I am trying to solve a question but I don't have any idea about how to solve it, could you please help me?
The log-ons to a network follow a Poisson process with an average of 180 counts per hour.
a) Probability that 5 counts occur in a minute?
b) Normal approximation (with correction) of the probability that
log-on counts to the network in a minute is between 1 and 5 (end-points are included).
For part (a) we have the following:
The mean ($\lambda$) is given as $180$ counts per HOUR, however, the question asks about the probability that $5$ counts in a MINUTE, in order to work with that we need both of them to have the same unit, the easiest conversion would be to convert the given mean to minutes; $180$ counts per hour is equal to $3$ counts per minute since we have to divide by $60$. Now since we fixed this we can apply the Poisson distribution:
$P(X=k)=\frac{\lambda^k e^{-\lambda}}{k!}$ where $\lambda$ is our mean .
Here we have $\lambda=3$ and $k=5$, applying our formula we get:
$P(X=k)=\frac{3^5 e^{-3}}{5!}=0.10081881...$
For part (b):
When approximating a Poisson distribution by the normal distribution we have the following: Poisson($\lambda$)~Normal($\mu=\lambda$, $\sigma^2=\lambda$), note that the larger $\lambda$ is the better the approximation. Here we will keep using $\lambda=3$ since they are still asking about the average in minutes.
For the sake of understanding and to avoid confusion the question in part (b) can be rephrased as:
The log-on per minute count to a network follow a normal distribution with $\mu=3$ and $\sigma^2=3$, find the probability that we get between $1$ and $5$ log-ons per minute.
This means we need $P(1<=X<=5)$, as mentioned in the question we need to do "correction", the more technical word here is continuity correction which yields our probability range to be $P(0.5<=X<=5.5)$ can transform this to the standard normal distribution to use the table, note that if $X$ is a normal distribution then $\frac{X-\mu}{\sigma}$ is a standard normal distribution $Z$ we here we need $P(\frac{0.5-3}{\sqrt 3}<=\frac{X-3}{\sqrt3}<=\frac{5.5-3}{\sqrt3}$) that is $P(\frac{-2.5}{\sqrt 3}<=Z<=\frac{2.5}{\sqrt 3}$), converting these to decimal values in order to use the table we get $P(-1.4437<=Z<=0.14437)$, since the standard distribution table only gives us $(P(X<=x)$ we need one last step and that is:
$P(1.4437<=Z<=1.4437)=P(Z<=1.4437)-P(Z<=-1.4437)$ Using the table that is equal to: $0.92507-0.07493=0.85014$
Note that the exact values aren't in the table (this is pretty normal and happens in most exercises), I used $P(Z<=-1.44)-P(Z<=-1.44)$ which is pretty close to the numbers given and the result doesn't change much.
I don't know what you have learned and haven't learned yet so if anything I used is unclear feel free to comment.
Some good sources for further reading and examples: http://wiki.stat.ucla.edu/socr/index.php/AP_Statistics_Curriculum_2007_Distrib_Poisson
http://wiki.stat.ucla.edu/socr/index.php/AP_Statistics_Curriculum_2007_Limits_Norm2Poisson