Question

n the 1992 presidential election, Alaska's 40 election districts averaged 2118 votes per district for President Clinton. The standard deviation was 578. (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 2118 a population mean or a sample mean? Select an answer Sample Mean Population Mean

c. Find the probability that a randomly selected district had fewer than 1960 votes for President Clinton.

d. Find the probability that a randomly selected district had between 2126 and 2364 votes for President Clinton.

e. Find the first quartile for votes for President Clinton. Round your answer to the nearest whole number.

Answer 1

Solution:

a) Since, the distribution of the votes per district for President Clinton was bell-shaped, therefore the distribution of X is normal.

X ~ N(2118, 578²)

b) 2118 is a population mean.

c) We have to find P(X < 1960).

We know that if X ~ N(μ, σ^{​​​​​​2}) then

We have μ = 2118 and σ^{​​​​​} = 578.

Using "NORM.S.DIST" function of excel we get,

P(Z < -0.2734) = 0.3923

The probability that a randomly selected district had fewer than 1960 votes for President Clinton is 0.3923.

d) We have to find P(2126 < X < 2364).

We have μ = 2118 and σ^{​​​​​} = 578.

Using "NORM.S.DIST" function of excel we get,

P(Z < 0.4256) = 0.6648 and P(Z < 0.0138) = 0.5055

The probability that a randomly selected district had between 2126 and 2364 votes for President Clinton is 0.1593.