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Question 1182545: In the 1992 presidential election, Alaska's 40 election districts averaged 2150 votes per district for President Clinton.
The standard deviation was 593.
(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.
Round all answers except part e. to 4 decimals.
a. What is the distribution of X?
X - N = 2150,593
b. Is 2150 a population mean or sample mean? Mean
c. Find the probability that a randomly selected district had fewer than 2142 votes for President Clinton.
d. Find the probability that a randomly selected district had between 2291 and 2561 votes for President Clinton.
e. Find the third quartile for votes President Clinton.
Round to the nearest whole number
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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.
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