Suppose a fair coin is tossed $900$ times. Find the probability of getting more than $475$ heads. Use the continuity correction.

My answer:

$n=900, p=1/2, q=1/2$

$\mu=900(1/2)=450, npq=\sigma^{2}=225,\sigma=15$


$P_B(X\geq475)=P_B(475 \leq x \leq 900)$

$=P_N(474.5 \leq 900.5)$








Then we get:



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?


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.

  • $\begingroup$ Thanks! I see how this works now and how to do it too. I see the difference where the limits are concerned. Thanks again! $\endgroup$ – carson Apr 28 '14 at 3:08

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 $1.63$.

  • $\begingroup$ Ok. I see how it works. Thanks for telling me about the rounding part. $\endgroup$ – carson Apr 28 '14 at 3:09

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