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August 1, 2014

August 1, 2014

Posted by **Felicia** on Tuesday, May 18, 2010 at 12:43pm.

of 0.84, 0.13, 0.02, and 0.01, respectively.

Graph this probability distribution. What is the

expected value for the random variable given the

number of accidents?

- Math -
**drwls**, Tuesday, May 18, 2010 at 2:02pm0.84*0 + 0.12*1 + 0.02*2 + 0.01*3 = 0.19

is the expected number of accidents (on average).

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