N pois

a) \(\displaystyle{P}{\left({N}{>}{2}\right)}={1}-{P}{\left({N}={0}\right)}-{P}{\left({N}={1}\right)}-{P}{\left({N}={2}\right)}\)

\(\displaystyle={1}-{e}^{{-{2.2}}}-{2.2}{e}^{{-{2.2}}}-{\frac{{{2.2}^{{2}}}}{{{2}!}}}{e}^{{-{2.2}}}\approx{0.3773}\)

b) Since the average number of crashes in one month is 2.2, the average number of crashes in two months is 4.4. Hence, if we say that \(\displaystyle{N}_{{1}}\) is the number of crashes in two months, we have that \(\displaystyle{N}_{{1}}\) pois. Thus

\(\displaystyle{P}{\left({N}_{{1}}{>}{4}\right)}={1}-{P}{\left({N}_{{1}}={0}\right)}-{P}{\left({N}_{{1}}={1}\right)}-{P}{\left({N}_{{1}}={2}\right)}-{P}{\left({N}_{{1}}={3}\right)}-{P}{\left({N}_{{1}}={4}\right)}\)

\(\displaystyle={1}-{\sum_{{{k}={0}}}^{{4}}}{\frac{{{4.4}^{{k}}}}{{{k}!}}}{e}^{{-{4.4}}}\approx{0.44882}\)

c) If we say that \(\displaystyle{N}_{{2}}\) marks the number of crashes in next three months, using the same argument as in a) have that \(\displaystyle{N}_{{2}}\). Hence

\(\displaystyle{P}{\left({N}_{{2}}{>}{5}\right)}={1}-{\sum_{{{k}={0}}}^{{5}}}{\frac{{{6.6}^{{k}}}}{{{k}!}}}{e}^{{-{6.6}}}\approx{0.64533}\)