According to the discrete distribution given:
1) There is a 0.18 = 18% probability that on a randomly selected day there were no accidents.
2) There is a 0.7 = 70% probability that there will be at least 2 accidents.
3) On any given day, the Wilmington police should expect to have 2.03 accidents.
The distribution is:
[tex]P(X = 0) = x[/tex]
[tex]P(X = 1) = 0.12[/tex]
[tex]P(X = 2) = 0.32[/tex]
[tex]P(X = 3) = 0.25[/tex]
[tex]P(X = 4) = 0.13[/tex]
Item 1:
The sum of the probabilities has to be 100% = 1, hence:
[tex]x + 0.12 + 0.32 + 0.25 + 0.13 = 1[/tex]
[tex]x + 0.82 = 1[/tex]
[tex]x = 0.18[/tex]
0.18 = 18% probability that on a randomly selected day there were no accidents.
Item 2:
This probability is:
[tex]P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.32 + 0.25 + 0.13 = 0.7[/tex]
0.7 = 70% probability that there will be at least 2 accidents.
Item 3:
The expected value is the sum of each outcome multiplied by it's respective probability, hence:
[tex]E(X) = 0.18(0) + 0.12(1) + 0.32(2) + 0.25(3) + 0.13(4) = 2.03[/tex]
On any given day, the Wilmington police should expect to have 2.03 accidents.
A similar problem is given at https://brainly.com/question/25653146
