The daily average concentration of pollutants in a stream follows a lognormal distribution with a mean of 50 mg/l and a standard deviation of 12 mg/l. It is desired to calculate the following probabilities: 1.) That the average concentration of pollutants in the stream on a given day will exceed 85 mg/l; and
2.) That a critical level of 95 mg/l will only be exceeded at the most one day in a given week (seven days). Assume that the pollutant concentrations between days are statistically independent.

1) probability that the average concentration of pollutants in the stream on a given day will exceed 85 mg/l is 0.0018

2) probability that a critical level of 95 mg/l will only be exceeded at the most one day in a given week (seven days) is 0.9993

Explanation:

Given the data in the question;

mean μ = 50 mg/l

standard deviation σ = 12 mg/l

we know that; x-score = x-μ /  σ

1) probability that the average concentration of pollutants in the stream on a given day will exceed 85 mg/l

p( x > 85 ) = P( Z > 85-50 / 12 )

= P( Z > 35/12 )

= P( Z > 2.9166)

= P( Z > 2.92)

= 1 - P( Z > 2.92)

from z-score table; P( Z > 2.92) = 0.9982

= 1 - 0.9982

p( x > 85 ) = 0.0018

Therefore, probability that the average concentration of pollutants in the stream on a given day will exceed 85 mg/l is 0.0018

2) probability that a critical level of 95 mg/l will only be exceeded at the most one day in a given week (seven days). Assume that the pollutant concentrations between days are statistically independent.

p( x > 95 ) = p( Z > 95-50 / 12 )

= p( Z > 45 / 12 )

= p( Z > 3.75 )

= 1 - p( Z > 3.75 )

from z-score table; p( Z > 3.75 ) = 0.9999

= 1 - 0.9999

= 0.0001

Now; p = 0.0001 and n = ( week) = 7

x = number of days × exceeds 99 mg/l

x ¬ Binomial ( n =7    p = 0.0001 )

p(x ≤ 1) = p(x=0) + (p=1)

= ¹∑  

=

= 7!/(0!(7-0)!)

= (1) ( 1 ) ( 0.9993 )

= 0.9993

Therefore,  probability that a critical level of 95 mg/l will only be exceeded at the most one day in a given week (seven days) is 0.9993

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explanation:

answer:

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