Question

16. The table below shows the population of California from 2010 to 2019.

Answer 1

The Solution:

The Regression Model that best fits the data given in the question is

`P(t)=(a)/(1+e^(-bt))`

From the Desmos plotter analysis attached above, we have that

`\begin{gathered} a=74.907\approx74.91 \\ b=0.01397\approx0.01 \end{gathered}`

So, by substituting the values of the parameters, we get the required logistic Regression Model is given below:

`P(t)=(74.91)/(1+e^(-0.01t))`

b. The model predicts that the population of California in 2025 will be:

`\begin{gathered} \text{From 2010 to 2025 is 25 years.} \\ So,\text{ t=25 years.} \\ S\text{ubstituting 25 for t in the regression model, we get} \end{gathered}`

`P(t)=(74.91)/(1+e^(-0.01(25)))=(74.91)/(1+0.77088)=(74.91)/(1.77088)=42.1127\approx42.1\text{ million people.}`

So, the population of California in 2025 will be 42.1 million people.

c. To find when the model predicts that the population of California will be 40 million,

we shall substitute 40 (in millions) for P in the model, and find t as below:

`40=(74.91)/(1+e^(-0.01t))`

Cross multiplying, we get

`\begin{gathered} 40(1+e^(-0.01t))=74.91 \\ \text{Dividing both sides by 40},\text{ we have} \\ 1+e^(-0.01t)=(74.91)/(40) \\ \\ 1+e^(-0.01t)=1.87275 \\ e^(-0.01t)=1.87275-1 \\ e^(-0.01t)=0.87275 \end{gathered}`

Taking the ln of both sides, we get

`\begin{gathered} \ln e^(-0.01t)=\ln 0.87275 \\ -0.01t=\ln 0.87275 \\ -0.01t=-0.136106 \\ \text{ Dividing both sides by -0.01, we get} \\ t=(-0.136106)/(-0.01)=13.61\approx14\text{ years} \end{gathered}`

So, 14 years from 2010 will be in the year 2024

d. According to the model, the carrying capacity for California's capacity is 74.9 million people.