Part A: Conjecture

Correct Answer: B.
Explanation:

• The mean and median are expected to be higher for dogs because their weights are generally larger than the cats' weights.
• The standard deviation is also expected to be higher for dogs because their weights vary more significantly, especially with a range from 15 to 101 pounds, compared to the cats, whose weights are in the narrower range of 3 to 23 pounds.

Part B: Compute the Mean and Standard Deviation

To compute the mean and standard deviation, we use the following formulas:

• Mean: $\mu = \frac{\sum x_i}{n}$, where $x_i$ represents each value and $n$ is the number of values.
• Standard deviation: $\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{n}}$, which measures the spread of the data.

1. Dogs' weights:

• Weights: 15, 22, 38, 46, 59, 65, 79, 101
• $n = 8$

Mean for dogs: $\mu = \frac{15 + 22 + 38 + 46 + 59 + 65 + 79 + 101}{8} = \frac{425}{8} = 53.1$

Standard deviation for dogs: $\sigma = \sqrt{\frac{(15-53.1)^2 + (22-53.1)^2 + \dots + (101-53.1)^2}{8}} = 28.4$ (Rounded to one decimal place)

2. Cats' weights:

• Weights: 3, 3, 6, 9, 11, 14, 18, 23
• $n = 8$

Mean for cats: $\mu = \frac{3 + 3 + 6 + 9 + 11 + 14 + 18 + 23}{8} = \frac{87}{8} = 10.9$

Standard deviation for cats: $\sigma = \sqrt{\frac{(3-10.9)^2 + (3-10.9)^2 + \dots + (23-10.9)^2}{8}} = 6.7$ (Rounded to one decimal place)

Summary of Results:

• Mean for dogs: 53.1
• Mean for cats: 10.9
• Standard deviation for dogs: 28.4
• Standard deviation for cats: 6.7

Do you want more details on how I calculated the standard deviation?

Follow-up Questions:

1. What does a higher standard deviation indicate about the variability of the data?
2. How would the results change if you added another dog or cat with a significantly higher weight?
3. How do the mean and median compare, and what does that tell us about the distribution?
4. What real-life situations might cause a large difference in weight ranges for animals of the same species?
5. How would the shape of the distribution change if the weights were distributed evenly?

Tip:

When comparing two sets of data, the standard deviation is a helpful measure to see how spread out the values are. A larger standard deviation means the data has more variability.