A small college has 1095 students. What is the approximate probability that more than five students were born on Christmas day? A small college has 1095 students. What is the approximate probability that more than five students were born on Christmas day? Assume that the birthrates are constant throughout the year and that each year has 365 days.

I tried doing
$X \sim Pn(3) $ and calculating $P(X\gt5)$. My calculation turned out to be 0.22....., which was wrong. (What was wrong with my approximation?). The solution given used a Normal approximation to get an answer of 0.0735. When I tried using a Normal approximation, I was still unable to get that answer. Here is how I attempted it.  

$ N\sim(3,1092/365)$.
$ P(X\gt5)=P(N\gt4.5) $  #continuity correction
$= P\left(Z\gt\frac{4.5-3}{\sqrt{\frac{1092}{365}}}\right)$
=$ P(Z&gt;0.86721)$.
=$0.193$.  

Any help is much appreciated.
poisson probability distribution $\lambda = 1095 / 365$ 
then calculate 

using

$P(N) = \frac{\lambda^n}{n!} \exp(-\lambda)$

1 - P(0) - P(1) - P(2) - P(3) - P(4) - P(5) = 

$1 - \exp(-\lambda)(1 + \lambda + \lambda^2 / 2! + \lambda^3 / 3! + \lambda^4/4! + \lambda^5/5!) = .084$

note this is an approximation,

also assuming that 5 is NOT included
N should be 5.5 not 4.5

P(X>5) = P(N >5.5) // continuity correction

N~(μ,σ^2)

N~(3,2.99) &lt;--- σ^2 should not be (1092/365)

μ = np(amount of students)(probability of person born on Christmas Day)

=1095 *(1/365)

=3

σ = sqrt(n*p(1-p))

= sqrt(1095*(1/365)(1-(1/365)))

= sqrt(3*(364/365))

= sqrt(2.99)

P(X > 5) = P(X > 5.5)

      = P(Z &gt;(5.5-3)/(sqrt(2.99)))         //Z =((X-μ)/σ)

      = P(Z &gt; 1.45)      //using the Standard Normal Probabilities table

      = 0.0735


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