# Thread: hospital diarrhea

1. ## hospital diarrhea

Any help offered in solving the below question will be much appreciated.
Thanks
Kingman

In a local hospital , on average 3.5 cases of diarrhea per week and each case is independent of the other similar cases.
Taking a week as consisting as 7 working days.
a) Show that the probability that the doctor sees at most 2 cases of diarrhea in a day is 0.986.
b) Using a suitable approximation, find the probability that, in a random sample of 100 days, there will be more than 96 days in which the doctor sees at most 2 cases of diarrhea in a day. (0.946)

2. Originally Posted by kingman
Any help offered in solving the below question will be much appreciated.
Thanks
Kingman

In a local hospital , on average 3.5 cases of diarrhea per week and each case is independent of the other similar cases.
Taking a week as consisting as 7 working days.
a) Show that the probability that the doctor sees at most 2 cases of diarrhea in a day is 0.986.
This is Poisson problem with parameter $\displaystyle \lambda=3.5$. However, we are told there are, on average, 3.5 cases per week (7 days); therefore, we want to consider the value $\displaystyle \lambda'=t\lambda=\frac{3.5}{7}$ as our parameter. Thus, the probability you seek is

$\displaystyle P(Y\leq\\2)=\sum_{y=0}^{2}\frac{(\lambda')^{y}e^{-\lambda'}}{y!}$.

3. Best. Title. Ever.

4. ## thanks

Thanks very much for answer to part a , but how about part b.
thanks very much
Kingman

5. Part B is confusing. It tells you to use the appropriate approximation to solve but then asks you a to find a different probability.

Can you double check to make sure you posted the question correctly? Maybe there were three parts.

a)solve by poisson
b)solve by approximation of poisson
c)solve the new probability

Maybe I'm missing something that someone else can point out

6. ## thaks

Thanks, apparently the question is referring to solving by approximation of poisson.