Poisson distribution problem including cdf The Question

The number of cracks in a section of highway that is significant
enough to require repair is assumed to follow a Poisson distribution.
(a) Let $Y$ be the number of cracks in $4$km, sketch the (CDF) Cumulative
Distribution Function and graph up to $ = 4.5$.
(b) If we should order the material to fix the cracks beforehand, how
many packages of the material (One package for one crack) shall we
order to ensure that all the cracks in $4$km can be fixed with at least
$95\%$ chance?

My Understanding
For part (a), I tried to compute the probabilities of $Y=0, 1, 2, 3, 4, 5$ respectively, by using the formula of $Pr(X=k)=\frac{e^{-}\mu^k}{k!}$ however, when I tried to sum these probabilities together , $F(Y)$ turns out to exceed $1$, where probability should not $\gt 1$, what's wrong with that? What is the appropriate way to find the probabilities and thus I can graph up to $y=4.5$?
For part (b), for my understanding, this Poisson distribution has infinite number of cracks, so I am quite doubt that how to ensure that all the cracks in $4$km can be fixed with at least $95\%$ chance?
 A: @callculus is correct that you do not have enough information to know
(or even estimate) the Poisson mean $\mu.$ I will use $\mu = 4.5$ for
illustration. Maybe that will give you enough information to make sense
of the problem in your question.
If the average number $X$ of cracks in $4$km of highway is distributed
$\mathsf{Pois}(\mu = 4.5),$ then $\sum_{k=0}^5 \frac{e^{-4.5}4.5^K}{k!} = 0.7029304 < 1.$ I used R to compute this in two ways: (a) by making a vector pdf of the six probabilities, and (b) by using the R procedure dpois, which is a Poisson PDF. Alternatively, you might use a calculator to sum these six terms.
k = 0:5;  pdf = exp(-4.5)*4.5^k/factorial(k)
sum(pdf)
[1] 0.7029304
sum(dpois(k, 4.5))
[1] 0.7029304

With $\mu = 4.5$ a graph of the CDF of this Poisson distribution is
as follows.
pdf = ppois(k, 4.5)
plot(k, pdf, type="s")  # 's' for 'stairstep'
 points(k, pdf, pch=19)


If you want to be 95% sure to have enough bags of material to
repair all of the cracks in $4$km of this highway, then you
need to find the number $r = 8$ at which the CDF first exceeds $0.95.$
You can use the R quantile function qpois (inverse CDF) for that.
Alternatively, using a calculator, you could sum the nine required terms to
verify this result.
qpois(.95, 4.5)
[1] 8
ppois(8, 4.5)
[1] 0.9597427

Here is an extended CDF plot that goes up to $P(X \le 9)\approx 1.$
The horizontal red line is at $0.95.$

R code for second figure:
k = 0:9;  pdf = ppois(k, 4.5)
plot(k, pdf, type="s")  # 's' for 'stairstep'
points(k, pdf, pch=19)
abline(h=.95, col="red")

A: Hint for part b: You have to set up the following inequality:
$$P(X\leq x)=\sum_{k=0}^x e^{-2}\cdot \frac{2^k}{k!}\geq 0.95,$$
where $X\sim Poi(2)$
From one of the previous exercises you know that $P(X\leq 5)>0.95$. So go on and try $x=4,3,...$ until $P(X\leq x)<0.95$.
