The number of parking tickets issued in a certain city... The number of parking tickets issued in a certain city on any given day has Poisson distribution with parameter $\mu = 50.$
Calculate the approximate probability that between $35$ and $80$ tickets are given on a day.  
I'm not sure how to approach this, but the wording in the problem gives me a hint that this is a normal distribution problem, that's all I know.
Any help will be appreciated.
 A: Its not a normal distribution problem, the problem clearly states that it is a Poisson distribution "tickets issued in a certain city on any given day has poisson distribution with parameter u = 50". See https://en.wikipedia.org/wiki/Poisson_distribution . 
The answers would be summing the pmf from $k = 35$ to $k = 80$.  One term for example, is the probability that exactly 35 tickets are given out on a day: this is $\frac{50^{35}e^{-50}}{35!}.$
The hint may be refering to the fact that the Poisson distribution can be approximated by a normal distribution in some situations, but this is not really necessary here.
A: It would be expected to use a normal approximation since the parameter value is so large ($>10$) and there are so many values to calculate.
However you should use a continuity correction, so if you mean $35\leq X\leq80$ this should be changed to $34.5<X<80.5$ when applying the normal.
A: If $X \sim \mathsf{Pois}(\mu = 50),$ then you seek
$$P(45 \le X \le 80) = e^{-50}\sum_{k=45}^{80} \frac{50^k}{k!} = 0.7789254.$$
This can be computed exactly using statistical software. Results from R, where dpois is a Poisson PDF (or PMF) and ppoisis a Poisson CDF, are as follows:
k = 45:80;  sum(dpois(k, 50))
## 0.7789254
diff(ppois(c(44,80), 50))
## 0.7789254

If you want to use $\mathsf{Norm}(\mu = 50, \sigma=\sqrt{50})$ as an approximation to $\mathsf{Pois}(50),$ then you can write
$$P(44.5 \le X \le 80.5) = P\left(\frac{44.5-50}{\sqrt{50}} \le 
\frac{X - \mu}{\sigma}  
\le \frac{80.5-50}{\sqrt{50}} \right)$$
$$ \approx P(-0.78 \le Z \le 4.13) = 0.7823,$$
from tables of the standard normal distribution $Z.$ (Depending on
rounding, etc., your answer from printed normal tables may differ slightly.
Generally speaking, one expects about two-place accuracy from a good
normal approximation.)
The plot below shows the not-quite-perfect fit of the normal density to the Poisson distribution.
The probability of interest is between the vertical dotted lines.

