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I was recently solving a quiz on Poisson distribution and I encountered this question

A call center receives an average of • 4.5 calls every 5 minutes. Each agent can handle one of these calls over the 5 minute period. If a call is received, but no agent is available to take it, then that caller will be placed on hold. Assuming that the calls follow a Poisson distribution, what is the minimum number of agents needed on duty so that calls are placed on hold at most 10% of the time?

I figured out this was a case of finding K while we are given the expected probability in the question itself.

what is the minimum number of agents needed on duty so that calls are
placed on hold at most 10% of the time?

I could not event determine the lambda and X to solve this question.
PS : The correct answer was 7 agents.A detailed explanation will be helpful.

Thanks

I believe you are being asked to focus on a typical 5-minute period of time.
Then the average number $X$ of incoming calls within 5 minutes has the distribution $X sim mathsfPois(lambda = 4.5).$ A bar plot of this distribution is shown below:

Poisson Model: Computations with the Poisson PDF $P(X = i) = e^-lambdafraclambda^ii!,$ where $lambda = 4.5,$ show that $P(X le 6) = 0.8311 < 0.9$ and
$P(X le 7) = 0.9134.$ Thus seven agents would suffice to serve
incoming customers at least 90% of the time.

Trial and error with Poisson CDF: Computations using
R statistical software (where the CDF of a Poisson distribution is denoted ppois) are as follows:

ppois(6, 4.5)
## 0.8310506
ppois(7, 4.5)
## 0.9134135

Using a Poisson quantile function: If you have such software available, you can use the inverse CDF
or quantile function qpois to get to the answer without
exploration.

qpois(.9, 4.5)
## 7

Note: Without knowing the context of this problem in your course, I suppose something like this is the approach you are expected to take. However, there
are some unstated assumptions involved. A complete analysis of such
a problem would involve finding the required number of servers $k$ required in an M/M/k queue (at steady state) that are sufficient to keep the average number of customers in the
system below $k$ 90% of the time. This would involve knowing the
exponential arrival rate $lambda = 4.5$ per five minutes or $0.9$ per
minute, and knowing the exponential rate (often denoted $mu)$ at
which each server can finish serving customers.

For starters: here are a few unstated assumptions: (a) No agents are busy at the beginning of the 5-minute period. (b) Each agent handles only one call within this period. (c) We are not concerned whether
agents are still handling calls from this 5-minute period ends and the next begins.