You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately $\sigma =64.3$. You would like to be $99\mathrm{\%}$ confident that your estimate is within 3.75 of the true population mean. How large of a sample size is required?
Do not round mid-calculation. Give your answer in whole people.
$$\mathrm{n}=$$
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Step 1 ：We are given that the population standard deviation $\sigma$ is 64.3, the desired margin of error $E$ is 3.75, and the confidence level is 99%.

Step 2 ：The Z-score corresponding to a 99% confidence level is approximately 2.576. This value can be found using a standard normal distribution table or a statistical calculator.

Step 3 ：We can use the formula for the sample size needed to estimate a population mean with a certain level of confidence: $n=(Z\ast \sigma /E{)}^2$.

Step 4 ：Substituting the given values into the formula, we get $n=(2.576\ast 64.3/3.75{)}^2$.

Step 5 ：Calculating the above expression, we find that $n$ is approximately 1950.6.

Step 6 ：Since we can't have a fraction of a person, we'll round up to the nearest whole number. Therefore, the required sample size is $\boxed{{\displaystyle 1951}}$.