Question

5. The table below shows the population of California from 2010 to 2019. Year Population (millions) 2010 37.3 2011 37.6 2012 38.0 2013 38.3 2014 38.6 2015 38.9 2016 39.2 2017 39.4 2018 39.5 2019 39.5 (a) Use a graphing calculator to build a logistic regression model that best fits this data, letting t = 0 in 2010. Round each coefficient to two decimal places. P t = (b) What does this model predict that the population of California will be in 2025? Round your answer to one decimal place. million people (c) When does this model predict that California's population will reach 40 million? Give your answer as a calendar year (ex: 2010). During the year (d) According to this model, what is the carrying capacity for California's population? million people

Answer 1

Given the table:

Year Population(millions)

2010 37.3

2011 37.6

2012 38.0

2013 38.3

2014 38.6

2015 38.9

2016 39.2

2017 39.4

2018 39.5

2019 39.5

Let's answer the following.

(a) Let's build a regression model that bets fits this data.

Let t = 0 in 2010.

To build a regression model, apply the formula:

y = mx + b

where m is the slope and b is the y-intercept of the regresion line.

To find the slope, m, apply the formula:

$m=\frac{n(\sum ^{}_{}xy)-\sum ^{}_{}x\sum ^{}_{}y}{n(\sum ^{}_{}x^2)-(\sum ^{}_{}x)^2}$

Since t = 0 in 2010, we have:

t = 1 in 2011

t = 2 in 2012

t = 3 in 2013

t = 4 in 2014

t = 5 in 2015

t = 6 in 2016

t = 7 in 2017'

t = 8 in 2018

t = 9 in 2019

$\sum ^{}_{}x=0+1+2+3+4+5+6+7+8+9=45$$\sum ^{}_{}y=37.3+37.6+38.0+38.3+38.6+38.9+39.2+39.4+39.5+39.5=386.3$$\sum ^{}_{}xy=1759.9$$\begin{gathered} \sum ^{}_{}x^2=0^2+1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2=285 \\ \\ \sum ^{}_{}y^2=37.3^2+37.6^2+38.0^2+38.3^2+38.6^2+38.9^2+39.2^2+39.4^2+39.5^2+39.5^2=14928.61 \end{gathered}$

Thus, to find the slope, we have:

Where n is the number of data = 10

$\begin{gathered} m=\frac{n(\sum^{}_{}xy)-\sum^{}_{}x\sum^{}_{}y}{n(\sum^{}_{}x^2)-(\sum^{}_{}x)^2} \\ \\ m=\frac{10(1759.9)-45\ast386.3}{10(285)-45^2}=0.26 \end{gathered}$

To find the y-intercept, we have:

$\begin{gathered} b=\frac{(\sum ^{}_{}y)(\sum ^{}_{}x^2)-\sum ^{}_{}x\sum ^{}_{}xy}{n(\sum ^{}_{}x^2)-(\sum ^{}_{}x)^2} \\ \\ b=\frac{(386.3)(485)-45\ast1759.9}{10(285)-(45)^2}=37.45 \end{gathered}$

Therefore, the regression model that best fits this data is:

$P_t=0.26t+37.45$

(b) To find the population of the model in 2025, substitute 15 for t and evalaute:

t = 15 in 2025

$\begin{gathered} P_{25}=0.26(15)+37.45 \\ \\ P_{25}=41.35 \end{gathered}$

Therefore, the population in 2025 will be 41.35 million people.

(c) When the model predicts the population will reach 40 million.

Substitute 40 for Pt and find t:

$\begin{gathered} 40=0.26t+37.45 \\ \\ 40-37.45=0.26t \\ \\ 2.55=0.26t \\ \\ t=9.81\approx10 \end{gathered}$

When t = 10, the year is 2020.

Therefore, the model predicts the population will reach 40 million by 2020.