Is there a way to test the "accuracy of a binomial survey"? I've been away from mathematics for a while and forgotten almost everything.
It doesn't come from a text book; I was given an assignment in my training for a job and wondered if I can use my mathematical knowledge. All I have to do is actually interpret the data and say "more than 50 percent of the people surveyed thinks "yes" to the question" but I'm taking a step further and trying to say how "accurate" is the result? Basically, here's what I am given

There's a supermarket that is experiencing a fall in revenues. A survey was conducted and it asked whether "the customer thinks the workers are unfriendly/unhelpful." out of a $100$ randomly chosen customers on the same day ($100$ different customers) $51\%$ answered "yes."
However, the total number of customers that visited the supermarket is expected to be around $485$. The total that visited the supermarket that month is $19700$. How confident are we to say that more than $50\%$ are not happy with the workers among all of those who visited the store a. that day b. that month?

I vaguely recalled Chi-squared ad $z$-test but I wasn't so sure; I tried the $z$-test with
$$z=\frac{p-\pi}{\sqrt{\frac{\pi(1-\pi)}{n}}}$$
where $p=0.51,\pi=0.5,n=100$. Thing is, I get $z=0.2$ and the $z$ table seems to tell me this is a very inaccurate result. In any case, my data and the question I ultimately want to answer is as above. Along the process of doing so, if no one would want to actually show me how to do this, can you please answer

What's the most apt test to answer a question like this? And why?

I think, as people start writing some answers, my senses will come back, some words and terms ringing a bell, reminding me of certain formulas, rules etc.
 A: The basic idea here is that you are making a statistical inference based on a sample from a finite population.  In the first case, your population comprises those customers who visited the store on the same day that the sample was taken.  In the second case, the population comprises all customers who visited in the same month as when the sample was taken.  
Therefore, the inference you make in each case answers a different question; in the first case, the inference about the proportion of customers who were unhappy with their service applies to the particular day on which the sampling took place.  In the second case, you could make a case--albeit not without serious reservation--that the inference might apply to the proportion of customers who were unhappy with their service in the entire month.
The second inference is more problematic because it is reasonable to assume that there exists day-to-day variation in the quality of service, and therefore, sampling customer satisfaction on a single day out of the month may lead to unreliable inferences.  The estimate of the true proportion of unsatisfied customers may remain unbiased, but by failing to consider this day-to-day variation in the sampling scheme, you could seriously underestimate the variance in the estimate.
That's the broad, conceptual overview.
The statistical details of how to conduct a hypothesis test for a proportion from a finite population is given here:
Estimating a Proportion for a Small, Finite Population (PSU.EDU)
Using this, your test statistic is $$Z = \frac{\hat p - p_0}{\sqrt{\frac{\hat p (1-\hat p)}{n} \cdot \frac{N-n}{N-1}}} \sim \operatorname{Normal}(0,1)$$
where $\hat p = x/n$ is the sample proportion of customers who are unsatisfied with their service; $n = 100$ is the number of customers sampled, $x$ is the number of customers out of $n$ who are unsatisfied.  $p_0$ is the null proportion, in your case $p_0 = 0.5$ for a one-sided hypothesis test $$H_0 : p = p_0 \quad \text{vs.} \quad H_a : p > p_0.$$  Finally $N$ is the total population size; in part $(a)$, this is $N = 485$.  Then, again for a one-sided test, you will reject $H_0$ at significance level $\alpha$ if $Z > z_{\alpha}^*$, where $z_{\alpha}^*$ is the $100(1-\alpha)\%$ percentile of the standard normal distribution; so for example, $z_{0.05}^* \approx 1.645$.
For the first part, your test statistic would be about $0.224$, which is greater than the value you computed, but still not large enough to provide sufficient evidence that the true proportion of unsatisfied customers that day was in fact more than $50\%$ ($p{\text{-value}} \approx 0.411266$).
I will leave it to you to do the hypothesis test for the second case, but it should be intuitively obvious what the conclusion would be.
Again, from a conceptual standpoint, we can see how the finite population correction factor $(N-n)/(N-1)$ adjusts for the standard error of the sample proportion in an intuitively meaningful way:  if the sample size $n$ that was collected from the population $N$ is relatively large, then we become increasingly confident that the variation represented in the sample is a reflection of the variation in the overall population, and the standard error of the sample proportion will decrease.  This in turn increases the magnitude of the test statistic, making it more likely to reject the null hypothesis, as we would expect.
