Question

In the 1992 presidential election, Alaska’s 40 election districts averaged 1956.8 votes per district for President Clinton. The standard deviation was 572.3. (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.a.State the approximate distribution of X.
b.Is 1,956.8 a population mean or a sample mean? How do you know?
c.Find the probability that a randomly selected district had fewer than 1,600 votes for President Clinton. Sketch the graph and write the probability statement.
d.Find the probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton.
e.Find the third quartile for votes for President Clinton.

Answer 1

Answer:

(a) The distribution of X is N (1956.8, 90.49²).

(b) The value 1956.8 is the population mean.

(c) The probability that a randomly selected district had fewer than 1,600 votes for President Clinton is 0.9996.

(d) The probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton is 0.6426.

(e) The third quartile for votes for President Clinton is 2018.3.

Step-by-step explanation:

The random variable X is defined the number of votes for President Clinton for an election district.

The information provided is:

$n=40\\\bar x=1956.8\\\sigma=572.3$

(a)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

$\mu_{x}=\mu = \bar x=1956.8$

And the standard deviation of the distribution of sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{572.3}{\sqrt{40}}=90.49$

Thus, the distribution of X is N (1956.8, 90.49²).

(b)

A mean of the distribution of sample mean is given by,

$\mu_{x}=1956.8$

The mean value, 1956.8 is the population mean of the sampling distribution of sample mean.

This is because the law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ($\bar x$) approaches the whole population mean ($\mu_{x}$).

Thus, the value 1956.8 is the population mean.

(c)

Compute the value of P (X < 1600) as follows:

$P(X$

                     $=P(Z$

Thus, the probability that a randomly selected district had fewer than 1,600 votes for President Clinton is 0.9996.

(d)

Compute the value of P (1800 < X < 2000) as follows:

$P(1800$

                                 $=P(-1.73$

Thus, the probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton is 0.6426.

(e)

The third quartile is the 75th percentile of a distribution.

Let x be the third quartile of votes.

Then, P (X < x) = 0.75.

⇒ P (Z < z) = 0.75

The value of z is:

z = 0.68.

*Use a z-table for the value.

Compute the value of x as follows:

$z=\frac{x-\mu_{x}}{\sigma_{x}}\\0.68=\farc{x-1956.8}{90.49}\\x=1956.8+(0.68\times 90.49)\\x=2018.3332\\x\approx 2018.3$

Thus, the third quartile for votes for President Clinton is 2018.3.