Is this answer key correct? (Probability - normal distributions) A popular news magazine wants to write an article on how much Americans know about  geography.  They devise a test that lists 100 cities in the US, all of them mentioned in the news magazine in the last year.  Each respondent must guess the state in which the city can be found.  Some examples were: (Los Angeles, Tuscon, Biloxi.)  Each correct answer earns one point, for a maximum of 100.  The random sample of 5000 people had a distribution of scores that was normally distributed with mean 62 and standard deviation 12.
a) What percentage of those sampled scored between 50 and 74 points?
ANS: Standardize first, and we can get –1 < x < 1.  Then, look at the table, the total area should be .8413– .1587 = .6826.  Or use 68. 95. 99.97 rule, 50 and 74 are 1 standard deviation from . So 68.5% of the data is between the points.

The only thing I'm getting tripped up on is the standardizing part. I thought we should find the variance by squaring the standard deviation and dividing by 100, getting a standard deviation of $\sqrt{1.44}$, and then standardizing with that standard deviation instead of 12. Is the 100 irrelevant when standardizing? If the question stated how many people were sampled, and that number was 100, then would I standardize with $\sqrt{1.44}$?

edit: Another part to this question while I'm here:
b) A score of 45 is considered a poor performance. What percentage of scores is below this score?
In this case after standardizing using the method the answer key used above, I need to find $P(Z \leq -1.42)$, and I'm getting $.0778$. Am I allowed to multiply by $100\%$ to get the percentage? Is that how it works?
 A: (a) The key provided looks right. Your instinct to divide the variance by 100 comes from the Central Limit Theorem, but it's important to consider what that Theorem addresses: it lets you handle probabilities involving a mean of the sample values. Here, you're not averaging the scores of the respondents; you're just asking what proportion of the respondents would score in a certain range.
Your italicized edit is closer; for the tactic you described to be appropriate, you would need a sample of 100 people and to be discussing the average of the sample participants' scores.
The key difference in the question is:
"What proportion of the sample scored..." 
which is equivalent to 
"What is the probability that an individual scored..." 
but rather different from
"If a random sample of 100 was drawn from this group of 5000, and their scores were averaged, what is the probability that their average would be..." 
The last of these is the one where dividing the variance by 100 would apply. 
(b) That is indeed how it works! 
