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Mathematics 10 Online
OpenStudy (anonymous):

PLEASE HELP!!!!!! I can't quite understand how to do this.

OpenStudy (anonymous):

The probability distribution of the number of accidents in a week at an intersection, based on past data, is given below. Find the expected number of accidents in one week. Round your answer to the nearest hundredth.

OpenStudy (anonymous):

number of accidents (n)| 0 | 1| 2| 3 Probability of n accidents| 0.75| 0.15| 0.08| 0.02

OpenStudy (anonymous):

what kinda distribution do ya use?

OpenStudy (anonymous):

That's where I am at a loss. I'm not quite sure. I'm getting confused because of the past problems I have been doing

OpenStudy (freckles):

since it is discrete \[E(\text{ accidents in one week })=\sum_{n=0}^{3} n \cdot p(n)\]

OpenStudy (anonymous):

probability mass function?

OpenStudy (freckles):

aka also known as mean value

OpenStudy (anonymous):

Oh! That makes sense! I was over complicating it again...

OpenStudy (freckles):

have you found the sum of the products that I mentioned?

OpenStudy (freckles):

that is have you computed: 0*.75+1*.15+2*.08+3*.02

OpenStudy (anonymous):

Yes, indeed I was working on that. cx

OpenStudy (anonymous):

dayum probability magics

OpenStudy (anonymous):

for my final answer I got 0.37

OpenStudy (freckles):

ok sounds awesome

OpenStudy (freckles):

that is what I have gotten too

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