Question

Suppose that an outbreak of cholera follows severe flooding in an isolated town of 2500 people. Initially (Day 0), 30 people are infected. Every day after, 18% of those still healthy fall ill. By what day will at least 87% of the population be infected?

Answer 1

On 11th day ( approx )

Explanation:

Since, if a population change with a constant rate,

Then final population,

`A=P(1-(r)/(100))^t`

Where,

P = initial population,

r = rate of change per period,

t = number of periods,

Given,

Initial population = 2500,

Out of which, infected population = 30,

So, healthy people, initially = 2500 - 30 = 2470

Every day after, 18% of those still healthy fall ill.

So, after x days the number of healthy people,

`P=2470(1-(18)/(100))^x=2470(1-0.18)^x=2470(0.82)^x----(1)`

Now, if 87% of 2470 are ill,

Then reaming healthy population = (100 - 87)% of 2470

= 13% of 2470

`=(13* 2470)/(100)`

`=(32110)/(100)`

= 321.10

If P = 321.10,

From equation (1),

`321.10 = 2470(0.82)^x`

`(321.10)/(2470)=0.82^x`

`\log((321.1)/(2470)) =x \log(0.82)`

`\implies x = (\log((321.1)/(2470)))/(\log(0.82))=10.2807315584`

Hence, by 11 day at least 87% of the population be infected.