Question

3.- In a certain desert region the average number of persons who become seriously ill each year from eating a certain poisonous plant each year is 3.2, determine what is the probability that at least 5 people will become seriously ill in two years (hint: this is a rare event and can be modeled using a Poisson distribution)

Answer 1

The probability that at least 5 people will become seriously ill in two years is 0.7649.

Explanation:

The question mentions that this problem can be modeled using the Poisson distribution so, we will use the formula:

P(X=x) = [(e^-λt)*(λt^x)]/x!

where λ = average number of occurrences per year

t = no. of years

x = number of people

We need to determine P(X≥5) so first we will calculate the probabilities at X=0,1,2,3,4 and subtract them from the total probability i.e. 1 to find P(X≥5).

We have λ = 3.2, t=2 years hence λt = (3.2)(2) = 6.4. So,

P(X=0) = [(e^(-6.4)*(6.4^0)]/0! = 0.00166

P(X=1) = [(e^(-6.4)*(6.4^1)]/1! = 0.01063

P(X=2) = [(e^(-6.4)*(6.4^2)]/2! = 0.03403

P(X=3) = [(e^(-6.4)*(6.4^3)]/3! = 0.07259

P(X=4) = [(e^(-6.4)*(6.4^4)]/4! = 0.011615

P(X≥5) = 1 - P(X<5)

= 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)]

= 1 - (0.00166 + 0.01063 + 0.03403 + 0.07259 + 0.011615)

= 1 - 0.23506

P(X≥5) = 0.7649

The probability that at least 5 people will become seriously ill in two years is 0.7649.

Answer 2

`P(X\geq 5)=1-P(X<5)=1-P(X\leq 4)=1-[P(X=0)+P(X=1)+P(X=2) +P(X=3) +P(X=4)]`

Using the pmf we can find the individual probabilities like this:

`P(X=0)=(e^(-6.4) 6.4^0)/(0!)=0.001662`

`P(X=1)=(e^(-6.4) 6.4^1)/(1!)=0.010634`

`P(X=2)=(e^(-6.4) 6.4^2)/(2!)=0.034029`

`P(X=3)=(e^(-6.4) 6.4^3)/(3!)=0.072595`

`P(X=4)=(e^(-6.4) 6.4^4)/(4!)=0.116151`

And replacing we got:

`P(X \geq 5) =0.76493`

Explanation:

Previous concepts

Let X the random variable that represent the number of people that will become sereiosly ill in two years. We know that
`X \sim Poisson(\lambda)`

The probability mass function for the random variable is given by:

`f(x)=(e^(-\lambda) \lambda^x)/(x!) , x=0,1,2,3,4,...`

And f(x)=0 for other case.

For this case the value for
`\lambda` would be:

`\lambda = 3.2 (ills)/(year) *2 years = 6.4`

For this distribution the expected value is the same parameter
`\lambda`

`E(X)=\mu =\lambda`

On this case we are interested on the probability of having at least two chocolate chips, and using the complement rule we have this:

`P(X\geq 5)=1-P(X<5)=1-P(X\leq 4)=1-[P(X=0)+P(X=1)+P(X=2) +P(X=3) +P(X=4)]`

Using the pmf we can find the individual probabilities like this:

`P(X=0)=(e^(-6.4) 6.4^0)/(0!)=0.001662`

`P(X=1)=(e^(-6.4) 6.4^1)/(1!)=0.010634`

`P(X=2)=(e^(-6.4) 6.4^2)/(2!)=0.034029`

`P(X=3)=(e^(-6.4) 6.4^3)/(3!)=0.072595`

`P(X=4)=(e^(-6.4) 6.4^4)/(4!)=0.116151`

And replacing we got:

`P(X \geq 5) =0.76493`