Question: A lab network consisting of $20$ computers was attacked by a computer virus, this virus enters each computer with probability $0.4$, independently of other computers. Find the probability that it entered at least $10$ computers.
Attempt:
$P(X\geq 10) = 1 - P(X\leq 9) = 1 - F(9)\,\to$ this is as far as I get, how do I calculate the cdf of this ? the solution to it is $0.2447$. We are not using R to calculate the answer, how do I do this using a calculator? I am studying for a test and any help on this is appreciated. Thanks.
Sure, seems like your logic is correct. Are you allowed to use a calculator? Seems strange, usually $$P(X\geq 10) = 1-P(X\leq 9) =1- \sum_{k = 0}^{9}\binom{20}{k}\left(\frac{4}{10}\right)^k\left(\frac{6}{10}\right)^{20-k}$$ is good enough.
For the first term of the sum, we have $$\binom{20}{0}\left(\frac{4}{10}\right)^0\left(\frac{6}{10}\right)^{20-0} = \left(\frac{6}{10}\right)^{20}=0.00003656.$$
For the second term we have $$\binom{20}{1}\left(\frac{4}{10}\right)^1\left(\frac{6}{10}\right)^{20-1}=\frac{20!}{1!19!}\cdot\frac{4}{10}\cdot\left(\frac{6}{10}\right)^{19}=0.000487488.$$
Continue for each term, add and subtract from $1$.
You might also be allowed to use a normal approximation.
We know $X\sim \text{Bin}(20, .4)$. Then using a normal approximation, we have \begin{align*} P(X\geq 10) &= 1-P(X\leq 9) \\ &\approx1-P\left[Z\leq \frac{9-20(.4)+.5}{\sqrt{20(.4)(.6)}}\right]\\ &=1-\Phi\left[\frac{9-20(.4)+.5}{\sqrt{20(.4)(.6)}}\right]\\ &=0.2467814 \end{align*} where $Z$ is a standard normal, and $\Phi$ is the usual cdf.