Suppose a batch of steel rods produced at a steel plant have a mean length of 194 millimeters, and a variance of 121. If 339 rods are sample

Question

Suppose a batch of steel rods produced at a steel plant have a mean length of 194 millimeters, and a variance of 121. If 339 rods are sampled at random from the batch, what is the probability that the mean length of the sample rods would differ from the population mean by more than 0.95 millimeters

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Delilah 2 months 2021-10-16T02:52:20+00:00 1 Answer 0 views 0

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    2021-10-16T02:53:33+00:00

    Answer:

    11.18% probability that the mean length of the sample rods would differ from the population mean by more than 0.95 millimeters

    Step-by-step explanation:

    To solve this question, we need 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, 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:

    Remembering that the standard deviation is the square root of the variance. So

    \mu = 194, \sigma = \sqrt{121} = 11, n = 339, s = \frac{11}{\sqrt{339}} = 0.5979

    If 339 rods are sampled at random from the batch, what is the probability that the mean length of the sample rods would differ from the population mean by more than 0.95 millimeters

    Either it differs by 0.95 millimeters or less, or it differs by more than 0.95 millimeters. The sum of these probabilities is 100.

    Probability that it differs by 0.95 millimeters or less

    pvalue of Z when X = 194+0.95 = 194.95 subtracted by the pvalue of Z when X = 194 – 0.95 = 193.05. So

    X = 194.95

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

    By the Central Limit Theorem

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

    Z = \frac{194.95 - 194}{0.5979}

    Z = 1.59

    Z = 1.59 has a pvalue of 0.9441

    X = 193.05

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

    Z = \frac{193.0.5 - 194}{0.5979}

    Z = -1.59

    Z = -1.59 has a pvalue of 0.0559

    0.9441 – 0.0559 = 0.8882

    88.82% probability that the mean length of the sample rods would differ from the population mean by less than 0.95 millimeters

    More than 0.95 millimeters

    p + 88.82 = 100

    p = 11.18

    11.18% probability that the mean length of the sample rods would differ from the population mean by more than 0.95 millimeters

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