Question

Suppose a rumor is going around a group of 191 people. Initially, only 38 members of the group have heard the rumor, but 3 days later 68 people have heard it. Using a logistic growth model, how many people are expected to have heard the rumor after 6 days total have passed since it was initially spread? (Round your answer to the nearest whole person.)

Answer 1

Answer:

106 people.

Step-by-step explanation:

Logistic equation:

The logistic equation is given by:

$P(t) = \frac{K}{1+Ae^{-kt}}$

In which

$A = \frac{K - P_0}{P_0}$

K is the carrying capacity, k is the growth/decay rate, t is the time and P_0 is the initial value.

Suppose a rumor is going around a group of 191 people. Initially, only 38 members of the group have heard the rumor.

This means that $K = 191, P_0 = 38$, so:

$A = \frac{191 - 38}{38} = 4.03$

Then

$P(t) = \frac{191}{1+4.03e^{-kt}}$

3 days later 68 people have heard it.

This means that $P(3) = 68$. We use this to find k.

$P(t) = \frac{191}{1+4.03e^{-kt}}$

$68 = \frac{191}{1+4.03e^{-3k}}$

$68 + 274.04e^{-3k} = 191$

$e^{-3k} = \frac{191-68}{274.04}$

$e^{-3k} = 0.4484$

$\ln{e^{-3k}} = \ln{0.4484}$

$-3k = \ln{0.4484}$

$k = -\frac{\ln{0.4484}}{3}$

$k = 0.2674$

Then

$P(t) = \frac{191}{1+4.03e^{-0.2674t}}$

How many people are expected to have heard the rumor after 6 days total have passed since it was initially spread?

This is P(6). So

$P(6) = \frac{191}{1+4.03e^{-0.2674*6}} = 105.52$

Rounding to the nearest whole number, 106 people.