The monthly worldwide average number of airplane crashes of commercial

Idilwsiw2 2021-11-12 Answered
The monthly worldwide average number of airplane crashes of commercial airlines is 2.2. What is the probability that there will be (a) more than 2 such accidents in the next month? (b) more than 4 such accidents in the next 2 months? (c) more than 5 such accidents in the next 3 more than 5 such accidents in the next 3 months? Explain your reasoning!

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Expert Answer

Cherry McCormick
Answered 2021-11-13 Author has 1421 answers
Define N as the random variable that marks the number of plane crashes in a certain month. We are going to use Poisson approximation since we are given that the average number of crashes is 2.2. Now, we have that
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
\(\displaystyle{P}{\left({N}_{{2}}{>}{5}\right)}={1}-{\sum_{{{k}={0}}}^{{5}}}{\frac{{{6.6}^{{k}}}}{{{k}!}}}{e}^{{-{6.6}}}\approx{0.64533}\)
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