Question 1182545
Here's how to solve this problem:

**a. What is the distribution of X?**

X ~ N(2150, 593)

This means X follows a normal distribution with a mean (μ) of 2150 and a standard deviation (σ) of 593.

**b. Is 2150 a population mean or sample mean?**

Population mean.  The problem states that 2150 is the *average* of *all* 40 election districts in Alaska.  Since it pertains to the entire population, it's the population mean.

**c. Find the probability that a randomly selected district had fewer than 2142 votes for President Clinton.**

To find this probability, we need to calculate the z-score and then use a standard normal distribution table (or calculator) to find the corresponding probability.

*   z = (x - μ) / σ
*   z = (2142 - 2150) / 593
*   z ≈ -0.0135

Now, we look up the probability associated with z = -0.0135.  P(X < 2142) = P(Z < -0.0135) ≈ 0.4946

**d. Find the probability that a randomly selected district had between 2291 and 2561 votes for President Clinton.**

We need to calculate two z-scores and find the area under the normal curve between them.

*   z₁ = (2291 - 2150) / 593 ≈ 0.2378
*   z₂ = (2561 - 2150) / 593 ≈ 0.7099

Now, find the probabilities associated with these z-scores: P(Z < 0.7099) and P(Z < 0.2378) and subtract to find the probability between the two Z-scores:
P(2291 < X < 2561) = P(0.2378 < Z < 0.7099) = P(Z < 0.7099) - P(Z < 0.2378) ≈ 0.7611-0.5937 ≈ 0.1674

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

The third quartile (Q3) is the value below which 75% of the data falls.  We need to find the z-score corresponding to 0.75 and then convert it to the corresponding number of votes.

*   Find z such that P(Z < z) = 0.75.  z ≈ 0.674

*   Now, convert the z-score to the corresponding value of X:
    X = μ + zσ
    X = 2150 + (0.674 * 593)
    X ≈ 2549.982

Rounding to the nearest whole number, the third quartile is approximately 2550 votes.