Respuesta :
Answer:
[tex]a) \simeq 0.3012[/tex] [tex]b) \simeq 0.0494[/tex] c) [tex]\simeq 0.2438[/tex]
Step-by-step explanation:
Rate of collision,
1.2 collisions every 4 months
or, [tex]\frac{1.2}{4}[/tex]
= 0.3 collisions per month
So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,
P(X =x) = [tex]\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!} [/tex]
for x ∈ N ∪ {0}
= 0 otherwise --------------------------------------(1)
here, [tex]\lambda = 0.3[/tex] collision / month
No collision over a 4 month period means no collision per month or X =0
Putting X = 0 in (1) we get,
P(X = 0) = [tex]\frac{e^{-0.3}\times {\0.3}^{0}}{0!} [/tex]
[tex]\simeq 0.7408182207[/tex] ------------------------------------(2)
Now, since we are calculating this for 4 months,
so, P(No collision in 4 month period)
=[tex]0.7408182207^{4}[/tex]
[tex]\simeq 0.3012[/tex] -----------------------------------------------------------(3)
2 collision in 2 month period means 1 collision per month or X =1
Putting X =1 in (1) we get,
P(X =1) = [tex]\frac{e^{-0.3}\times {\0.3}^{1}}{1!} [/tex]
[tex]\simeq 0.2222454662[/tex] ------------------------------------(4)
Now, since we are calculating this for 2 months, so ,
P(2 collisions in 2 month period)
=[tex]0.2222454662^{2}[/tex]
[tex]\simeq 0.0494[/tex] -----------------------------------------(5)
1 collision in 6 months period means
[tex]\frac{1}{6}[/tex] collision per month
Now, P(1 collision in 6 months period)
= P( X = 1/6] (which is to be estimated)
=[tex]\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}[/tex]
= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]
[tex]\simeq 0.6543894283[/tex]-------------------------------------------(6)
So,
P(1 collision in 6 month period)
= [tex]0.6543894283^{6}[/tex]
[tex]\simeq 0.0785267444[/tex] ------------------------------------------------(7)
So,
P(No collision in 6 months period)
= [tex](P(X =0)^{6}[/tex]
[tex]\simeq 0.1652988882[/tex] ---------------------------------(8)
so,
P(1 or fewer collision in 6 months period)
= (8) + (7 ) = 0.0785267444 +0.1652988882
[tex]\simeq 0.2438[/tex] ---------------------------------------------(9)
Using the Poisson distribution, it is found that there is a:
a) 0.3012 = 30.12% probability of having no collisions occur over a four-month period.
b) 0.0988 = 9.88% probability of having exactly two collisions in a two-month period.
c) 0.4628 = 46.28% probability of having one or fewer collisions in a six-month period.
What is the Poisson distribution?
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
- [tex]\mu[/tex] is the mean in the given interval.
Item a:
Mean of 1.2 collisions every four months, hence [tex]\mu = 1.2[/tex].
The probability is P(X = 0), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.2}1.2^{0}}{(0)!} = 0.3012[/tex]
0.3012 = 30.12% probability of having no collisions occur over a four-month period.
Item b:
Two months, hence [tex]\mu = 0.5(1.2) = 0.6[/tex].
The probability is P(X = 2), hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 2) = \frac{e^{-0.6}0.6^{2}}{(2)!} = 0.0988[/tex]
0.0988 = 9.88% probability of having exactly two collisions in a two-month period.
Item c:
Six months, hence [tex]\mu = 1.5(1.2) = 1.8[/tex].
The probability is:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
In which:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-1.8}1.8^{0}}{(0)!} = 0.1653[/tex]
[tex]P(X = 1) = \frac{e^{-1.8}1.8^{1}}{(1)!} = 0.2975[/tex]
Hence:
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.1653 + 0.2975 = 0.4628[/tex]
0.4628 = 46.28% probability of having one or fewer collisions in a six-month period.
More can be learned about the Poisson distribution at https://brainly.com/question/13971530
