Question

(a) Census data for the United States between 1790 and 1950 are given in Table. Construct a logistic population model using the data from 1790,1850 , and 1910 .

(b) Construct a table comparing actual census population with the population predicted by the model in part (a). Compute the error and the percentage error for each entry pair.

Short answer

So, the error and the percentage error is $P\left(t\right)=\frac{197.274}{1+49.21e^{-0.0313t}}$

Step-by-step solution

Definition of percentage

Percentage is a relative value indicating hundredth parts of any quantity.

Use the given table

(a) Consider the following table

Construction of a logistic population model

The object is to construct a logistic population model using the data from 1790,1850 , and 1910 .

The solution of the logistic equation is given by,

$P\left(t\right)=\frac{a{P}_0}{b{P}_0+\left(a-b{P}_0\right)e^{-at}}$

Here,$P\left(0\right)=3.929\left(Year1790\right),P\left(60\right)=23.192(Year1850)$, and$P\left(120\right)=91.972\left(Year1910\right)$.

Plug$P\left(0\right)=3.929$in equation (1), we have

$\begin{array}{rcl}P\left(t\right)& =& \frac{3.929a}{3.929b+\left(a-3.929b\right)e^{-at}}\\ & =& \frac{{\displaystyle \frac{a}{b}}}{1+\left({\displaystyle \frac{1}{3.929}}{\displaystyle \frac{a}{b}}-1\right)e^{-at}}\end{array}$

Let $\begin{array}{rcl}c& =& \frac{a}{b}\end{array}$then $\begin{array}{rcl}P\left(t\right)& =& \frac{c}{1+\left({\displaystyle \frac{c}{3.929}}-1\right)e^{-at}}\end{array}$.

Substitute the value of (p) in the above function

For $P\left(60\right)=23.192$, this implies that,

$\begin{array}{rcl}23.192& =& \frac{c}{1+\left({\displaystyle \frac{c}{3.929}}-1\right)e^{-60a}}\\ c& =& 23.192+23.192\left(\frac{c}{3.929}-1\right)e^{-60a}\end{array}$......(2)

For $\begin{array}{rcl}P\left(120\right)& =& 91.972\end{array}$implies that,

$\begin{array}{rcl}c& =& 91.972+91.972\left(\frac{c}{3.972}-1\right)e^{-120a}\\ & =& 91.972+91.972\left(\frac{c}{3.972}-1\right){\left(e^{-60a}\right)}^2......\left(3\right)\end{array}$

From equation (2) and (3), we have

$\begin{array}{rcl}c& =& 91.972+91.972\left(\frac{c}{3.972}-1\right){\left[\frac{c-23.192}{23.192\left({\displaystyle \frac{c}{3.929}}-1\right)}\right]}^2\\ c-91.972& =& 91.972\left(\frac{c-3.972}{3.972}\right)\times \frac{1}{\left({\displaystyle \frac{c}{3.929}}-1\right)}\end{array}$

So,$\begin{array}{rcl}c& =& 197.274\end{array}$. Use a calculator

Thus, the non-zero solution for c is 197.274.

Substitute the value of c

Plug $c=197.274$in (2), we have

$\begin{array}{rcl}{e}^{-60a}& =& \frac{197.274-23.192}{23.192\left({\displaystyle \frac{197.274}{3.929}}-1\right)}\\ -60a& =& \ln \left(\frac{197.274-23.192}{23.192\left({\displaystyle \frac{197.274}{3.929}-1}\right)}\right)\\ a& =& -\frac{1}{60}\ln \left(\frac{197.274-23.192}{23.192\left({\displaystyle \frac{197.274}{3.929}-1}\right)}\right)\\ & =& 0.0313395\\ & \approx & 0.0313\end{array}$

Thus, the value of a is $\begin{array}{rcl}a& \approx & 0.0313\end{array}$.

Plug the values in the function

Plug $a=0.0313,c=197.274$in$P\left(t\right)=\frac{c}{1+\left({\displaystyle \frac{c}{3.929}}-1\right)e^{-at}}$.

$\begin{array}{rcl}P\left(t\right)& =& \frac{197.274}{1+\left({\displaystyle \frac{197.274}{3.929}}-1\right)e^{-0.0313t}}\\ & =& \frac{197.274}{1+\left({\displaystyle 49.21}\right)e^{-0.0313t}}\end{array}$

Thus,$\begin{array}{rcl}P\left(t\right)& =& \frac{197.274}{1+\left({\displaystyle 49.21}\right)e^{-0.0313t}}\end{array}$