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The time of occurrence of the first accident during rush hour traffic at a major intersection is uniformly distributed between the three hour interval 4pm to 7pm. let X=the amount of time (hours) it takes for the first accident to occur.

Assume Ramon has kept track of the times for the first accidents to occur for 40 different days. let C=the total cumulative time. Then C follows which distribution?

Select one: A) U(0,3) B) Exp(13) C) N(60, 5.477) D) N(1.5, 0.01875)

Stated correct answer: N(60,5.477)

Can I please get a hint as to why the stated answer is correct? i dont recall that format in the text, but im guessing the correct answer is normal distribution with a mean of 60 and std dev of 5.477

But what path do I use to get those values?

Can someone please explain how to arrive at the value of square root of 30 ?!?!?

N S
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    I would choose "none of the above." The stated "correct" answer is only an approximation (albeit a pretty good one). How can I be so sure? Because an $N(60, 5.477)$ (really, $N(60,\sqrt{30})$) distribution has almost $10^{-27}$ chance of being negative or exceeding $120$, which obviously is impossible. A fully correct answer is given at https://stats.stackexchange.com/questions/41467. The analysis needed to obtain it should help you appreciate why people focus on approximations! – whuber Jan 10 '18 at 23:49
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    @whuber to be honest, I can live with that normality approximation more easily than with the assumption that "time until first accident of the afternoon" would be uniformly distributed. Something very weird is going on at that intersection! – Geoffrey Brent Jan 11 '18 at 01:32
  • what are the formulas that yield 60 and (30)^0.5 ? – N S Jan 11 '18 at 01:57
  • im guesing that the 40*1.5=60 for mean – N S Jan 11 '18 at 01:57
  • but how did you get that std dev ? – N S Jan 11 '18 at 01:57
  • @Geoffrey That is an excellent point--thank you for making it. The concern I had in writing my first comment was pedagogical: here, I think, we get a glimpse into why so many people leave their first statistics course with so many misconceptions about the role and ubiquity of the Normal distribution. It would be so easy to improve the question, and avoid sowing such misconceptions, simply by asking "which distribution best describes C"? – whuber Jan 11 '18 at 14:05
  • WHERE DOES THE SQUARE ROOT OF 30 COME FROM?!?!?!?!?! – N S Jan 11 '18 at 15:00
  • or a more accurate question: how do you get the value of square root of 30 ? – N S Jan 11 '18 at 15:00
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    @N S What is the variance of a Uniform[0,3] random variable? Once you have that, you're well on your way. – Mark L. Stone Jan 11 '18 at 16:16
  • variance = (1/12)*(b-a)^2 = 30 if i convert 3 hours to 180 minutes. but then where does the 60 come from? mean = (1/2)*(a+b) = 90 – N S Jan 11 '18 at 18:59
  • 40 different days. And stick with hours. So what is the variance of C? – Mark L. Stone Jan 11 '18 at 19:02
  • variance = (1/12)*(b-a) = (1/12)*(3-0) = 0.75 -> 0.75*40 = 30 – N S Jan 11 '18 at 19:20
  • mean = (1/2)*(0+3) = 1.5 -> 1.5*40 = 60 – N S Jan 11 '18 at 19:21
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    @whuber The question would definitely be improved by wording it as you suggest. Many multiple-choice exams I've seen have some sort of "choose the best answer" boilerplate, which I suspect is the low-effort way to cover the authors against issues such as this - even if all of them are wrong, one is clearly the least wrong. – Geoffrey Brent Jan 11 '18 at 21:41
  • @GeoffreyBrent - sounds like what we hope for from model comparison, actually! – jbowman Jan 11 '18 at 22:16

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