The population p of a small community on the outskirts of a city grows rapidly over a 20-year period: t05101520p1002004509502000
As an engineer working for a utility company, you must forecast the population 5 years into the future in order to anticipate the demand for power. Employ an exponential model and linear regression to make this prediction. (Round the final answer to the nearest whole number.)
1 answer:
Answer:
The population of the small community, 5 years into the future, after the initial 20-year period = 4268.
Step-by-step explanation:
t | 0 | 5 | 10 | 15 | 20
p | 100 | 200 | 450 | 950 | 2000
The exponential function will look like
p = aeᵏᵗ
where a and k are constants.
Take the natural logarithms of both sides
In p = In aeᵏᵗ
In p = In a + In eᵏᵗ
In p = In a + kt
In p = kt + In a.
We then use linear regression to fit the data of In p against t to obtain k and In a.
t | 0 | 5 | 10 | 15 | 20
p | 100 | 200 | 450 | 950 | 2000
In p | 4.605 | 5.298 | 6.109 | 6.856 | 7.601
In p = kt + In a.
y = mx + b
m = k and b = In a
Performing a linear regression analysis on the now-linear relationship between In p and t and also plotting a graph of the variables.
The regression equation obtained is
y = 0.151x + 4.584
The first attached image shows the equations necessary for the estimation of the linear regression parameters.
The second attached image shows the use of regression calculator and the plot of the function In p versus t.
Comparing
y = 0.151x + 4.584
With
In p = kt + In a.
y = In p
k = 0.151
x = t
In a = 4.584
a = 97.905
The exponential function relating p and t,
p = aeᵏᵗ now becomes
p = 97.905 e⁰•¹⁵¹ᵗ
So, to predict the population 5 years into the future, that is 5 years after the 20 year period.
we need p at t=25 years.
0.151 × 25 = 3.775
p(t=25) = 97.905 e³•⁷⁷⁵ = 4268.41 = 4268.
Hope this Helps!!!
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