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Suppose a large shipment of laser printers contained 22% defectives. If a sample of size 276 is selected, what is the probability that the sample proportion will differ from the population proportion by less than 6%

User Colyn
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Final answer:

To calculate the probability that the sample proportion will differ from the population proportion by less than 6%, we can use the formula for the standard error of the proportion and the properties of the standard normal distribution. In this case, the probability is approximately 98.9%.

Step-by-step explanation:

To calculate the probability that the sample proportion will differ from the population proportion by less than 6%, we can use the formula for the standard error of the proportion:

SEp = √[(p)(1-p)/n]

In this case, the population proportion is 22% (or 0.22) and the sample size is 276. Plugging these values into the formula, we get:

SEp = √[(0.22)(1-0.22)/276] ≈ 0.020

Next, we need to find the z-score corresponding to a 6% difference from the population proportion. We can use a standard normal distribution table or a calculator to find that the z-score for a 6% difference is approximately 2.17.

Finally, we can use the z-score and the standard error to calculate the probability.

P(|p - P| < 0.06) = P(-2.17 < (p - P)/SEp < 2.17)

This probability can be approximated by finding the area under the standard normal curve between -2.17 and 2.17. Using a standard normal distribution table or a calculator, we find that this probability is approximately 0.989 or 98.9%.

User Long Nguyen Duc
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