Question #236617

The article ”The statistics of Phytoxic Air Pollutants” suggests the log-normal distribution as a model for SO2 concentration above a certain forest. Suppose the parameter values are µ = 1.9 and σ = 0.9.

(a) What are the mean value and standard deviation of concentration?

(b) What is the probability that concentration is at most 10? Between 5 and 10? 


1
Expert's answer
2021-09-17T01:13:32-0400

Solution:

μ=1.9,σ=0.9\mu=1.9,\sigma=0.9

(a) Mean and standard deviation of log-Normal distribution:

Mean=eμ+σ2/2=e1.9+0.92/2=e2.305=10.024=e^{\mu+\sigma^2/2}=e^{1.9+0.9^2/2}=e^{2.305}=10.024

Variance=(eσ21)e2μ+σ2=(e0.921)e2(1.9)+0.92=125.394=({e^{\sigma}}^2-1)e^{2\mu+\sigma^2}=({e^{0.9}}^2-1)e^{2(1.9)+0.9^2}=125.394

S.D=variance=125.394=11.197=\sqrt{variance}=\sqrt{125.394}=11.197

(b) XN(10.024,11.197)X\sim N(10.024,11.197)

P(X10)=P(z1010.02411.197)=P(z0.002)=0.5008P(X\le10)=P(z\le \dfrac{10-10.024}{11.197})=P(z\le0.002)=0.5008

P(5X10)=P(510.02411.197z1010.02411.197)P(5\le X\le10)=P(\dfrac{5-10.024}{11.197}\le z\le \dfrac{10-10.024}{11.197})

=P(0.448z0.002)=P(z0.002)P(z0.45)=P(z0.002[1P(z0.45)]=0.50081+0.67364=0.17444=P(-0.448\le z\le0.002)=P(z\le 0.002)-P(z\le -0.45) \\=P(z\le 0.002-[1-P(z\le 0.45)] \\=0.5008-1+0.67364 \\=0.17444


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