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A manufacturing process is designed to produce ball bearings with a diameter of 0.620 cm. the standard deviation of all ball bearing diameters is 0.009 cm. every day a random sample of 40 ball bearings is selected and their diameters are measured. if the mean diameter from this sample falls outside of a given range, then the management will stop production to make adjustments. for the range 0.619 cm to 0.621 cm, find the probability that the management will stop production unnecessarily. (hint: find the probability that a sample’s mean will have a value inside this range when the population mean actually is 0.620 cm, and then use complements to find the probability that a sample mean’s value is outside that range.) a. 0.4822 b. 0.5178 c. 0.0885 d. 0.9115

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Final answer:

To find the probability that the management will stop production unnecessarily, we can find the probability that a sample's mean will have a value inside a given range when the population mean is actually 0.620 cm. By using the standard deviation of the population, the sample size, and the z-score formula, we can calculate the probability associated with the range limits and then subtract it from 1 to find the probability of the sample mean falling outside the range. In this case, the probability is 0.4822.

Step-by-step explanation:

To find the probability that the management will stop production unnecessarily, we need to find the probability that a sample's mean will have a value inside the given range when the population mean is 0.620 cm.

To do this, we can use the standard deviation of the population, which is 0.009 cm, and the sample size, which is 40, to find the standard error of the mean (SEM). The formula for SEM is σ/√n, where σ is the standard deviation and n is the sample size. So, SEM = 0.009/√40 = 0.001424 cm.

Next, we can use the z-score formula to standardize the range limits. The z-score formula is (x - μ)/SEM, where x is the value, μ is the population mean, and SEM is the standard error of the mean. For the lower limit of 0.619 cm, the z-score is (0.619 - 0.620)/0.001424 = -0.7028. For the upper limit of 0.621 cm, the z-score is (0.621 - 0.620)/0.001424 = 0.7028.

Now, we can use a standard normal distribution table or calculator to find the probability associated with the z-scores. The probability of a z-score being less than -0.7028 is 0.2411, and the probability of a z-score being less than 0.7028 is 0.7589.

Lastly, we can subtract the probability of the sample mean falling inside the range from 1 to find the probability of it falling outside the range. So, the probability that the management will stop production unnecessarily is 1 - (0.7589 - 0.2411) = 0.4822.

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