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April 4th, 2012, 04:13 PM  #1 
Newbie Joined: Apr 2012 Posts: 2 Thanks: 0  probability sample mean
The number of people/car passing through a certain intersection between 8:00  9:00 A.M., is a random variable H (for Humans), with mean 4 people, and variance 2 people. If a random sample of 30 cars is chosen, at this intersection, during this time period, what is the probability that the average number of people/car will be at least 5? I tried doing it by using CLT so: P(H>5) = P((xbar  4)/(sqrt 2 / sqrt 30) > (5  4)/(sqrt 2 / sqrt 30)) = P (Z>3.87) I'm not sure if this is the right approach. sqrt = square root / = divide > greater than 
April 5th, 2012, 12:33 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,762 Thanks: 697  Re: probability sample mean
I haven't checked the arithmetic, but you seem to have the right idea.


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