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A manufacturer produces a commodity where the length of the commodity has approximately normaldistribution with a mean of 15.7 inches and standard deviation of 1.4 inches. If a sample of 45 items are chosenat random, what is the probability the sample's mean length is greater than 15.7 inches? Round answer to fourdecimal places

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Final answer:

To find the probability, calculate the z-score using the formula, where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. The z-score of 0 indicates that the sample mean is the same as the population mean. Look up the area under the normal distribution curve for z = 0, subtract it from 1, and find the probability.

Step-by-step explanation:

To find the probability that the sample's mean length is greater than 15.7 inches, we need to calculate the z-score for the given mean and standard deviation.

First, determine the z-score using the formula: z = (x - μ) / (σ / √n), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, x = 15.7 inches, μ = 15.7 inches, σ = 1.4 inches, and n = 45. Plugging these values into the z-score formula:

z = (15.7 - 15.7) / (1.4 / √45) = 0 / 0.2082 = 0

The z-score of 0 indicates that the sample mean is the same as the population mean. To find the probability that the sample's mean is greater than 15.7 inches, we can look up the area under the normal distribution curve for z = 0, which is 0.5, and subtract it from 1:

Probability = 1 - 0.5 = 0.5

Therefore, the probability that the sample's mean length is greater than 15.7 inches is 0.5.

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