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Interpret the variability of the following two datasets, representing the number of leaves on two different species of growing pothos plants: Species 1: {14,18,19,22,17} Species 2: {22,19,18,24,23} Which dataset is more variable? (1 point) Responses Species 1, because it has a higher mean absolute deviation Species 1, because it has a higher mean absolute deviation Species 1, because it has a lower mean absolute deviation Species 1, because it has a lower mean absolute deviation Species 2, because it has a higher mean absolute deviation Species 2, because it has a higher mean absolute deviation Species 2, because it has a lower mean absolute deviation Species 2, because it has a lower mean absolute deviation

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To interpret the variability of the two datasets, we can calculate the mean absolute deviation (MAD) for each species. The MAD measures how much the values in a dataset deviate from the mean of that dataset.

Dataset for Species 1: {14, 18, 19, 22, 17}

  1. Calculate the mean: \[ \text{Mean} = \frac{14 + 18 + 19 + 22 + 17}{5} = \frac{90}{5} = 18 \]
  2. Calculate the absolute deviations from the mean:
    • |14 - 18| = 4
    • |18 - 18| = 0
    • |19 - 18| = 1
    • |22 - 18| = 4
    • |17 - 18| = 1
  3. Calculate the mean absolute deviation: \[ \text{MAD} = \frac{4 + 0 + 1 + 4 + 1}{5} = \frac{10}{5} = 2 \]

Dataset for Species 2: {22, 19, 18, 24, 23}

  1. Calculate the mean: \[ \text{Mean} = \frac{22 + 19 + 18 + 24 + 23}{5} = \frac{106}{5} = 21.2 \]
  2. Calculate the absolute deviations from the mean:
    • |22 - 21.2| = 0.8
    • |19 - 21.2| = 2.2
    • |18 - 21.2| = 3.2
    • |24 - 21.2| = 2.8
    • |23 - 21.2| = 1.8
  3. Calculate the mean absolute deviation: \[ \text{MAD} = \frac{0.8 + 2.2 + 3.2 + 2.8 + 1.8}{5} = \frac{10.8}{5} = 2.16 \]

Conclusion:

  • Species 1 has a mean absolute deviation of 2.
  • Species 2 has a mean absolute deviation of 2.16.

Since the mean absolute deviation for Species 2 (2.16) is higher than that for Species 1 (2), Species 2 is more variable. Therefore, the correct response is:

Species 2, because it has a higher mean absolute deviation.