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anonymous
 4 years 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?
anonymous
 4 years 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?

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kropot72
 4 years ago
Best ResponseYou've already chosen the best response.1a) The binomial distribution can be used to find the required probability as follows: \[P(exactly\ 20\ defective)=\left(\begin{matrix}200 \\ 20\end{matrix}\right)0.1^{20}(10.1)^{180}=you\ can\ calculate\]

kropot72
 4 years ago
Best 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.

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.1@hager Please ask if you need more explanation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thnx so much I'm trying to work on it but still i didn't get it

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.1What result do you get when you calculate using the numbers for the binomial distribution that I gave above?

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.1The result of the calculation is 0.0936

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.1The 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thnx so much i really appreciate that!!!

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.1I am looking at it. Please wait.

kropot72
 4 years ago
Best ResponseYou've already chosen the best response.1When 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.
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