Increased research and discussion have focused on the

Step 1 of 2:

Sporadic outbreaks of E.coli have occurred in Colorado at a rate of approximately 2.4 per 100,000 for a period of two years.

Here \(\lambda=2.4\) 

Our goal is:

a). We need to find the probability that at least 5 cases of E.coli will be observed.

b). If 100,000 cases from Colorado are reviewed for this year and the number of E.coli

     cases exceeded 5, would you suspect that the state's mean E.coli has changed?    

     Explain.

a). Here if this has not changed and if 100,000 cases from Colorado are reviewed for this year.

Now we have to find the probability that at least 5 cases of E.coli will be observed.

We know the average rate \lambda\).

\(\lambda=2.4\) 

\(\mathrm{P}(\mathrm{X}=\mathrm{k})=\frac{e^{-\lambda} \lambda^{k}}{k !}\)

Where k is the actual number of successes that result from the experiment.

e is approximately equal to 2.71828.

Now we have to find the probability that at least 5 cases is

\(P(X \geq 5)=1-P(X<5)\)

We know that \lambda\) is 2.4.

\(P(X \geq 5)=1-P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)\)

Where, \(P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)\) is obtained from Excel by using the function 

“=poisson(x,\(\lambda\),false)”

Then the table is given below.

\(\begin{array}{|l|l|} \hline x & p(x<5) \\ \hline 0 & 0.090718 \\ \hline 1 & 0.217723 \\ \hline 2 & 0.261268 \\ \hline 3 & 0.209014 \\ \hline 4 & 0.125408 \\ \hline \text { Total } & 0.904131 \\ \hline \end{array}\)

Now,

\(\begin{aligned} P(X \geq 5) & =1-P(X<5) \\ & =1-0.9041 \\ & =0.0959 \end{aligned}\)

Therefore, the probability that at least 5 cases of E.coli will be observed is 0.0959.