In a certain area, regulations require that the chlorine level in wastewater discharges be less than 100 $\mu$/L. In a sample of 85 wastewater specimens, the mean chlorine concentration was 98 $\mu$g/L and the standard deviation was 20 /L. Let $\mu$ represent the mean chlorine level. A test is made of $$H_0:\mu \ge 100$$$$ H_1:\mu \lt 100$$ Find the p-value.
I'm not really sure how to do it because my teacher is hard to understand. But this is my attempt from learning by what he wrote.
$n = 100, \mu = 98, \sigma = 20$
Since n is sufficiently large, we can find the Z-score. $$z = \frac{\sqrt{n}(X- \mu_0)}{\sigma} = -.922$$ $$p\{Z<-.922\} \implies 1-p\{z < .922\} \implies 1-\phi(.922)=.1782$$ $\phi(.922)$ was obtained from using a table
Assuming a 95%level of significance, the it is not plausible because $.1782 > .05$ which means we reject the null hypothesis. There is strong evidence against it.
Even with some confusion about notation and terminology, you are on the right track. Here is a corrected version of what you have.
You want $Z = \frac{\sqrt{n}(\bar X - \mu_0)}{\sigma} = \frac{\sqrt{85}(98-100)}{20} = -0.922.$
From normal tables $P(Z < -0.922) \approx 0.1783.$ This is the P-value. Because the P-value exceeds 5% we do not to reject.
(Your 95% might be for a confidence interval, but 5% is a reasonable significance level for a test of hypothesis.)