In the 1992 presidential election, Alaska's 40 election districts averaged 1898 votes per district for President Clinton. The standard deviation was 599. (There are only 40 election districts in Alaska.) The distribution of the votes per district for President Clinton was bell-shaped. Let X = number of votes for President Clinton for an election district. (Source: The World Almanac and Book of Facts) Round all answers except part e. to 4 decimal places.
a. What is the distribution of X? X ~ N(,)
b. Is 1898 a population mean or a sample mean? Select an answer Population Mean Sample Mean
c. Find the probability that a randomly selected district had fewer than 1904 votes for President Clinton.
d. Find the probability that a randomly selected district had between 2006 and 2197 votes for President Clinton.
e. Find the third quartile for votes for President Clinton. Round your answer to the nearest whole number.

Since the population of interest is the "40 districts of Alaska" and this value corresponds to the average number of votes for Clinton in all of the Alaskan districts, then it is the population mean.

c. Symbolically:

P(X<1904)

To find this probability you have to use the standard normal tables, so first is to standardize or "transform" the value of X into a value of Z using: Z=(X-μ)/δ ~N(0;1)

P(X<1904)= P(Z<(1904-1898)/599)= P(Z<0.01) = 0.50399

d. Symbolically:

P(2006≤X≤2197)= P(X≤2197)-P(X≤2006)

P(Z≤(2197-1898)/599)-P(X≤(2006-1898)/599)= P(Z≤0.50)-P(X≤0.18)= 0.69146-0.57142= 0.12004

e. The third quantile is the value of the variable that separates the bottom 75% of the distribution from the top 25%, symbolically:

P(X≤x₀)=0.75

To find this value you have to look in the body of the Z-table the probability of 0.75 and then it's corresponding Z-value (z₀), then using that Z-value you have to "reverse" the standardization and clear the value of x₀ from the formula of Z:

z₀= 0.67

z₀= (x₀-μ)/δ

δ*z₀= x₀-μ

x₀= (δ*z₀)+μ

x₀= (599*0.67)+1898

x₀= 2299.33

I hope this helps!