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April 4th, 2012, 04:13 PM   #1
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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
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April 5th, 2012, 12:33 PM   #2
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Re: probability sample mean

I haven't checked the arithmetic, but you seem to have the right idea.
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