Math  /  Data & Statistics

QuestionShown to the right is a certain population, in billions, for seven selected years from 1950 through 2006. Using a graphing utility's logistic regression option, we obtain the logistic growth model shown below for population, f(x)\mathrm{f}(\mathrm{x}), in billions, x years after 1949. How well does the function model the data for 2006? f(x)=11.821+3.81e0.027xf(x)=\frac{11.82}{1+3.81 e^{-0.027 x}} \begin{tabular}{|c|c|} \hline X, Number of Years after 1949 & y, Population (billions) \\ \hline 1(1950)1(1950) & 2.6 \\ \hline 11(1960)11(1960) & 3.0 \\ \hline 21(1970)21(1970) & 3.7 \\ \hline 31(1980)31(1980) & 4.5 \\ \hline 41(1990)41(1990) & 5.3 \\ \hline 51(2000)51(2000) & 6.1 \\ \hline 57(2006)57(2006) & 6.5 \\ \hline \end{tabular}
For 2006 , the function \square the population to one decimal place. accurately predictś slightly underestimates slightly overestimates

Studdy Solution

STEP 1

What is this asking? Given a logistic growth model and some real-world population data, how well does the model predict the population for the year 2006? Watch out! Make sure to correctly plug in the right value for xx based on the number of years after 1949.

STEP 2

1. Calculate the model's prediction.
2. Compare the prediction with the actual value.

STEP 3

Since the model uses xx as the number of years after 1949, and we're looking at the year 2006, we need to find the difference: 20061949=572006 - 1949 = \textbf{57}.
So, our xx value is **57**.

STEP 4

Now, let's substitute x=57x = \textbf{57} into our logistic growth model: f(x)=11.821+3.81e0.027xf(x) = \frac{11.82}{1 + 3.81 e^{-0.027 x}} f(57)=11.821+3.81e0.02757f(\textbf{57}) = \frac{11.82}{1 + 3.81 e^{-0.027 \cdot \textbf{57}}}

STEP 5

First, we'll calculate the exponent: 0.02757=-1.539-0.027 \cdot \textbf{57} = \textbf{-1.539}.
Then, we calculate the exponential: e-1.5390.2145e^{\textbf{-1.539}} \approx \textbf{0.2145}.

STEP 6

Now, we plug this back into the denominator: 1+3.810.21451+0.8182=1.81821 + 3.81 \cdot \textbf{0.2145} \approx 1 + \textbf{0.8182} = \textbf{1.8182}.

STEP 7

Finally, we divide the numerator by the denominator: 11.821.81826.50\frac{11.82}{\textbf{1.8182}} \approx \textbf{6.50}.
So, our model predicts a population of approximately **6.50** billion for 2006.

STEP 8

The table shows that the actual population in 2006 was **6.5** billion.

STEP 9

Our model predicted **6.50** billion, and the actual value is **6.5** billion.
These values are essentially the same when rounded to one decimal place!

STEP 10

The function accurately predicts the population to one decimal place.

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