Using continuity correction for normal distribution Suppose a fair coin is tossed $900$ times. Find the probability of getting more than $475$ heads. Use the continuity correction.
My answer:
$n=900, p=1/2, q=1/2$
$\mu=900(1/2)=450, npq=\sigma^{2}=225,\sigma=15$
$Z=(X-\mu)/\sigma$
$P_B(X\geq475)=P_B(475 \leq x \leq 900)$
$=P_N(474.5 \leq 900.5)$
$Z=(X-\mu)/\sigma$
$=(474.5-450)/15=1.63$
$Z=(X-\mu)/\sigma$
$(900.5-450)/15=30.03$
$=P_N(0<z<1.63)+P_N(0,z,30.03)$
$.4484+.5000$
$.9484$
Then we get:
$1-.9484$
$.0516$
I was just trying this problem to see what it would be like after reading about the topic. I wanted to know what I did wrong. The answer says $.0446$. Can someone help me with this?
 A: Your definition of the parameters is correct. $\mu=np=900\cdot0.5=450$. $\sigma^2 = np(1-p)=900\cdot0.5\cdot0.5=225$. 
Now, the moment you perform normal approximation, you need to first understand that the possible range is $(-\infty, \infty)$ and not just $[0,900]$. That is why it is an approximation. By the way, the probabilities beyond 900 are very small. So, simply put,
\begin{align}\text{Pr}(X>475) &= \text{Pr}(X>475.5)\\&=1-\text{Pr}(X<475.5)\\&=1-\text{Pr}\begin{pmatrix}Z<\frac{475.5-450}{\sqrt{225}}\end{pmatrix}
\\&=1-\text{Pr}(Z<1.7)\\&=1-0.9554\\&=0.0446
\end{align}
Hence, your mistakes are that your upper limit was defined incorrectly. With regards to your lower limit, you have to be more liberal ie add the case of $475.5$, not remove and cut to $474.5$. It was interesting because you added the half at the upper limit $(900.5)$ but did not do so at your lower bound.
By the way just fyi, your conversion to $Z$ ie normalization was good.
A: You used $4.75-\dfrac12$ and whoever came up with $0.0446$ used $475+\dfrac12$ since it said strictly more than $475$ rather than $475$ or more.  That would result in $0.0446$.
A nitpicking point: Rounding should be done LAST, except when you know what effect it will have on your answer.  I got $0.0512$ where you had $0.0516$ because I didn't round $1.633333$ to $1.63$.
