Question 1168930
To determine the required sample size, we can use the formula for the margin of error when the population standard deviation is known:
$E = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$
where:
$E$ is the desired margin of error (within 2)
$z_{\alpha/2}$ is the critical value from the standard normal distribution corresponding to the desired confidence level
$\sigma$ is the population standard deviation (69.7)
$n$ is the required sample size

We are given a 99.5% confidence level. The significance level $\alpha$ is:
$\alpha = 1 - 0.995 = 0.005$
The critical value $z_{\alpha/2}$ corresponds to the $1 - \alpha/2 = 1 - 0.005/2 = 1 - 0.0025 = 0.9975$ percentile of the standard normal distribution.

We need to find the z-score such that the area to the left is 0.9975. Using a standard normal distribution table or a calculator, we find the z-score:
$z_{0.9975} \approx 2.807$

Now, we can plug the values into the margin of error formula and solve for $n$:
$E = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$
$2 = 2.807 \times \frac{69.7}{\sqrt{n}}$

Now, solve for $\sqrt{n}$:
$\sqrt{n} = \frac{2.807 \times 69.7}{2}$
$\sqrt{n} = \frac{195.6479}{2}$
$\sqrt{n} = 97.82395$

Square both sides to find $n$:
$n = (97.82395)^2$
$n \approx 9569.52$

Since the sample size must be a whole number, we round up to the nearest integer to ensure the desired margin of error is met.
$n = 9570$

Final Answer: The final answer is $\boxed{9570}$