I have a dataset of 133 points for which I have an expected value of an event $X$ occurring $68.12$ times.
I just ran the dataset and $X$ ended up occurring $84$ times. I am pretty sure my expected value is wrong and something statistically significant is going on here but I want to run a statistical test of some sort to prove it.
Is there any test I can run for statistical significance in this case without standard deviation?
With some reading between the lines, it seems to me your null hypothesis is that binomial success probability is $p = 68.12/133 = 0.5122.$ Now you have observed $X = 84$ and wonder if it came from $\mathsf{Binom}(n=133,p=0.5122).$ Your estimated success probability is $\hat p = 84/133 = 0.6316$ (which does seem a lot larger than $0.5122).$
Then it seems you want to test $H_0: p = .5122$ against $H_1: p \ne .5122.$ If that is the case, the hypothetical standard error $SD(\hat p)$ is $\sigma_{\hat p} = \sqrt{p(1-p)/n} = .0433.$
Using a normal approximation to binomial for your test, you have $$Z = \frac{\hat p - .5122}{.0433} = \frac{.6116 - .5122}{.0433}= 2.91.$$ Because $|Z| > 1.96,$ you reject $H_0$ at the 5% level of significance. So it seems your hunch that you have the wrong success probability was correct.
Another approach is to note that if $X \sim \mathsf{Binom}(133, .5122),$ then $P(X \ge 84) = 0.00365.$ So the P-value for a one-sided test is 0.00356. Double that for a two-sided test. Either way, the P-value is much less than 5%, so you conclude the true success probability isn't .5122.