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anonymous
 one year ago
Data downloaded from a permanent count station yielded that the evening peak hour traffic
averaged 1915 (vph) and had a standard deviation of 430 (vph). The data appear to follow a normal distribution. Previous analyses found that the roadway has a capacity of 2275 (vph).
a.) What is the probability that the hourly traffic will exceed capacity?
anonymous
 one year ago
Data downloaded from a permanent count station yielded that the evening peak hour traffic averaged 1915 (vph) and had a standard deviation of 430 (vph). The data appear to follow a normal distribution. Previous analyses found that the roadway has a capacity of 2275 (vph). a.) What is the probability that the hourly traffic will exceed capacity?

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shaik0124
 one year ago
Best ResponseYou've already chosen the best response.0since the data is normal distribution median is 1915 ( avg is equal to median when data is equally spaced)

shaik0124
 one year ago
Best ResponseYou've already chosen the best response.0in normal distribution mean ,median and mode all are equal so the next hour traffic will be 1915+430 =2345

shaik0124
 one year ago
Best ResponseYou've already chosen the best response.0consider this set s={2,3,4,5} difference beween each number is 1 i.e,equally spaced now mean =(2+3+4+5)/4 =3.5 meadian =3+4)/2 =3.5

shaik0124
 one year ago
Best ResponseYou've already chosen the best response.0dou have answer choices to this question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, but I think I got the answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444369978024:dw So we want to probability that x > 2275. P(x>2275) z = xu/o = 22751915/430 = 0.84 So now we have P(z>0.84) = 1P(z<0.84) = 1 0.7995 We get 0.7995 from something called the z score table. Here's a video with examples. https://www.youtube.com/watch?v=mai23vW8uFM
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