Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 128 millimeters, and a stand

Question

Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 128 millimeters, and a standard deviation of 8 millimeters. If a random sample of 41 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by greater than 0.9 millimeters

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Caroline 1 week 2021-10-10T04:15:29+00:00 1 Answer 0

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    2021-10-10T04:17:09+00:00

    Answer:

    47.16% probability that the sample mean would differ from the population mean by greater than 0.9 millimeters

    Step-by-step explanation:

    To solve this question, we have to understand the central limit theorem and the normal probability distribution.

    Normal probability distribution:

    Problems of normally distributed samples can be solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

    Central limit theorem:

    The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

    In this problem, we have that:

    \mu = 128, \sigma = 8, n = 41, s = \frac{8}{\sqrt{41}} = 1.25

    If a random sample of 41 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by greater than 0.9 millimeters

    Either the difference is 0.9 millimeters or less, or it is more than 0.9 millimeters. The sum of the probabilities of these events is 100. Initially, i will find the probability that they differ by less than 0.9 millimeters.

    Probability that they differ by less than 0.9 millimeters.

    Pvalue of Z when X = 128 + 0.9 = 128.9 subtracted by the pvalue of Z when X = 128 – 0.9 = 127.1. So:

    X = 128.9

    Z = \frac{X - \mu}{\sigma}

    By the Central Limit Theorem

    Z = \frac{X - \mu}{s}

    Z = \frac{128.9 - 128}{1.25}

    Z = 0.72

    Z = 0.72 has a pvalue of 0.7642

    X = 127.1

    Z = \frac{X - \mu}{s}

    Z = \frac{127.1 - 128}{1.25}

    Z = -0.72

    Z = -0.72 has a pvalue of 0.2358

    0.7642 – 0.2358 = 0.5284

    52.84% probability that they differ by less than 0.9 millimeters

    52.84 + p = 100

    p = 47.16

    47.16% probability that the sample mean would differ from the population mean by greater than 0.9 millimeters

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