Let $X$ denote the number of drivers who travel between a particular origin and destination during a designated time period. Since it has a Poisson distribution with parameter $\lambda=20$. Then the pmf of $X$ is
\[
p(x;\lambda)=\frac{e^{-\lambda}\lambda^x}{x!}=\frac{e^{-20}20^x}{x!},\quad x=0,1,2,\ldots
\]
(a)
The probability that the number of drivers will be at most $10$ is
\[
P(X\leqslant 10)=\sum_{x=0}^{10}p(x;20)=\sum_{x=0}^{10}\frac{e^{-20}20^x}{x!}=e^{-20}\sum_{x=0}^{10}\frac{20^x}{x!}\approx 0.010812
\]
Do calculation with Calculator
>> sum(20^x/(x!),x,0,10)
in> sum(20^x/(x!),x,0,10)
out> 5245469.677249
------------------------
>> exp(20)
in> exp(20)
out> 485165195.409790
------------------------
>> 5245469.677249/485165195.409790
in> 5245469.677249/485165195.409790
out> 0.010812
(b)
The probability that the number of drivers will be exceed $20$ is
\[
\begin{split}
P(X\geqslant 20)&=1-P(X\leqslant 20)=1-\sum_{x=0}^{20}p(x;20)\\
&=1-\sum_{x=0}^{20}\frac{e^{-20}20^x}{x!}\\
&=1-e^{-20}\sum_{x=0}^{20}\frac{20^x}{x!}\\
&\approx=1-0.559093\\
&=0.440907
\end{split}
\]
(c)
The probability that the number of drivers will be between $10$ and $20$ inclusive is
\[
\begin{split}
P(10\leqslant X\leqslant 20)&=P(X\leqslant 20)-P(X\leqslant 9)\\
&=\sum_{x=0}^{20}p(x;20)-\sum_{x=0}^{9}p(x;20)\\
&=\sum_{x=0}^{20}\frac{e^{-20}20^x}{x!}-\sum_{x=0}^{9}\frac{e^{-20}20^x}{x!}\\
&=e^{-20}\sum_{x=10}^{20}\frac{e^{-20}20^x}{x!}\\
&\approx 0.554097
\end{split}
\]
The probability that the number of drivers will be strictly between $10$ and $20$ is
\[
\begin{split}
P(10 < X < 20)&=P(X\leqslant 19)-P(X\leqslant 10)\\
&=\sum_{x=11}^{19}p(x;20)\\
&=\sum_{x=11}^{19}\frac{e^{-20}20^x}{x!}\\
&=e^{-20}\sum_{x=11}^{19}\frac{e^{-20}20^x}{x!}\\
&\approx 0.459446
\end{split}
\]
(d)
By proposition, the expected value or mean number of $X$ is $\mu=E(X)=\lambda=20$. And the variance is $V(X)=\lambda=20$. Hence, the standard deviation is $\sigma=\sqrt{V(X)}=\sqrt{20}\approx 4.472136$.
The probability that the number of drivers will be exceed the mean number by more than two standard deviations is
\[
\begin{split}
P(X\geqslant\mu+2\sigma)&=P(X\geqslant 20+2\times 4.472136)=P(X\geqslant 28.944272)\\
&=P(X\geqslant 29)=1-P(X < 29)=1-P(X\leqslant 28)\\
&=1-\sum_{x=0}^{28}\frac{e^{-20}20^x}{x!}\\
&\approx 1-0.965666\\
&=0.034334
\end{split}
\]
