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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since the data is normal distribution
median is 1915 ( avg is equal to median when data is equally spaced)
in normal distribution mean ,median and mode all are equal
so the next hour traffic will be 1915+430 =2345
consider this set
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
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dou have answer choices to this question
No, but I think I got the answer.
so how did u solve
So we want to probability that x > 2275. P(x>2275)
z = x-u/o = 2275-1915/430 = 0.84
So now we have P(z>0.84) = 1-P(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