title: How does the pnorm aspect of work with z scores & x-values?
body: <p>My professor assigned us some homework questions regarding normal distributions. We are using R studio to calculate our values instead of the z-tables.</p>
  
  <p>One question asks about something about meteors where the mean (μ) = 4.35, standard deviation (σ) = 0.59 and we are looking for the probability of x>5.
  <strong>I already figured out the answer with 1-pnorm((5-4.35)/0.59) ~ 0.135.</strong></p>
  
  <p><strong>However</strong>, I am currently having some difficulty trying to understand what pnorm calculates. </p>
  
  <p>Originally, I just assumed that z scores were the only arguments needed. So I proceeded to use pnorm(z-score) for most of the normal curvature problems. 
  The help page for pnorm accessed through ?pnorm() indicates that the usage is:
  <em>pnorm(q, mean = 0, sd = 1, lower.tail = TRUE, log.p = FALSE)</em>. </p>
  
  <p>My professor also says that I am ignoring the mean and sd by just using pnorm(z-score). I feel like it is just easier to type in one value instead of the whole set of arguments. So I experimented and found that
  <strong>1-pnorm((5-4.35)/0.59) = 1-pnorm(5,4.35,0.59)</strong></p>
  
  <p>So it looks like <strong>pnorm(z-score) = pnorm (x,μ,σ)</strong>. </p>
  
  <blockquote>
    <p>Is there a reason that using the z-score allows to skip the mean and
    standard deviation in the pnorm function?</p>
  </blockquote>
  
  <p>I have also noticed that trying to add μ,σ arguments with the z-score gives the wrong answer <strong>(ex: pnorm(z-score,μ,σ).</strong></p>
  
  <pre><code>&gt; 1-pnorm((5-4.35)/0.59)
  [1] 0.1352972
  &gt; pnorm(5,4.35,0.59)
  [1] 0.8647028
  &gt; 1-pnorm(5,4.35,0.59)
  [1] 0.1352972
  &gt; 1-pnorm((5-4.35)/0.59,4.35,0.59)
  [1] 1
  </code></pre>