The cost of 5 gallons of ice cream has a variance of 64 with a mean of 34 dollars during the summer. What is the probability that the sampl

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The cost of 5 gallons of ice cream has a variance of 64 with a mean of 34 dollars during the summer. What is the probability that the sample mean would differ from the true mean by less than 1.1 dollars if a sample of 38 5-gallon pails is randomly selected

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Thiên Ân 4 years 2021-08-11T18:31:17+00:00 1 Answers 71 views 0

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  1. Ladonna
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    2021-08-11T18:32:56+00:00
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    Answer:

    0.5587 = 55.87% probability that the sample mean would differ from the true mean by less than 1.1 dollars.

    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 normal distributions can be solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

    Central Limit Theorem

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

    For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

    The cost of 5 gallons of ice cream has a variance of 64 with a mean of 34 dollars during the summer.

    This means that \sigma = \sqrt{64} = 8, \mu = 34

    Sample of 38

    This means that n = 38, s = \frac{8}{\sqrt{38}}

    What is the probability that the sample mean would differ from the true mean by less than 1.1 dollars ?

    P-value of Z when X = 34 + 1.1 = 35.1 subtracted by the p-value of Z when X = 34 – 1.1 = 32.9. So

    X = 35.1

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

    By the Central Limit Theorem

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

    Z = \frac{35.1 - 34}{\frac{8}{\sqrt{38}}}

    Z = 0.77

    Z = 0.77 has a p-value of 0.77935

    X = 32.9

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

    Z = \frac{32.9 - 34}{\frac{8}{\sqrt{38}}}

    Z = -0.77

    Z = -0.77 has a p-value of 0.22065

    0.77935 – 0.22065 = 0.5587

    0.5587 = 55.87% probability that the sample mean would differ from the true mean by less than 1.1 dollars.

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