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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?  

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  1. shaik0124
    • one year ago
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    since the data is normal distribution median is 1915 ( avg is equal to median when data is equally spaced)

  2. shaik0124
    • one year ago
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    in normal distribution mean ,median and mode all are equal so the next hour traffic will be 1915+430 =2345

  3. shaik0124
    • one year ago
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    consider 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

  4. shaik0124
    • one year ago
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    dou have answer choices to this question

  5. anonymous
    • one year ago
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    No, but I think I got the answer.

  6. shaik0124
    • one year ago
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    so how did u solve

  7. anonymous
    • one year ago
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    |dw:1444369978024:dw| 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

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