The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 30,393 miles, with a standard deviation of

Question

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 30,393 miles, with a standard deviation of 2876 miles. What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

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Charlie 2 months 2021-10-15T16:25:32+00:00 1 Answer 0 views 0

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    2021-10-15T16:26:49+00:00

    Answer:

    52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

    Step-by-step explanation:

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

    Normal probability distribution:

    Problems of normally distributed samples are 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, a large sample size 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 = 30393, \sigma = 2876, n = 37, s = \frac{2876}{\sqrt{37}} = 472.81

    What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

    This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 – 339 = 30054. So

    X = 30732

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

    By the Central Limit Theorem

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

    Z = \frac{30732 - 30393}{472.81}

    Z = 0.72

    Z = 0.72 has a pvalue of 0.7642.

    X = 30054

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

    Z = \frac{30054 - 30393}{472.81}

    Z = -0.72

    Z = -0.72 has a pvalue of 0.2358

    0.7642 – 0.2358 = 0.5284

    52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

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