Determining sample proportion probability using population proportion and a sample of size Suppose 40% of all seniors have a computer at home and a sample of 64 is taken. What is the probability that more than 30 of those in the sample have a computer at home?"
My attempt:
n=64
0.4x64=25.6
p=?
x=??
A>30=??
Don't have an idea of what equation would be appropriate to determine proportion so as to use s=root p(1−p)/n
 A: Given an event $E$ with probability $p$ of occurring, where $0 < p < 1$ 
and $q = (1-p)$, and given $n$ independent trials of the event, 
the probability
that the event $E$ occurs exactly $k$ times out of $n$ is 
$\binom{n}{k}p^k q^{(n-k)}.$
This means that the probability of at least $r$ successes in $n$ trials is
$$\sum_{k=r}^n \binom{n}{k}p^k q^{(n-k)}.\tag1$$
All that you have to do is plug in the appropriate values for the variables in
equation (1) above:
$$r = 31, n = 64, p = 0.4.$$
A: The number $X$ of seniors in a random sample of size $n = 64,$ who will have computers at home has $X \sim \mathsf{Binom}(n = 64, p = 0.4).$ You seek
$P(X > 30) = 1 - P(X\le 30).$ You may be expected to
solve this problem using software to get an exact answer
or using a normal approximation to get a useful approximate answer.
Using R statistical software, where pbinom is a binomial DCF and dbinom is a binomial PDF, you could obtain the answer $0.1062776$ in either of two ways, as shown below. [The second method is a software implementation of the sum displayed in @user's (+1) Answer.]
1 - pbinom(30, 64, 0.4)
[1] 0.1062776
sum(dbinom(31:64, 64, 0.4))
[1] 0.1062776

Normal approximation begins by finding $ \mu = E(X) = np = 64(.4) = 25.6$ and $\sigma = \sqrt{64(.4)(.6)} = 3.9192.$ Then $X \stackrel{aprx}{\sim} 
\mathsf{Norm}(\mu=25.6, \sigma = 3.9192).$
Using a continuity correction, you would find
$P(X > 30.5) = 1 - P(X < 30.5) \approx 0.1056.$ This could be evaluated in R, where pnorm is a normal CDF:
1 - pnorm(30.5, 25.6, 3.9192)
[1] 0.1056032

Often, a normal approximation of a binomial probability
is accurate to two decimal places, provided $n$ is large
enough that $np$ and $n(1-p)$ both exceed $5.$ (Roughly speaking, this ensures that there is very little normal
probability below $9$ or above $n.)$
Standardizing and using a printed standard normal table
is done as follows:
$$P(X > 30.5) =
P\left(\frac{X=\mu}{\sigma} > \frac{30.5 - 25.6}{3.9192}\right)\\
\approx P(Z > 1.25) =  0.1056,$$
where $Z$ is standard normal and the final answer
is obtained from a printed table. Sometimes this method
is a little less precise because printed table give
z-scores only to two decimal places.
The plot below shows the good fit of the normal distribution to this particular binomial distribution.
x = 10:40; PDF= dbinom(x, 64, .4)
hdr = "BINOM(64, 0.4) with Normal Approx."
plot(x, PDF, type="h", lwd=2, main=hdr)
 curve(dnorm(x, 25.5, 3.9192), add=T, col="red", lwd=2)
 abline(v=30.5, col="blue", lty="dotted", lwd=2)
 abline(h=0, col="green2")


