Use the magnitudes​ (Richter scale) of the earthquakes listed in the data set below. Find the mean and median of this data set. Is the magni

Question

Use the magnitudes​ (Richter scale) of the earthquakes listed in the data set below. Find the mean and median of this data set. Is the magnitude of an earthquake measuring 7.0 on the Richter scale an outlier​ (data value that is very far away from the​ others) when considered in the context of the sample data given in this data​ set? Explain.

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Adela 5 months 2021-08-19T09:08:16+00:00 1 Answers 21 views 0

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    2021-08-19T09:09:37+00:00

    Answer:

    \bar x = 1.4896

    Median = 1.525

    7 is an outlier

    Step-by-step explanation:

    Given

    See comment for dataset

    Solving (a): The mean

    The mean is calculated as:

    \bar x = \frac{\sum x}{n}

    So, we have:

    \bar x = \frac{2.53 +1.33 +0.64 +2.55 +0.67 +0.57 +............+2.03 +1.78}{50}

    \bar x = \frac{74.48}{50}

    \bar x = 1.4896

    Solving (b): The median

    First, we sort the data:

    0.01, 0.01, 0.08, 0.14, 0.24, 0.35, 0.57, 0.64, 0.65, 0.67,

    0.69, 0.74, 0.81, 0.82, 0.92, 0.99, 1.01, 1.04, 1.15, 1.16, 1.33,

    1.41, 1.45, 1.48, 1.52, 1.53, 1.58, 1.61, 1.62, 1.73, 1.77, 1.78, 1.84,

    1.98, 2.01, 2.02, 2.03, 2.06, 2.06, 2.31, 2.41, 2.53, 2.55, 2.59,

    2.66, 2.74, 2.74, 2.77, 2.77, 2.91

    The median position is:

    Median = \frac{n + 1}{2}

    Median = \frac{50 + 1}{2}

    Median = \frac{51}{2}

    Median = 25.5th

    This means that the median is the average of the 25th and the 26th item.

    So:

    Median = \frac{1.52+1.53}{2}

    Median = \frac{3.05}{2}

    Median = 1.525

    Solving (c): Is 7 an outlier

    Yes; 7 is an outlier.

    Because the range of the dataset (0.01 to 2.92) is far from 7.

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