Consignment with 15 defective products

The problem

Probability of defective part is 10%.
What probability that 15 products will be defective from 200 products.

Solution

The solution can be found by formula of local theorem of Moivre-Laplace

$$\\ P_{n}(k) \approx \frac{\varphi (x)}{\sqrt{npq}} \\ \\ x = \frac{k-np}{\sqrt{npq}} \; ; \; \varphi(x) = \frac{1}{\sqrt{2\pi }}e^{-\frac{x^{2}}{2}}$$

So we could just put our values to the formula

p = 0.1
q = 0.9
k = 15
n = 200

$$\\ x = \frac{15-200\cdot 0.1}{\sqrt{200\cdot 0.1\cdot 0.9}} = \frac{15-20}{\sqrt{18}} = -1.178511 \\ \\ P_{200}(15) \approx \frac{\varphi(-1.178511) }{\sqrt{18}} \approx 0.047$$

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