Probability question dealing with standard normal variables and De Moivre Laplace theorem To examine the accuracy of an algorithm that selects random numbers from the set $\{1, 2, …, 40\},$ $100,000$ numbers are selected and there are $3500$ ones. Given that the expected number of ones is $2500,$ is it fair to say that this algorithm is not accurate?
The picture here is the steps I have so far
I uploaded this question on chegg, but I am still very confused with how the last three lines are determined. How do we know that $Z$ can be between $-3.8$ and $3.8.$ Any hints would be incredibly appreciated!!

Let the number of 1's in $n=100,000$ draws with probability $p=1/40$ of 1 at each draw be $X \sim \mathsf{Binom}(n,p).$ In R, $P(X \ge 3000) = 1 - P(X \le 2999) \approx 0.$
1 - pbinom(2999, 100000, 1/40)
[1] 0

As mentioned in the question, $\mu_X=E(X) = np = 2500.$ Also, $\sigma_X = SD(X) = \sqrt{np(1-p)} = 49.3710.$
sqrt(100000*(1/40)*(39/40))
[1] 49.37104

Thus $3000$ is $\frac{3000-2500}{49.27} = 8.7152$ standard deviations above the mean. Binomial distributions with such large $n$ are approximately normal, and the Empirical Rule
states that there is almost no probability of having an
observation more than 3 standard deviations above the mean.
(3000 - 2500)/57.3710
[1] 8.715205

A normal approximation to $P(X \ge 3000) = P(X \ge 2999.5)\approx 0$
can be obtained directly in R by using a normal PDF:
1 - pnorm(2999.5, 2500, 57.3719)
[1] 0

Ordinarily, you could also come close to a normal approximation by standardizing and using printed normal tables, as follows:
$$P(X \ge 3000) = P(X > 2999.5)\\
= P\left(\frac{X-\mu_x}{\sigma_x} > 
\frac{2999.5-2500}{57.3790}\right)\\
\approx P(Z > 8.7043) \approx 0.$$
where the final result is far beyond tabled values,
and so must be taken as $0.$
(2999.5-2500)/57.3790
[1] 8.705275

Your posted handwritten work seems mainly in the
right direction, except that I can't read the
last few lines.
