Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found th

Question

Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let X be the random variable representing the time it takes her to complete one review. Assume X is normally distributed. Let X be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews. Complete the distributions.
A. Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs.
B. P(_____) = _______.
C. Find the 95th percentile for the mean time to complete one month’s reviews.
D.The 95th Percentile =________.

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Diễm Kiều 3 days 2021-07-21T21:05:09+00:00 1 Answers 0 views 0

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    2021-07-21T21:06:36+00:00

    Answer:

    a) P(3.5 \leq X \leq 4.25) = 0.7492

    b) The 95th percentile is 4.4935 hours.

    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 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 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 reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours.

    This means that \mu = 4, \sigma = 1.2

    16 reviews.

    This means that n = 16, s = \frac{1.2}{\sqrt{16}} = 0.3

    A. Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs.

    This is the pvalue of Z when X = 4.25 subtracted by the pvalue of Z when X = 3.5. So

    X = 4.25

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

    By the Central Limit Theorem

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

    Z = \frac{4.25 - 4}{0.3}

    Z = 0.83

    Z = 0.83 has a pvalue of 0.7967

    X = 3.5

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

    Z = \frac{3.5 - 4}{0.3}

    Z = -1.67

    Z = -1.67 has a pvalue of 0.0475

    0.7967 – 0.0475 = 0.7492

    So

    P(3.5 \leq X \leq 4.25) = 0.7492

    C. Find the 95th percentile for the mean time to complete one month’s reviews.

    This is X when Z has a pvalue of 0.95, so X when Z = 1.645.

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

    1.645 = \frac{X - 4}{0.3}

    X - 4 = 0.3*1.645

    X = 4.4935

    The 95th percentile is 4.4935 hours.

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