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Assume that police estimate that 12​% of drivers do not wear their seatbelts. They set up a safety​ roadblock, stopping cars to check for seatbelt use. They stop 50 cars during the first hour. a. Find the​ mean, variance, and standard deviation of the number of drivers expected not to be wearing seatbelts. Use the fact that the mean of a geometric distribution is μ= 1 p and

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JeanaShupp

Answer: mean = 8.33, variance =61.11 , standrad deviation = 7.82

Step-by-step explanation:

For geometric distribution,

Mean = [tex]\dfrac1p[/tex]

variance = [tex]\dfrac{1-p}{p^2}[/tex]

Standard deviation[tex]=\sqrt{Variance}[/tex] , where p = probability of success in each trial.

Given: p= 0.12

Mean = [tex]\dfrac1{0.12}=8.33[/tex]

Variance = [tex]\dfrac{1-0.12}{0.12^2}=\dfrac{0.88}{0.0144}=61.11[/tex]

Standard deviation = [tex]\sqrt{61.11}\approx7.82[/tex]

Hence,  mean = 8.33, variance =61.11 , standrad deviation = 7.82

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