Environmental agencies around the world regulate dust pollutants on the basis of
ID: 3176851 • Letter: E
Question
Environmental agencies around the world regulate dust pollutants on the basis of mass, not chemistry, and most governments focus on the particles easiest to catch and quantify; those that are 10 micrometers across (the PM-10 fraction). Federal law prohibits PM-10 concentrations in air from exceeding 150 micrograms per cubic meter (mu g/m^3) over any 24-hour period. If the managers of an industrial park know that they are creating PM-10 concentrations that average 120 (mu g/m^3) with a standard deviation of 15 (mu g/m^3) assuming the PM-10 concentrations are normally distributed, find the following probabilities: That a day's PM-10 concentration is below 140 (mu g/m^3). That a day's PM-10 concentration is between 100 and 140 (mu g/m^3). That a day's PM-10 concentration is above the federal acceptable limit of 150 (mu g/m^3).Explanation / Answer
We are given that mean = 120 and standard deviation = 15 , please keep the z tables handy for this .
a) P(X<140) , we need to first calculate the z scores using the formula as
Z = (X - mean)/SD , no we get
(140-120)/15 = 1.33
now we need to look for P(Z<1.33) in the z tables
P ( Z<1.33 )=0.9082
b) for X between 100 and 140 ,
so z score for 140 si 1.33 as calculated above
for 100
Z = 100-120 /15 = -1.33
so we need to find P(-1.33<Z<1.33)
To find the probability of P (1.33<Z<1.33), we use the following formula:
P (1.33<Z<1.33 )=P ( Z<1.33 )P (Z<1.33 )
We see that P ( Z<1.33 )=0.9082.
P ( Z<1.33 ) can be found by using the following fomula.
P ( Z<a)=1P ( Z<a )
After substituting a=1.33 we have:
P ( Z<1.33)=1P ( Z<1.33 )
We see that P ( Z<1.33 )=0.9082 so,
P ( Z<1.33)=1P ( Z<1.33 )=10.9082=0.0918
At the end we have:
P (1.33<Z<1.33 )=0.8164
c) here we need to find P(X>150) , using the same concept weget
Z = (150 - 120) /15 = 2
P(Z>2)
P ( Z>2 )=1P ( Z<2 )=10.9772=0.0228
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