91Ó°ÊÓ

Given in the question that, Alaska’s 40 election districts averaged 1,956.8 votes per district for President Clinton.

We have to state the approximate distribution of$X$

$X$is a random variable that represents the number of votes cast in each of Alaska's $40$election districts for President Clinton. $X$has a bell-shaped distribution, with a mean of $1956.8$and a standard deviation of $572.3$. We know that if a continuous random variable $X$has a bell-shaped curve with a mean of $\mathrm{μ}$and a standard deviation of $s$, it is normally distributed and is denoted as $X~N(\mathrm{μ},s)$. As a result, the random variable $X$ in this situation is denoted as:$X~N(1956.8,572.3)$

The standard deviation was 572.3. 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.

We have to find whether 1,956.8 is the population mean or a sample mean.

The population mean is the average determined from the observations that correspond to each and every unit of the population. The sample mean is the average determined from the observations that belong to a subset of the population. There are a total of 40 districts in Alaska in this situation. The total votes from all of these districts were added together and the result was $1956.8$dollars. As a result, it is the population mean value.

We have to 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.

We know that, $X~N(1956.8,572.3)$

The probability that a randomly selected district would vote for President Clinton with less than 1,600 votes is stated as:

$$\begin{gathered} P(x<1600)=P\left(\frac{x-\mathrm{μ}}{\mathrm{σ}}<\frac{1600-1956.8}{572.3}\right) \\ =P(z<-0.6235) \end{gathered}$$

The shaded zone in the graph below depicts the needed probability. As a result, it can be written:

$$\begin{gathered} P(z<-0.6235)=0.5-P(-0.6235≤z≤0) \\ =0.5-P(0≤z≤0.6235) \\ =0.5-0.2335 \\ =0.2665 \end{gathered}$$

We have to find the probability that a randomly selected district had between 1,800 and 2,000 votes for President Clinton.

The probability that a randomly selected district would vote for President Clinton between 1,800 and 2,000 times is stated as:

$$\begin{gathered} P(1800<x<2000)=P\left(\frac{1800-1956.8}{572.3}<\frac{x-\mathrm{μ}}{\mathrm{σ}}<\frac{2000-1956.8}{572.3}\right) \\ =P(-0.2740<z<0.0755) \end{gathered}$$

The shaded zone in the graph below depicts the needed probability. As a result, it can be written:

$$\begin{gathered} P(-0.2740<z<0.0755)=P(-0.2740≤z≤0)+P(0≤z≤0.0755) \\ =P(0≤z≤0.2740)+P(0≤z≤0.0755) \\ =0.1080+0.0301 \\ =0.1381 \end{gathered}$$

We have to find the third quartile for votes for President Clinton.

The probability that an individual's proportion of fat calories is smaller than $k$is determined by:

$P(x<k)=P\left(\frac{x-\mathrm{μ}}{\mathrm{σ}}<\frac{k-1956.8}{572.3}\right)=P\left(z<z_1\right)=0.75$

We get the following results from the normal distribution tables:

$z_1=0.6745$

In the equation $x=\mathrm{μ}+z\mathrm{σ}$replace $x$with $k$and $z$with $z_1$:

$k=1956.8+0.6745*572.3=2342.82$