It is given that number of accidents on a particular highway is average 4.4 per year.
a. Let X be the number of accidents on a particular highway.
X follows Poisson distribution with mean μ =4.4
The probability function of X , Poisson distribution is given by;
P(X=k) = ![\frac{e^{-4.4} (4.4)^{k}}{k!}](https://tex.z-dn.net/?f=%20%5Cfrac%7Be%5E%7B-4.4%7D%20%284.4%29%5E%7Bk%7D%7D%7Bk%21%7D%20%20%20%20)
b. Probability that there are exactly four accidents next year, X=4
P(X=4) = ![\frac{e^{-4.4} (4.4)^{4}}{4!}](https://tex.z-dn.net/?f=%20%5Cfrac%7Be%5E%7B-4.4%7D%20%284.4%29%5E%7B4%7D%7D%7B4%21%7D%20%20%20%20)
P(X=4) = 0.1917
Probability that there are exactly four accidents next year is 0.1917
c. Probability that there are more that three accidents next year is
P(X > 3) = 1 - P(X ≤ 3)
= 1 - [ P(X=3) + P(X=2) + P(X=1) + P(X=0)]
P(X=3) = ![\frac{e^{-4.4} (4.4)^{3}}{3!}](https://tex.z-dn.net/?f=%20%5Cfrac%7Be%5E%7B-4.4%7D%20%284.4%29%5E%7B3%7D%7D%7B3%21%7D%20%20%20%20)
P(X=3) = 0.1743
P(X=2) = ![\frac{e^{-4.4} (4.4)^{2}}{2!}](https://tex.z-dn.net/?f=%20%5Cfrac%7Be%5E%7B-4.4%7D%20%284.4%29%5E%7B2%7D%7D%7B2%21%7D%20%20%20%20)
P(X=2) = 0.1188
P(X=1) = ![\frac{e^{-4.4} (4.4)^{1}}{1!}](https://tex.z-dn.net/?f=%20%5Cfrac%7Be%5E%7B-4.4%7D%20%284.4%29%5E%7B1%7D%7D%7B1%21%7D%20%20%20%20)
P(X=1) = 0.054
P(X=0) = ![\frac{e^{-4.4} (4.4)^{0}}{0!}](https://tex.z-dn.net/?f=%20%5Cfrac%7Be%5E%7B-4.4%7D%20%284.4%29%5E%7B0%7D%7D%7B0%21%7D%20%20%20%20)
= 0.0122
Using these probabilities into above equation
P(X > 3) = 1 - P(X ≤ 3) = 1 - [ P(X=3) + P(X=2) + P(X=1) + P(X=0)]
= 1 - (0.1743 + 0.1188 + 0.054 + 0.0122)
P(X > 3) = 1 - 0.3593
P(X > 3) = 0.6407
Probability that there are more than three accidents next year is 0.6407
Answer:
C
Reason:
It doesn’t pass through the orgin
Answer:
5 HOURS
Step-by-step explanation:
352.21-200.16=152.21
152.21÷30.41=5
B^2-6b+8 turns into b^2-2b-4b+8 due to the sum product pattern. which then turns into b(b-2) - 4(b-2) and then its (b-4)(b-2)