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A manufacturer knows that their items have lengths that are approximately normally distributed, with a mean of 9.1 inches, and standard deviation of 1.8 inches. If 38 items are chosen at random, what is the probability that their mean length is greater than 8.8 inches?

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Final answer:

To find the probability, we can use the central limit theorem and the standard normal distribution. First, calculate the standard error of the mean. Next, calculate the z-score. Finally, use a standard normal distribution table or calculator to find the probability and interpret the result.

Step-by-step explanation:

We can solve this problem by using the central limit theorem and the standard normal distribution. First, we need to calculate the standard error of the mean (SEM). The SEM is equal to the standard deviation divided by the square root of the sample size. In this case, the standard deviation is 1.8 inches and the sample size is 38, so the SEM is 1.8 / √38. Next, we can calculate the z-score using the formula z = (sample mean - population mean) / SEM. Plugging in the values, we get z = (8.8 - 9.1) / (1.8 / √38).

Now, we can use a standard normal distribution table or a calculator to find the probability associated with the z-score. We are interested in the probability that the sample mean is greater than 8.8 inches, so we need to find the area to the right of the z-score.

Finally, we interpret the probability. Since the z-score represents the number of standard deviations a sample mean is away from the population mean, the probability tells us the likelihood of obtaining a sample mean greater than 8.8 inches. By finding the area to the right of the z-score, we can determine the probability that the mean length is greater than 8.8 inches.

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