Suppose cattle in a large herd have a mean weight of 1312lbs and a standard deviation of 54lbs.What is the probability that the mean weight

Question

Suppose cattle in a large herd have a mean weight of 1312lbs and a standard deviation of 54lbs.What is the probability that the mean weight of the sample of cows would differ from the population mean by greater than 9lbs if 117 cows are sampled at random from the herd

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Audrey 2 weeks 2021-09-13T23:54:55+00:00 1 Answer 0

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    2021-09-13T23:56:45+00:00

    Answer:

    7.18% probability that the mean weight of the sample of cows would differ from the population mean by greater than 9lbs if 117 cows are sampled at random from the herd

    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:

    \mu = 1312, \sigma = 54, n = 117, s = \frac{54}{\sqrt{117}} = 4.99

    Either the mean weight would differ by 9 lbs or less, or it would differ by greater than 9 lbs. The sum of the probabilities of these events is 100%.

    Probability it differs by 9 lbs or less

    pvalue of Z when X = 1312 + 9 = 1321 subtracted by the pvalue of Z when X = 1312 – 9 = 1303. So

    X = 1321

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

    By the Central Limit Theorem

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

    Z = \frac{1321 - 1312}{4.99}

    Z = 1.80

    Z = 1.80 has a pvalue of 0.9641

    X = 1303

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

    Z = \frac{1303 - 1312}{4.99}

    Z = -1.80

    Z = -1.80 has a pvalue of 0.0359

    0.9641 – 0.0359 = 0.9282

    92.82% probability that the mean weight of the sample differs from the mean by 9 lbs or less.

    What is the probability that the mean weight of the sample of cows would differ from the population mean by greater than 9lbs if 117 cows are sampled at random from the herd

    p + 92.82 = 100

    p = 7.18%

    7.18% probability that the mean weight of the sample of cows would differ from the population mean by greater than 9lbs if 117 cows are sampled at random from the herd

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