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$70\%$ of items are defective. You randomly select $20$ items. Find the probability that the number of defective items is exactly $14$.

I have $n$ as $20$, $x$ as $14$, $p$ as $.7$ and $q$ as $.3$.

So, I get ${20\choose 14} (0.7)^{14} (0.3)^6$ however I am not sure about the answer. Cannot determine the standard deviation and mean.

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  • $\begingroup$ Have you tried anything? $\endgroup$ – user243664 Jun 11 '15 at 13:42
  • $\begingroup$ Have you looked at the binomial distribution? $\endgroup$ – James Jun 11 '15 at 13:47
  • $\begingroup$ If the number of items is large enough then you are on the right track (with binomial distribution) for an estimation. $\endgroup$ – drhab Jun 11 '15 at 13:47
  • $\begingroup$ 20C14 x (0.7)^14 x (0.3)^6 however I am not sure about the answer. Cannot detemine the standard deviation and mean. $\endgroup$ – Jon Jun 11 '15 at 13:47
  • $\begingroup$ @Jon that is correct. For a $\mathrm{Bin}(n,p)$ distribution the mean is $np$ and variance $np(1-p)$, so the standard deviation is $(np(1-p))^{\frac12}$, but that isn't relevant to answering this question. $\endgroup$ – Math1000 Jun 11 '15 at 13:50
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You can alternatively use the Normal approximation to the Binomial Distribution.

Where:

$\mu=np=(20)(0.7)=14$, $\sigma= \sqrt{npq}\approx 2.05$ and use the following:

$P(a \leq X \leq b) \approx \phi(\frac{b+ \frac{1}{2}-\mu}{\sigma})- \phi(\frac{a-\frac{1}{2}-mu}{\sigma})$

However since we are looking to find an exact result, i.e $P(X=14)$ we simply put 14 in for both $a$ and $b$

So:

$P(X=14) \approx \phi (0.24) - \phi(-0.24) $

$= \phi(0.24)-(1- \phi(0.24))$

$=2\phi(0.24)-1$

(From looking in the Normal table)

$2(0.5948)-1=0.1896$

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This is just the probability that a binomial distribution with parameters $n=20$, $p=\frac7{10}$ takes value $14$:

$$\binom{20}{14}\left(\frac7{10}\right)^{14}\left(\frac3{10}\right)^6 \approx 0.1916$$

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