From the question, let's compute the requested values step-by-step.

Given Data:

Dogs' weights: 14, 21, 36, 45, 55, 64, 78, 102
Cats' weights: 3, 3, 6, 8, 11, 14, 18, 28

Step 1: Compute Mean (Average)

Formula for Mean:
$\text{Mean} = \frac{\text{Sum of all weights}}{\text{Number of animals}}$

Dogs' Mean: $\text{Sum of dogs' weights} = 14 + 21 + 36 + 45 + 55 + 64 + 78 + 102 = 415$
$\text{Mean} = \frac{415}{8} = 51.875$

Cats' Mean: $\text{Sum of cats' weights} = 3 + 3 + 6 + 8 + 11 + 14 + 18 + 28 = 91$
$\text{Mean} = \frac{91}{8} = 11.375$

Step 2: Compute Standard Deviation

Formula for Standard Deviation:
$\text{SD} = \sqrt{\frac{\sum{(x - \text{mean})^2}}{n}}$

Dogs' Standard Deviation:

1. Compute deviations: $(x - \text{mean})$: $14 - 51.875, \, 21 - 51.875, \, 36 - 51.875, \, \dots, \, 102 - 51.875$ Square these deviations and sum them up: $\text{Sum of squared deviations} = 1487.109375$
2. Divide by $n = 8$, then take the square root: $\text{SD} = \sqrt{\frac{1487.109375}{8}} = 13.60$

Cats' Standard Deviation:

1. Compute deviations: $(x - \text{mean})$: $3 - 11.375, \, 3 - 11.375, \, 6 - 11.375, \, \dots, \, 28 - 11.375$ Square these deviations and sum them up: $\text{Sum of squared deviations} = 382.875$
2. Divide by $n = 8$, then take the square root: $\text{SD} = \sqrt{\frac{382.875}{8}} = 6.90$

Step 3: Analyze Median and Spread

Dogs' Median:

• Sort weights: 14, 21, 36, 45, 55, 64, 78, 102
• Middle values: 45, 55
• Median = $\frac{45 + 55}{2} = 50$

Cats' Median:

• Sort weights: 3, 3, 6, 8, 11, 14, 18, 28
• Middle values: 8, 11
• Median = $\frac{8 + 11}{2} = 9.5$

Final Conjectures:

1. Dogs have a higher mean (51.875 vs. 11.375), median (50 vs. 9.5), and standard deviation (13.60 vs. 6.90).
2. The spread in dogs’ weights is higher because their weights vary more widely.

Let me know if you want further clarification or deeper delving into any part!
Here are five related questions for you to think about:

1. What causes the larger spread in the dogs' weights?
2. How might a larger spread influence decision-making in a veterinary context?
3. Could outliers affect these calculations significantly?
4. How is variance related to standard deviation?
5. Why is the median less affected by outliers than the mean?

Tip: Always double-check your computation steps for accuracy, especially for mean and standard deviation!