Let's solve the given problem step by step.

Data Set:

$41.6, 68, 58.6, 68, 46.6, 68, 30.7, 32.2, 20.7, 47.2$

1. Arithmetic Mean

The arithmetic mean is calculated by summing all the values and then dividing by the number of values.

$\text{Mean} = \frac{\sum \text{data points}}{\text{number of data points}}$

$\text{Sum of data points} = 41.6 + 68 + 58.6 + 68 + 46.6 + 68 + 30.7 + 32.2 + 20.7 + 47.2 = 482.6$

$\text{Number of data points} = 10$

$\text{Mean} = \frac{482.6}{10} = 48.26$

2. Median

The median is the middle value when the data points are arranged in ascending order. If there is an even number of data points, the median is the average of the two middle values.

First, let's arrange the data in ascending order:

$20.7, 30.7, 32.2, 41.6, 46.6, 47.2, 58.6, 68, 68, 68$

Since there are 10 data points, the median will be the average of the 5th and 6th values:

$\text{Median} = \frac{46.6 + 47.2}{2} = \frac{93.8}{2} = 46.9$

3. Mode

The mode is the value that appears most frequently in the data set.

From the data set:

$68 \text{ appears 3 times, which is the most frequent.}$

So,

$\text{Mode} = 68$

Final Answers:

• Mean = 48.26
• Median = 46.9
• Mode = 68

Would you like more details or have any questions? Here are five related questions:

1. How would the mean change if an additional value of 50 was added to the data set?
2. Can a data set have more than one mode? If so, how is it called?
3. How do you calculate the median for an odd number of data points?
4. What is the difference between the mean and the median in a skewed distribution?
5. How can outliers affect the mean and median?

Tip: The median is less sensitive to outliers compared to the mean, making it a better measure of central tendency for skewed distributions.