# 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?