What is probability that more than 30% of owners pay to walk their dog? This equation comes from Edgenuity's course of Statistics, and I am taking the course as a high school senior. I understand how to find the z-score and the standard deviation of the sampling distribution of $\hat{\rho}_1 - \hat{\rho}_2$.

In an urban area, $31\%$ of dog owners pay for someone to walk their dogs. Morgan asks a random sample of $50$ dog owners in this urban area if they pay for someone to walk their dog. What is the probability that more than $30\%$ of the owners in the sample pay for someone to walk their dog?

I found the standard deviation of the sampling distribution of $\hat{\rho}_1 - \hat{\rho}_2$. For the variables $n_1$ and $n_2$, I substituted $50$, which calculated to approx. $0.0921$. This already gives me a red flag, because my $z$-score would be very large (larger than what I am used to in these problems).
My guess with the observed value and mean value for the z-score equation would be to multiply $0.31$ with $50$—finding the mean—and multiply $0.30$ by $50$ to find the observed value.
I likely was incorrect in my standard deviation substitution (and perhaps also in calculating the observed value and the mean in the $z$-score formula). How do I correct these errors and find the correct answer?
 A: You have $X \sim \mathsf{Binom}(n=50, p = 0.31).$ Because $\hat p = X/n,$ you seek
$P(\hat p > 0.3) = P(X > 15) = 1 - P(X \le 15) = 0.4921939.$
This exact value can be found using R statistical software, where pbinom is a binomial CDF, as follows.
1 - pbinom(15, 50, 0.31)
[1] 0.4921939

An approximate value can be found using a normal distribution with the same
mean and standard deviation as $X.$ The mean is $\mu_X = E(X) = np = 50(.31) = 15.5,$
$\sigma_X^2 = Var(X) = np(1-p) = 10.695,$ and $\sigma_X= SD(X) = 3.2703.$
Then
$$P(X > 15) = P(X > 15.5) = 
P\left(\frac{X-\mu_X}{\sigma_X} > \frac{15.5 - 15.5}{3.2703} = 0\right) \approx
P(Z > 0),$$
where $Z$ has a standard normal distribution. I will leave the rest to you.
One can usually expect about two decimal places of accuracy from a normal
approximation when $p$ is not far from $1/2$ and both $np > 5, n(1-p) > 5).$
Here is a bar plot of the distribution $\mathsf{Binom}(n = 50, p=.31)$ along
with the density function of $\mathsf{Norm}(\mu = 15.5, \sigma = 3.2703).$
The vertical dotted line is at $x = 15.5.$

R code for figure:
x = 0:50;  PDF = dbinom(x, 50, .31)
hdr = "PDF of BINOM(50, 0.31) with density of NORM(15.5, 3.27)"
plot(x, PDF, type = "h", lwd=2, main=hdr)
 curve(dnorm(x, 15.5, 3.27), add=T, col="blue", lwd=2)
 abline(v = 15.5, col="red", lwd=2, lty="dotted")
 abline(h=0, col="green2")
 abline(v=0, col="green2")

