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veronika’s five test scores are 59, 80, 95, 88, and 93. if the outlier …

Question

veronika’s five test scores are 59, 80, 95, 88, and 93. if the outlier of 59 is removed, what is the mean absolute deviation of the remaining four test scores?
○ 4.5
○ 5
○ 10.8
○ 36

Explanation:

Step1: Find the mean of the remaining scores

First, identify the remaining scores after removing 59: 80, 95, 88, 93.
The formula for the mean $\bar{x}$ is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 4$ and $x_{i}$ are the scores.
$\sum_{i=1}^{4}x_{i}=80 + 95+88 + 93=356$
$\bar{x}=\frac{356}{4}=89$

Step2: Calculate the absolute deviations

For each score, find $|x_{i}-\bar{x}|$:

  • For 80: $|80 - 89|=9$
  • For 95: $|95 - 89|=6$
  • For 88: $|88 - 89|=1$
  • For 93: $|93 - 89|=4$

Step3: Find the mean of these absolute deviations

The formula for the mean absolute deviation (MAD) is $MAD=\frac{\sum_{i = 1}^{n}|x_{i}-\bar{x}|}{n}$
$\sum_{i = 1}^{4}|x_{i}-\bar{x}|=9 + 6+1 + 4=20$
$MAD=\frac{20}{4}=5$

Answer:

5