1. Due to the vagaries of human activity and weather, Greensboro\'s daily air qu
ID: 3062324 • Letter: 1
Question
1. Due to the vagaries of human activity and weather, Greensboro's daily air quality varies from day to day and can be approximated by a normal distribution function. In 2013, ozone levels in Guilford County have achieved a daily average of 53 ppm with a standard deviation of 8 ppm. a) According to EPA guidelines, moderate or better air quality is defined by a daily average of County will experience moderate or better air quality. IlIlustrate your answer with a (carefully hand drawn if you like) picture of your work ozone that is below 45 ppm. Calculate the number of days that Guilford According to EPA guidelines, ozone levels above 65 ppm are considered hazardous and people are discouraged from spending time outside. What percentage of the days can Guilford County expect to have such days? b) Health officials have learned through research that harmful effect of ozone on human and plant life is somewhat cumulative in effect. This means that one day of unhealthy ozone levels places a moderate level of stress on living organisms, but three consecutive days places a disproportionate and increasing level of stress on these organisms. As such, calculate the probability of three consecutive days in which Greensboro's air will average above the 65 ppm standard. Why is this answer so c) different from your answer in b). Municipalities that experience more than 20 days in a year in which their air quality exceeds the 65 ppm threshold are considered to be in violation of the air quality standard. According to your evidence, was Guilford County in violation? Why or why not? d)Explanation / Answer
Question1
Mean level of ozone levels in Guilford County = 53 ppm
Standard deviation of ozone level in the county = 8 ppm
if x is the daily average ozone
Pr(x < 45 ppm ; 53 ppm ; 8 ppm) = ?
Z = (45 - 53)/8 = -1
Pr(x < 45 ppm ; 53 ppm ; 8 ppm) = Pr(Z < -1) = 0.158655
Number of days will experience moderate or better air quality = 365 * 0.158655 = 58 days
(b)
if x is the daily average ozone
Pr(x > 65 ppm ; 53 ppm ; 8 ppm) = ?
Z = (65 - 53)/8 = 1.5
Pr(x > 65 ppm ; 53 ppm ; 8 ppm) = Pr(Z > 1.5) = 1 - Pr(Z < 1.5) = 1 - 0.9332 = 0.0668
Percentage of days Guiford country expect to have these days = 0.0668 * 365 = 24.38 days
(c) Here Pr(on any one day there is hazadous level of ozone) = 0.0668
Pr(Hazardous level of ozone on three consecutive days) = 0.06683 = 0.0003
It is different because here we are talking about an event happening for three consecutive days.
(d) Here as expected number of days where there are hazardous level of air quality standard = 24.38 days
so here Pr(Number of days > 20 ; 24.38) = POISSON (Number of such days > 20 ; 24.38) = 0.78
SO we can say that there is 80% chance that guilford county is in violation of standards as it has expected number of days exceeds 20.
(e) Here Z value for year 2009
Z = (65 - 55)/6 = 10/6 = 1.67
Pr(Hazardous days) = 1 - Pr(Z < 1.67) = 1 - 0.9522 = 0.0478
Expected number of days are = 0.0478 * 365 = 17.443
so yes the average daily nuber of days where ozone levels exceed 65 ppm are lesser in year 2009 and it is more by (24.38 - 17.44= 6.94 days
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