To answer your question, we'll need to:

1. Plot a frequency histogram of the given data (weights of M&M candies).
2. Calculate the mean and median of the data.
3. Determine which measure of central tendency (mean or median) better describes the weights of the M&Ms.

Let me first calculate the mean and median and then create a histogram for you.

Step 1: Calculate the Mean and Median

Mean:

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

$\text{Mean} = \frac{\sum \text{weights}}{n}$

Median:

The median is the middle value of the sorted data. If there is an odd number of values, it’s the middle number; if there is an even number of values, it’s the average of the two middle numbers.

Now, I will compute these values based on the data provided. numpy as np

M&M candy weight data

weights = [ 0.87, 0.88, 0.82, 0.90, 0.90, 0.84, 0.84, 0.91, 0.94, 0.86, 0.86, 0.86, 0.88, 0.87, 0.89, 0.91, 0.86, 0.87, 0.93, 0.88, 0.83, 0.95, 0.87, 0.93, 0.91, 0.85, 0.91, 0.86, 0.89, 0.87, 0.84, 0.88, 0.88, 0.89, 0.82, 0.83, 0.90, 0.88, 0.84, 0.93, 0.81, 0.90, 0.88, 0.92, 0.85, 0.84, 0.84, 0.86, 0.87, 0.88, 0.82, 0.90, 0.90, 0.84, 0.84, 0.91, 0.94, 0.86, 0.86, 0.86, 0.88, 0.87, 0.89, 0.91, 0.86, 0.87, 0.93, 0.88, 0.83, 0.95, 0.87, 0.93, 0.91, 0.85, 0.91, 0.86, 0.89, 0.87, 0.84, 0.88, 0.88, 0.89, 0.82, 0.83, 0.90, 0.88, 0.84, 0.93, 0.81, 0.90, 0.88, 0.92, 0.85, 0.84, 0.84, 0.86 ]

Calculating the mean and median

mean_weight = np.mean(weights) median_weight = np.median(weights)

mean_weight, median_weightThe calculations for the central tendencies are as follows:

• Mean weight of the M&Ms: $0.8756$ grams
• Median weight of the M&Ms: $0.875$ grams

Step 2: Histogram

Now, I will generate a frequency histogram to visually analyze the distribution of the M&M weights. matplotlib.pyplot as plt

Plotting the histogram

plt.figure(figsize=(8, 6)) plt.hist(weights, bins=10, color='skyblue', edgecolor='black', alpha=0.7) plt.title('Frequency Histogram of M&M Candy Weights') plt.xlabel('Weight (grams)') plt.ylabel('Frequency') plt.grid(True) plt.show()The histogram above shows the frequency distribution of the M&M candy weights. From this plot, we can observe the shape of the distribution and make some conclusions.

Shape of the Distribution:

The distribution appears to be approximately symmetric with a slight tendency towards a normal distribution. The data points are concentrated around the mean (0.875 grams), with a peak near this value.

Measure of Central Tendency:

Both the mean (0.8756 grams) and the median (0.875 grams) are very close to each other, suggesting that the distribution is symmetric. In this case, both measures describe the data similarly. However, since the distribution is close to normal, the mean provides a good summary of the data.

Conclusion:

• The mean is a good representation of the central tendency of the candy weights due to the symmetric nature of the distribution.

Would you like more details on any of the calculations, or would you like to explore further?

Related Questions:

1. How do outliers affect the mean and median?
2. What are some methods for improving the accuracy of the mean in data analysis?
3. How do you determine the number of bins for a histogram?
4. What is the significance of the skewness in a distribution?
5. How can you check if a dataset follows a normal distribution?

Tip:

If you encounter a skewed distribution, the median might be a more reliable measure of central tendency than the mean because it’s less influenced by extreme values.