The weight of full-grown German Shepherds is normally distributed. The population standard deviation for the weights of full-grown German Shepherds is $5.3 \mathrm{lbs}$. What is the minimum sample size needed to estimate $\mu$ if you desire a Margin of Error of $1.5 \mathrm{lbs}$. with a $99 \%$ Level of Confidence?

Step 1 ：The problem is asking for the minimum sample size needed to estimate the population mean (μ) with a certain margin of error and level of confidence. This is a problem of sample size determination for estimating a population mean in statistics.

Step 2 ：The formula for the sample size n is given by: \(n = (Z*σ/E)^2\) where: Z is the z-score, which depends on the desired level of confidence. For a 99% level of confidence, the z-score is approximately 2.576 (this value can be found in z-score tables). σ is the population standard deviation, which is given as 5.3 lbs. E is the desired margin of error, which is given as 1.5 lbs.

Step 3 ：We can substitute the given values into the formula to find the minimum sample size. Z = 2.576, σ = 5.3, E = 1.5

Step 4 ：By substituting the values into the formula, we get \(n = (2.576*5.3/1.5)^2\)

Step 5 ：Solving the equation gives n = 83

Step 6 ：Final Answer: The minimum sample size needed to estimate the population mean with a margin of error of 1.5 lbs and a 99% level of confidence is \(\boxed{83}\)