To find the mean and median of the numbers you've provided, let's follow the steps:

Step 1: List of numbers

You have the following numbers:

1.09, 2.81, 0.14, 1.21, 1.42, 0.96, 0.01, 2.23, 1.03, 0.74, 0.74, 0.58, 1.36, 2.63, 1.14, 1.78, 1.24, 0.17, 1.68, 0.89, 0.75, 1.17, 1.56, 1.71, 0.12, 1.68, 0.23, 2.62, 1.24, 1.67, 1.63, 1.86, 1.99, 2.49, 2.86, 0.94, 2.48, 1.15, 1.67, 0.49, 2.59, 2.54, 0.46, 1.15, 0.56, 0.32, 2.89, 2.71, 1.79, 0.42.

Step 2: Mean Calculation

The mean is the sum of all the numbers divided by the number of values.

$\text{Mean} = \frac{\text{Sum of all values}}{\text{Total number of values}}$

Step 3: Median Calculation

The median is the middle value when the data is arranged in ascending order. If the number of values is odd, it's the middle one. If even, it's the average of the two middle values.

Now, let me calculate both values.

I'll proceed with the calculation. numpy as np

List of numbers provided by the user

numbers = [ 1.09, 2.81, 0.14, 1.21, 1.42, 0.96, 0.01, 2.23, 1.03, 0.74, 0.74, 0.58, 1.36, 2.63, 1.14, 1.78, 1.24, 0.17, 1.68, 0.89, 0.75, 1.17, 1.56, 1.71, 0.12, 1.68, 0.23, 2.62, 1.24, 1.67, 1.63, 1.86, 1.99, 2.49, 2.86, 0.94, 2.48, 1.15, 1.67, 0.49, 2.59, 2.54, 0.46, 1.15, 0.56, 0.32, 2.89, 2.71, 1.79, 0.42 ]

Calculate mean

mean_value = np.mean(numbers)

Calculate median

median_value = np.median(numbers)

mean_value, median_valueThe calculations are as follows:

• Mean: 1.3918 (rounded to 4 decimal places, as requested)
• Median: 1.24

Would you like a detailed explanation of how I arrived at these values or have any questions? Here are 5 related questions you might find interesting:

1. How does the mean change if a few extreme values are removed from the data?
2. What is the difference between the mean and median when dealing with skewed data?
3. How do you calculate the mode, and what does it represent in this dataset?
4. Can you explain the impact of outliers on the mean and median?
5. What are other measures of central tendency, and how do they compare with the mean and median?

Tip: For large datasets, using the median is often a better representation of the "center" because it is less affected by extreme values (outliers).