1. Uniform random variable question

Suppose that 10^6 people arrive at a service station at times that are independent random variables, each of which is uniformly distributed over (0,10^6). Let N denote the number that arrive in the first hour. Find an approximation for $P(N = i).$

2. Hello,

Consider the event : the customer arrives at the first hour, and its complementary : the customer doesn't arrive at the first hour.

Then we have a binomial distribution.
You can approximate it by a Poisson distribution (see here : Binomial distribution - Wikipedia, the free encyclopedia )

3. I guess I'm just confused by the wording of the question. Is the probability that the customer arrives at the first hour $= \frac{1}{10^6}?$

4. Originally Posted by Anonymous1
I guess I'm just confused by the wording of the question. Is the probability that the customer arrives at the first hour $= \frac{1}{10^6}?$
Yes (depending on what 10^6 represents...minutes or hours)

And n=10^6

5. Thanks.

So,

$P(N = i) = e^{-1} \frac{1^i}{i!}$

I'm only worried because my answer seems to simple. I don't need to use a summation from 0 to i, do I?

6. Originally Posted by Anonymous1
Thanks.

So,

$P(N = i) = e^{-1} \frac{1^i}{i!}$

I'm only worried because my answer seems to simple. I don't need to use a summation from 0 to i, do I?
No you don't. You're asked the probability for a given i

And it's rather $P(N=i)~{\color{red}\approx}~e^{-1}\cdot\frac{1^i}{i!}~\left[=\frac{e^{-1}}{i!}\right]$