The standard deviation is as applicable here as anywhere else: it gives useful information about the dispersion of the data. In particular, the sd divided by the square root of the sample size is one standard error: it estimates the dispersion of the sampling distribution of the mean. Let's calculate:
$$3.2\% / \sqrt{10000} = 0.032\% = 0.00032.$$
That's tiny--far smaller than the $\pm 0.50\%$ precision you seek.
Although the data are not Normally distributed, the sample mean is extremely close to Normally distributed because the sample size is so large. Here, for instance, is a histogram of a sample with the same characteristics as yours and, at its right, the histogram of the means of a thousand additional samples from the same population.
It looks very close to Normal, doesn't it?
Thus, although it appears you are bootstrapping correctly, bootstrapping is not needed: a symmetric $100 - \alpha\%$ confidence interval for the mean is obtained, as usual, by multiplying the standard error by an appropriate percentile of the standard Normal distribution (to wit, $Z_{1-\alpha/200}$) and moving that distance to either side of the mean. In your case, $Z_{1-\alpha/200} = 2.5758$, so the $99\%$ confidence interval is
$$\left(0.977 - 2.5758(0.032) / \sqrt{10000},\ 0.977 + 2.5758(0.032) / \sqrt{10000}\right) \\ = \left(97.62\%, 97.78\%\right).$$
A sufficient sample size can be found by inverting this relationship to solve for the sample size. Here it tells us that you need a sample size around
$$(3.2\% / (0.5\% / Z_{1-\alpha/200}))^2 \approx 272.$$
This is small enough that we might want to re-check the conclusion that the sampling distribution of the mean is Normal. I drew a sample of $272$ from my population and bootstrapped its mean (for $9999$ iterations):
Sure enough, it looks Normal. In fact, the bootstrapped confidence interval of $(97.16\%, 98.21\%)$ is almost identical to the Normal-theory CI of $(97.19\%, 98.24\%)$.
As these examples show, the absolute sample size determines the accuracy of estimates rather than the proportion of the population size. (An extreme but intuitive example is that a single drop of seawater can provide an accurate estimate of the concentration of salt in the ocean, even though that drop is such a tiny fraction of all the seawater.) For your stated purposes, obtaining a sample of $10000$ (which requires more than $36$ times as much work as a sample of $272$) is overkill.
R code to perform these analyses and plot these graphics follows. It samples from a population having a Beta distribution with a mean of $0.977$ and SD of $0.032$.
set.seed(17)
#
# Study a sample of 10,000.
#
Sample <- rbeta(10^4, 20.4626, 0.4817)
hist(Sample)
hist(replicate(10^3, mean(rbeta(10^4, 20.4626, 0.4817))),xlab="%",main="1000 Sample Means")
#
# Analyze a sample designed to achieve a CI of width 1%.
#
(n.sample <- ceiling((0.032 / (0.005 / qnorm(1-0.005)))^2))
Sample <- rbeta(n.sample, 20.4626, 0.4817)
cat(round(mean(Sample), 3), round(sd(Sample), 3)) # Sample statistics
se.mean <- sd(Sample) / sqrt(length(Sample)) # Standard error of the mean
cat("CL: ", round(mean(Sample) + qnorm(0.005)*c(1,-1)*se.mean, 5)) # Normal CI
#
# Compare the bootstrapped CI of this sample.
#
Bootstrapped.means <- replicate(9999, mean(sample(Sample, length(Sample), replace=TRUE)))
hist(Bootstrapped.means)
cat("Bootstrap CL:", round(quantile(Bootstrapped.means, c(0.005, 1-0.005)), 5))
Best Answer
Rather than representing problem in the bootstrap, this feature is sometimes used to estimate the bias in your original estimator, see for example chapter 10 of Bradley Efron and Robert Tibshirani (1993) "An Introduction to the Bootstrap". Chapman & Hall/CRC.