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- anonymous

A manufacturer of batteries knows that 10% of the batteries produced by a particular production line are defective. A sample of 200 batteries is removed for testing.
a) What is the probability that exactly 20 batteries in the sample are defective?
b) What is the probability that at least 20 batteries in the sample are defective?

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- anonymous

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- anonymous

help p/zzzzzzz

- kropot72

a) The binomial distribution can be used to find the required probability as follows:
\[P(exactly\ 20\ defective)=\left(\begin{matrix}200 \\ 20\end{matrix}\right)0.1^{20}(1-0.1)^{180}=you\ can\ calculate\]

- kropot72

@hager I also used the hypergeometric distribution assuming a production run of 1000 batteries containing 100 defectives and a sample size of 200. The result was very close to the probability calculated using the binomial distribution.

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- kropot72

@hager Please ask if you need more explanation.

- anonymous

thnx so much I'm trying to work on it but still i didn't get it

- kropot72

What result do you get when you calculate using the numbers for the binomial distribution that I gave above?

- anonymous

i couldn't do it

- anonymous

320

- kropot72

The result of the calculation is 0.0936

- anonymous

how

- kropot72

The calculation of the binomial coefficient is as follows:
\[\left(\begin{matrix}200 \\20 \end{matrix}\right)=\frac{200!}{20!180!}=1.613588\times 10^{27}\]
Then continuing the calculation:
\[1.613588\times 10^{27}\times (0.1)^{20}\times (1-0.1)^{180}=0.0936\]
So the probability of exactly 20 batteries in the sample being defective = 0.0936

- anonymous

thnx so much i really appreciate that!!!

- kropot72

You're welcome :)

- anonymous

help for b?

- kropot72

I am looking at it. Please wait.

- kropot72

When the sample size is large (200 batteries in this case) and the probability of a defective is neither small nor near 1, the binomial distribution can be approximated by the normal distribution with mean np where n is the sample size and p is the probability of a defective.
In this case np = 200 * 0.1 = 20
Therefore the probability of at least 20 batteries in the sample being defective is half the area under the distribution curve = 0.5.

- anonymous

thnx

- kropot72

You're welcome :)

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