Question

In the 1992 presidential election, Alaska's 40 election districts averaged 2079 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 2079 a population mean or a sample mean?

c. Find the probability that a randomly selected district had fewer than 2041 votes for President Clinton.

d. Find the probability that a randomly selected district had between 2083 and 2369 votes for President Clinton.

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

Answer 1

Answer: a. The distribution of X is X ~ N(2079, 599^2). This means that the number of votes for President Clinton in each election district follows a normal distribution with a mean of 2079 and a standard deviation of 599.

b. 2079 is a sample mean. It is the average number of votes per district calculated from the 40 election districts in Alaska.

c. To find the probability that a randomly selected district had fewer than 2041 votes for President Clinton, we need to calculate the z-score and then find the corresponding probability from the standard normal distribution table.

z = (2041 - 2079) / 599 ≈ -0.0634

Using a standard normal distribution table or calculator, the probability corresponding to a z-score of -0.0634 is approximately 0.4744.

So, the probability that a randomly selected district had fewer than 2041 votes for President Clinton is approximately 0.4744.

d. To find the probability that a randomly selected district had between 2083 and 2369 votes for President Clinton, we need to calculate the z-scores for both values and then find the difference in probabilities.

For 2083:

z = (2083 - 2079) / 599 ≈ 0.0067

For 2369:

z = (2369 - 2079) / 599 ≈ 0.0484

Using the standard normal distribution table or calculator, the probability corresponding to a z-score of 0.0067 is approximately 0.5026, and the probability corresponding to a z-score of 0.0484 is approximately 0.5198.

Now, find the difference in probabilities:

Probability = 0.5198 - 0.5026 ≈ 0.0172

So, the probability that a randomly selected district had between 2083 and 2369 votes for President Clinton is approximately 0.0172.

e. To find the third quartile for votes for President Clinton, we need to find the z-score corresponding to the 75th percentile of the standard normal distribution.

Using a standard normal distribution table or calculator, the z-score corresponding to the 75th percentile is approximately 0.6745.

Now, find the value of X at the third quartile:

X = 2079 + 0.6745 * 599 ≈ 2117.53

Rounding to the nearest whole number, the third quartile for votes for President Clinton is 2118.