In the 1992 presidential election Alaska's 40 election districts averaged 1955 votes per district for President Clinton. The standard deviation was 556. (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-NO
b. Is 1955 a population mean or a sample mean? Select an answer
C. Find the probability that a randomly selected district had fewer than 1911 votes for President Clinton.
d. Find the probability that a randomly selected district had between 1868 and 2105 votes for President Clinton
e. Find the 95th percentile for votes for President Clinton, Round your answer to the nearest whole number.

a) X = N(1955,556)

b) 1955 is the population mean

c) P(X <1911) = 0.4685

d) P(1868 < X <2105) = 0.1685

e) 95th percentile = 2870 (nearest whole number)

Step-by-step explanation:

a) Average number of votes per district, μ = 1955

The standard deviation,

The distribution of X will take the form

Therefore the distribution of X is N(1955,556)

b)1955 is the average of all the votes president Clinton had per district(i.e. the mean of all the values of the population) and not the mean of collected samples.

1955 is the population mean

c) Probability that a randomly selected district had fewer than 1911 votes for president Clinton

z=(1911-1955)/556

z=-0.079

From the probability distribution table

P(z<-0.079)=0.4685

Therefore, P(X <1911) = 0.4685

d)probability that a randomly selected district had between 1868 and 2105 votes for President Clinton

P(1868 < X <2105) = 0.6063 - 0.4378

P(1868 < X <2105) = 0.1685

e)  Find the 95th percentile for votes for President Clinton

P95=1955+1.645*556=2870