To solve the problem, let's calculate the mean, median, and standard deviation for the provided data:

Data set: $x = \{26.5, 23.5, 23.8, 28.9, 16.5, 5.9, 4.9, 7.8, 19.4\}$

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1. Mean (Average):

The mean is calculated using the formula: $\text{Mean} = \frac{\sum x_i}{n}$ Where $\sum x_i$ is the sum of the data points and $n$ is the number of data points.

$\sum x_i = 26.5 + 23.5 + 23.8 + 28.9 + 16.5 + 5.9 + 4.9 + 7.8 + 19.4 = 156.7$ $n = 9$ $\text{Mean} = \frac{156.7}{9} = 17.41$

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2. Median:

The median is the middle value of the ordered dataset. First, sort the data in ascending order: $\{4.9, 5.9, 7.8, 16.5, 19.4, 23.5, 23.8, 26.5, 28.9\}$ The dataset has 9 elements ($n$ is odd), so the median is the 5th value: $\text{Median} = 19.4$

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3. Standard Deviation (Population):

The standard deviation is calculated using: $\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{n}}$ Where $\mu$ is the mean.

1. Subtract the mean from each data point, square the result, and sum the squares: $(x_i - \mu)^2 = (26.5 - 17.41)^2 + (23.5 - 17.41)^2 + \dots + (19.4 - 17.41)^2$

$(26.5 - 17.41)^2 = 82.59,\; (23.5 - 17.41)^2 = 37.01, \dots$

The sum of squares is: $\sum (x_i - \mu)^2 = 563.77$

2. Divide by $n$ and take the square root: $\sigma = \sqrt{\frac{563.77}{9}} = \sqrt{62.64} = 7.91$

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Final Results:

1. Mean: $17.41$
2. Median: $19.4$
3. Standard Deviation: $7.91$

Let me know if you'd like detailed steps or further explanations! 😊

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Related Questions:

1. What happens to the mean if the value $28.9$ is removed from the dataset?
2. How does the range of the data affect the standard deviation?
3. Can the median be calculated for a dataset with even $n$?
4. What is the variance for the given data?
5. How would adding a constant value to each $x$ affect the mean and standard deviation?

Tip: The standard deviation measures how spread out data points are around the mean. A higher standard deviation means greater variability.