant contributions of streams 1 and 2 are correlated with a correlation coefficient of rho=.4
What is the probability that stream Z will be contaminated?
Mean of Z = 20+0.4*10 =24
Variance of Z = Var(X)+Var(Y)+2*r*sxsy
=64+49+2*0.4*8*7
=157.8
Standard deviation of Z = sqrt(157.8)
=12.5619
X value for 24, z =(24-24)/12.5619
=0.00
P( x >24) = P( z>0.00)
=0.50
probability that stream Z will be contaminated=0.50
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