# Central Limit Theorem Approximation

• May 12th 2010, 07:05 PM
Janu42
Central Limit Theorem Approximation
The Internal Revenue Services uses a letter-opening machine to remove checks from envelopes. The machine has a probability p=0.03 of failure for any envelope. On a certain day, n=500 envelopes are independently handled by the machine. Let Y be the number of envelopes for which the machine fails to remove a check. Using the Central Limit Theorem (in this case DeMoivre's Theorem) to approximate P{Y$\displaystyle \leq$10}
• May 12th 2010, 08:21 PM
Anonymous1
Quote:

Originally Posted by Janu42
The Internal Revenue Services uses a letter-opening machine to remove checks from envelopes. The machine has a probability p=0.03 of failure for any envelope. On a certain day, n=500 envelopes are independently handled by the machine. Let Y be the number of envelopes for which the machine fails to remove a check. Using the Central Limit Theorem (in this case DeMoivre's Theorem) to approximate P{Y$\displaystyle \leq$10}

In a case like this $\displaystyle Bin(n,p) \approx Normal(np, np(1-p)).$

$\displaystyle P(Y \leq 10) = P\Big(\frac{Y - np}{\sqrt{np(1-p)}} \leq \frac{10 - np}{\sqrt{np(1-p)}}\Big) \approx 1- \Phi\Big(\frac{\color{red}{10.5}\color{black}{} - np}{\sqrt{np(1-p)}}\Big).$

You have n and p, now plug in and look up in the tables.
• May 12th 2010, 08:51 PM
matheagle
1 you should use 10.5 instead of 10.
2 I might use a Poisson approximation since p is near 0.
• May 12th 2010, 09:58 PM
matheagle
$\displaystyle P(Bin \le 10) \approx P(Normal\le 10.5)=P\left(Z\le {10.5-np\over {\sqrt{np(1-p)}}}\right)$

The 10.5 is the continuity correction, where we are overlaying a curve (normal) over the rectangles (binomial).