MATH SOLVE

4 months ago

Q:
# 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?

Accepted Solution

A:

Answer:(a) more than 2 such accidents in the next month [tex]\approx 0.3773[/tex](b) more than 4 such accidents in the next 2 months [tex]\approx 0.44882[/tex](c) more than 5 such accidents in the next 3 more than 5 such accidents in the next 3 months [tex]\approx 0.64533[/tex]Step-by-step explanation:Let N be the Random variable that marks the number of crashes in certain month.Now let us use Poisson distribution since we are given with average number of crashes that is N \sim Pois(2.2)(A) more than 2 such accidents in the next monthProbability(more than 2 such accidents in the next month)=P(N>2)P(N>2)=1-P(N=0)-P(N=1)-P(N=2)=>[tex]1-e^-{2.2}-2.2e^{-2.2}-\frac{2.2^2}{2!}e^{2.2}[/tex]=> [tex]\approx 0.3773[/tex]B) more than 4 such accidents in the next 2 monthssince the average number of crashes in 1 month is 2.2, the average number of crashes in two months is 4.4. hence, if we say that [tex]N_1[/tex] is the number of crashes in 2 months, we have that [tex]N\sim[/tex]Pois(4.4)Thus,Probability(more than 4 such accidents in the next 2 months)=P([tex]N_1>4[/tex])=[tex]1-P(N_1=0)-P(N_1=1)-P(N_1=2)=P(N_1=3)-P(N_1=4)[/tex][tex]1-\sum_{k=1}^{4} \frac{4.4^{k}}{k !} e^{-4.4}[/tex]=> [tex]\approx 0.44882[/tex]C) more than 5 such accidents in the next 3 more than 5 such accidents in the next 3 monthsIf we say that [tex]N_2[/tex] marks the number of crashes in the next 3 months , using the same argument as in (a) we have that a [tex]N\sim[/tex]Pois(6.6)HenceP([tex]N_2>5[/tex])=[tex]1-\sum_{k=0}^{5} \frac{6.6^{k}}{k !} e^{-6.6}[/tex]=>[tex]\approx 0.64533[/tex]