According to records, the amount of precipitation in a certain city on a November day has a mean of inches, with a standard deviation of inc

Question

According to records, the amount of precipitation in a certain city on a November day has a mean of inches, with a standard deviation of inches. What is the probability that the mean daily precipitation will be inches or less for a random sample of November days (taken over many years)

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Lệ Thu 5 years 2021-08-27T07:51:40+00:00 1 Answers 13 views 0

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  1. RobertKer
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    2021-08-27T07:52:55+00:00
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    Answer:

    The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is mean amount of inches of rain and \sigma is the standard deviation.

    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.

    In this question:

    Mean \mu, standard deviation \sigma

    n days:

    This means that s = \frac{\sigma}{\sqrt{n}}

    Applying the Central Limit Theorem to the z-score formula.

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

    What is the probability that the mean daily precipitation will be of X inches or less for a random sample of November days?

    The probability that the mean daily precipitation will be of X inches or less for a random sample of n November days is the p-value of Z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}, in which \mu is mean amount of inches of rain and \sigma is the standard deviation.

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