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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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kropot72Best ResponseYou've already chosen the best response.1
a) The binomial distribution can be used to find the required probability as follows: \[P(exactly\ 20\ defective)=\left(\begin{matrix}200 \\ 20\end{matrix}\right)0.1^{20}(10.1)^{180}=you\ can\ calculate\]
 one year ago

kropot72Best ResponseYou've already chosen the best response.1
@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.
 one year ago

kropot72Best ResponseYou've already chosen the best response.1
@hager Please ask if you need more explanation.
 one year ago

hagerBest ResponseYou've already chosen the best response.0
thnx so much I'm trying to work on it but still i didn't get it
 one year ago

kropot72Best ResponseYou've already chosen the best response.1
What result do you get when you calculate using the numbers for the binomial distribution that I gave above?
 one year ago

kropot72Best ResponseYou've already chosen the best response.1
The result of the calculation is 0.0936
 one year ago

kropot72Best ResponseYou've already chosen the best response.1
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: \[1.613588\times 10^{27}\times (0.1)^{20}\times (10.1)^{180}=0.0936\] So the probability of exactly 20 batteries in the sample being defective = 0.0936
 one year ago

hagerBest ResponseYou've already chosen the best response.0
thnx so much i really appreciate that!!!
 one year ago

kropot72Best ResponseYou've already chosen the best response.1
I am looking at it. Please wait.
 one year ago

kropot72Best ResponseYou've already chosen the best response.1
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.
 one year ago
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