Let X be the number of burglaries in a week. X follows Poisson distribution with mean of 1.9
We have to find the probability that in a randomly selected week the number of burglaries is at least three.
P(X ≥ 3 ) = P(X =3) + P(X=4) + P(X=5) + ........
= 1 - P(X < 3)
= 1 - [ P(X=2) + P(X=1) + P(X=0)]
The Poisson probability at X=k is given by
P(X=k) =
Using this formula probability of X=2,1,0 with mean = 1.9 is
P(X=2) =
P(X=2) =
P(X=2) = 0.2698
P(X=1) =
P(X=1) =
P(X=1) = 0.2841
P(X=0) =
P(X=0) =
P(X=0) = 0.1495
The probability that at least three will become
P(X ≥ 3 ) = 1 - [ P(X=2) + P(X=1) + P(X=0)]
= 1 - [0.2698 + 0.2841 + 0.1495]
= 1 - 0.7034
P(X ≥ 3 ) = 0.2966
The probability that in a randomly selected week the number of burglaries is at least three is 0.2966
The answer: Jamal has spent $7.50 which leaves him with $7.50 left. He should have enough to buy sour straws, I mean how expensive are they???
Answer:
ill give you a idea to answer the question
Answer: 172.5% OR 1.725 as decimal
Step-by-step explanation:
138-80=58
80=100% OR 1
58=x%
x=58/80
x=72.5% OR 0.725
58=72.5
1+0.725=1.725
80x1.725=138
Hopefully I somehow helped :) -KJ
I'm sorry. I can't explain how I got it. But use
https://goo.gl/3CQjcC
MathPapa
It helps explain it and tells you the answer. Sprry, but hope it helps. It's a website and app.