HELP PLS! Suppose the year of car accidents per day in Wilmington NC has been recorded for the past year. The table below is a probability distribution table representing the data collected.
1) what is the probability that on a randomly selected day there were no accidents?

2) find the probability there will be at least 2 accidents

3) on any given day how many accidents should the Wilmington police expect to have.

HELP PLS! Suppose The Year Of Car Accidents Per Day In Wilmington NC Has Been Recorded For The Past Year.

Answers

Answer 1

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.

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