# Central Limit Theorem Approximation

#### 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}

#### Anonymous1

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.

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#### matheagle

MHF Hall of Honor
1 you should use 10.5 instead of 10.
2 I might use a Poisson approximation since p is near 0.

• Anonymous1

#### matheagle

MHF Hall of Honor
$$\displaystyle P(Bin \le 10) \approx P(Normal\le 10.5)=P\left(Z\le {10.5-np\over {\sqrt{np(1-p)}}}\right)$$

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