Given that the weights of this dog breed are approximately normally distributed with a mean (μ) = 52 pounds and a standard deviation (σ) = 6.6 pounds, we can answer these questions by using the standard normal distribution (also called the Z-distribution).

We'll calculate the Z-scores for the values, which tells us how many standard deviations away a given value is from the mean. The Z-score is given by the formula:

$Z = \frac{X - \mu}{\sigma}$

Once we have the Z-scores, we can use standard normal distribution tables (or a graphing calculator) to find the corresponding percentages.

a) Find the percentage of dogs that weigh less than 52 pounds:

Since 52 pounds is the mean, the Z-score for 52 pounds is:

$Z = \frac{52 - 52}{6.6} = 0$

From the Z-table (or calculator), a Z-score of 0 corresponds to 50%. So, 50% of the dogs weigh less than 52 pounds.

b) Find the percentage of dogs that weigh less than 50 pounds:

Now, for X = 50 pounds, the Z-score is:

$Z = \frac{50 - 52}{6.6} = \frac{-2}{6.6} \approx -0.303$

Using a Z-table or calculator, a Z-score of -0.303 gives a cumulative probability of approximately 38.1%. So, 38.1% of the dogs weigh less than 50 pounds.

c) Find the percentage of dogs that weigh more than 50 pounds:

To find the percentage of dogs weighing more than 50 pounds, we subtract the cumulative probability for 50 pounds from 100%:

$100\% - 38.1\% = 61.9\%$

Thus, 61.9% of the dogs weigh more than 50 pounds.

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Let me know if you need further clarifications!

Here are some related questions:

1. What percentage of dogs weigh more than 52 pounds?
2. How would the results change if the standard deviation were 5 pounds?
3. What is the weight corresponding to the 90th percentile for this dog breed?
4. How can you calculate the percentage of dogs weighing between 50 and 55 pounds?
5. If the weight of a dog is 60 pounds, what is its Z-score?

Tip: Always check the Z-score table for exact values when approximating percentages from Z-scores!