In practice, a very large sample---supposedly from a normal population---can 'fail' a normality test
either (a) because the population is not even close to normal or (b) the sample is from a very nearly normal population, but has some unimportant quirk that leads to 'failure' of standard normality tests.
Many statistical computer programs (including R) will not perform Shapiro-Wilk normality tests
for sample larger than a few thousand.
However, you can (a) make an empirical CDF (ECDF) of the data and compare it with
the CDF of an appropriate actual normal CDF, (b) make a normal probability plot
(normal quantile-quantile plot) to see if it is approximately linear, or (c) test
randomly chosen, moderately large subsamples of the huge sample to see if they, 'pass' the test as normal.
Simulated precisely-normal sample. Consider the following sample x of size $n = 100\,000$ from $\mathsf{Norm}(\mu = 50, \sigma = 7).$
set.seed(2021)
x = rnorm(10^5, 50, 7)
summary(x); length(x); sd(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
18.38 45.30 50.01 50.03 54.77 83.19
[1] 100000 # sample size
[1] 7.036199 # sample SD
Matching ECDF to CDF. In the plot below, the ECDF of x is shown along with the
CDF (broken blue) of $\mathsf{Norm}(50, 7).$
hdr="ECDF (black) with NORM(50,7) CDF (blue)"
plot(ecdf(x), main=hdr)
curve(pnorm(x, 50, 7), add=T, lwd=3, lty="dashed", col="blue")
[If you do not know the mean $\mu$ and variance $\sigma,$ you could
approximate $\mu$ as mean(x) and $\sigma$ as sd(x),$ which are
excellent approximations for such a large sample, to get a very
similar plot.]
Normal Q-Q plot. A normal probability plot of x has points that fall very close to
a straight reference line (except possibly for a few points near the ends of the plot).
qqnorm(x)
qqline(x, col="red", lwd=2)
[This method works whether or not you know the population mean and standard deviation.]
Tests on subsamples of moderate size.
Kolmogorov-Smirnov test of match to particular normal distribution: In R, the procedure ks.test will test to see if a sample was taken at
random from a particular normal population---with specified mean and SD.
We show Kolmogorov-Smirnov tests for three sub-samples of size $1000.$
All three 'pass' the test; that is, the test fails to reject the null
hypothesis that data are from $\mathsf{Norm}(50, 7).$
set.seed(1234)
ks.test(sample(x,1000), pnorm, 50, 7)$p.val
[1] 0.8436103
ks.test(sample(x,1000), pnorm, 50, 7)$p.val
[1] 0.7222755
ks.test(sample(x,1000), pnorm, 50, 7)$p.val
[1] 0.1864064
Shapiro-Wilk test to unspecified normal distribution. In R, the procedure shapiro.test will test a sample of size $n_1 \le 5000.$
Here are P-values of three samples of size $n_1=1000,$ all of which fail to
reject the null hypothesis of normality at the 5% level. [The null hypothesis of a Shapiro-Wilk test is that data match some normal distribution---mean and SD not specified.]
set.seed(1030)
shapiro.test(sample(x, 1000))$p.val
[1] 0.5004599
shapiro.test(sample(x, 1000))$p.val
[1] 0.6135974
shapiro.test(sample(x, 1000))$p.val
[1] 0.9206584
Addendum. When data are not exactly normal. Suppose our data are heights of students (inches) whose heights come 50:50 from two distributions:
$\mathsf{Norm}(65, 4)$ and $\mathsf{Norm}(69, 4).$ [In reality, distributions of almost all human populations are mixtures of several distributions.]
set.seed(2021)
x1 = rnorm(50000, 65, 4)
x2 = rnorm(50000, 69, 4)
y = sample(c(x1,x2))
summary(y)
Min. 1st Qu. Median Mean 3rd Qu. Max.
46.93 63.93 67.02 67.02 70.08 85.17
length(y); sd(y)
[1] 100000 # sample size
[1] 4.493029 # sample SD
For many practical
purposes this mixture distribution is essentially $\mathsf{Norm}(67, 4.5).$
(a) The histogram of $100\,000$ students is well matched by its density estimator (dashed brown); both "look normal." (b) More precisely, the ECDF is similar to the CDF of
$\mathsf{Norm}(67, 4.5).$ (c) Also, a normal Q-Q plot is essentially linear.
As is often the case, formal tests give a mixed message. Based on a sub-sample of $5000$ students, the K-S test (known for its poor power even for moderately large samples) does not reject the null hypothesis that the population
is $\mathsf{Norm}(67, 4.5).$ However, the Shapiro-Wilk test rejects the null
hypothesis that y1 is normal at all---regardless of parameters.
set.seed(123)
y1 = sample(y, 5000)
ks.test(y1, pnorm, 67, 4.5)$p.val
[1] 0.1272236
shapiro.test(y1)$p.val
[1] 0.01925218
[For what it's worth, the K-S test does reject the entire sample y of size $100\,000$ at the 2% level.]
Note: If you somehow know that the target distribution is normal and have a huge sample, then you may get a good result by estimating $\mu$ by $\bar X$ and $\sigma$ by the sample standard deviation $S,$ and using the CDF of $\mathsf{Norm}(\bar X, S).$ That may be about as good as
using the ECDF of the sample.
Simulation showing estimation of $P(X \le 55)$ for $X \sim \mathsf{Norm}(50, 7)$ with samples of size $10\,000:$
set.seed(1776)
m = 10^5; P.55 = p.55 = numeric(m)
for(i in 1:m)
{x = rnorm(10^4,50,7)
P.55[i] = pnorm(55, mean(x),sd(x))
p.55[i] = mean(x <= 55)
}
mean(p.55) # usimg ECDF
[1] 0.7624912
2*sd(p.55)/sqrt(m) # 95% marg of sim err
[1] 2.693748e-05
mean(P.55) # estimating mean & SD
[1] 0.7624899
2*sd(p.55)/sqrt(m) # 95% marg of sim err
[1] 2.693748e-05
pnorm(55,50,7) # exact
[1] 0.7624747
