Deadly disease and probability to die NHS records show that of patients suffering from a certain disease, 75% die of it.
In a specific hospital, there are 100 patients. What is the probability that 25 to 65 will die?
Clearly this is an example of use of binomial distribution.
I don't know, however, how to deal with the range 25 to 65. If the question was for exactly 25 patients to die, it would be
$P = \binom {100}{25}0.75^{25}.0.25^{75}$
but I don't know how to take into account the range.
Thank you.
 A: There are a couple of approaches we could take.
First, we can find the exact answer using the binomial distribution. Let $X\sim \mathscr{B}(100, 0.75)$. Then, the question is what is $P[25 \le X \le 65]$? The key to a problem like this is that the possible values of $X$ are mutually exclusive. If $X = 25$, then exactly 25 patients died, so we know that 26 did not die, 27 did not die, etc. Because of this mutual exclusivity, we can separate the probability of the union of each possible value of $X$ into into the sum of the probabilities of each possible value of $X$:
$$ P[25 \le X \le 65] $$
$$ = P[X = 25 \, \cup \, X = 26 \, \cup \, ... \, \cup \, X = 65] $$
$$ = P[X = 25] + P[X = 26] + ... + P[X = 65] $$
$$ = \sum_{x=25}^{65} \binom{100}{x}0.75^{x}\cdot 0.25^{100-x} . $$
$$ \approx 0.016 . $$
We could also use the normal approximation to the binomial distribution. Essentially, this says that under certain conditions on $n$ and $p$, $\mathscr{B}(n,p)$ can be approximated by $\mathscr{N}\left(np,np(1-p)\right)$.
My go-to rule of thumb is $np > 10$ and $n(1-p) > 10$, which both hold in this situation, so I'd feel comfortable using the approximation, but different practitioners have different preferences.
In this case, the mean of our normal approximation is $ \mu = np = 100\cdot 0.75 = 75 $, and the standard deviation is $\sqrt{100*0.75*0.25} = \sqrt{18.75}$.
We can then convert 25 and 65 into Z-scores and calculate the probability using the standard normal cdf from there:
$$ P[25 \le X \le 65] $$
$$ P\left[\frac{25-75}{\sqrt{18.75}} \le Z \le \frac{65-75}{\sqrt{18.75}}\right] $$
$$ \approx 0.010 . $$
A: I feel like I've got this completely wrong, because if I'm right then it's quite elementary maths and someone would have answered it by now. I'll give my answer anyway and look forward to the reason why I'm wrong.
I will interpret the question as:
NHS records show that of patients suffering from a certain disease, $75$% die of it. In a specific hospital, there are $100$ patients with this disease. What is the probability that $25$ to $65$ (inclusive) will die from this disease?
Otherwise the question doesn't contain enough information to answer it.
Let $X$ be the random variable, "number of people who will die from the disease in the specific hospital".
Then assuming $(*)$ the four following assumptions are true:

*

*Each outcome is either a "success" or a "failure": here I define success to mean the person dies. Not because I'm sadistic, but because that's how Binomial distribution works. We need to define terms. We could instead define "success" as the person not dying from the disease if we wanted: that would lead to different calculations, but the same answer of course.

*The probability of "success", $p = 0.75$, is constant: it does not vary from person to person;

*The "trials", namely seeing whether or not someone will die from the disease, are independent from one another;

*The number of trials, $100$, is fixed/constant (i.e. it is $100$ and cannot be anything else/doesn't change);

then we may model $X$, the number of people who will die from the disease in the specific hospital, by a Binomial distribution  $X \sim B(100, 0.75). $
We then need to calculate $p(25 \leq X \leq 65)$. Any decent calculator should allow you to calculate this via the cumulative Binomial distribution function, either directly, or by using:
$$p(X \leq 65) - p(X \leq 24) = p(25 \leq X \leq 65).$$
I got: $p(25 \leq X \leq 65) = 0.0164\ (3sf).$
$(*)$ Some assumptions are more reasonable than others: only the 4th assumption is clearly a reasonable assumption. The 1st assumption is not a bad one, but doesn't take into consideration that the cause of death can be more than one disease, which is in fact often the case. The second assumption may or may not be reasonable: some people may fight the disease better than others - or not: perhaps this disease is immune system-independent - we don't know. I'm not sure how reasonable the 3rd assumption is. Anyway, the question ignores all of this for the sake of you learning how the Binomial distribution works.
