Continous random exercise

# Continous random exercise - probability of passing the car...

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Section A, LT 9 Date: 11 th Oct 2011 5.34 Solution Capacity = 10 customers per 15 minute period Average Arrival = 7 customers per 15 minute period X = number of customers arriving per 15 minute period a) Probability that 10 customers will arrive in 15 minute period = P(10) = P (7,10) = (e^-7*7^10)/10! = .07098 (calculation) = .0710 (from table) b) Probability that 10 or few customers will arrive in a particular 15 minutes P(X<=10) = P(0)+P(1)+P(2)+P(3)+P(4)+P(5) +P(6) +P(7) +P(8) +P(9) +P(10) = . 0009+.0064+.0223+.0521+.0912+.1277+.1490+.1490+.1304+.1014+.0710 = .9015 (from table) c) Probability that more than 10 customers will arrive during a particular 15 minute period P (X>10) = 1- P(X<=10) = 1-.9015 = .0985 5.36 Solution Mean = 3 times per nightly shift a) Probability that no cars pass through = P(X=0) = .0498 (from table) b) Probability that no cars will pass through is just .0498, meaning 4.98%. Hence,

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Unformatted text preview: probability of passing the car is 95.02%. Therefore, agency claim should not be believed. c) Probability that exactly one patrol car will pass through the neighborhood on each of four consecutive nights = P (one patrol car, 4 consecutive nights)^independence = (P(one patrol car per night))^4 = ((3^1)*(e^-3))/1! = .1494^4 = .000498 = .0005 5.38 Solution Mean = 200*.5% = 1 a) Probability of four or more car batteries in a random sample of 200 to be defective = P(X>=4) = 1-P(X<4) = 1- (P(0)+P(1)+P(2)+P(3)) = 1- (.3679+.3679+.1839+.0613) = 1-.9810 = .0190 b) No 5.42 Solution N =10, r =4 and n =3 a) Probability distribution of X(success) = (4CX*((10-4)C(3-X)))/10C3 b) Mean = 3*(4/10) = 3*.4 = 1.2 Variance = 3*(4/10)*(1-(4/10))*((10-3)/(10-1)) = 1.2*.6*.7777 = .5600 5.44 Solution N =10, r =9 and n =3 P(X=3) =( 9C3*(1C0))/10C3 = (84*1)/120 = .7000...
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## This note was uploaded on 01/16/2012 for the course MBA 101 taught by Professor Wormer during the Spring '08 term at Indian Institute Of Management, Ahmedabad.

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Continous random exercise - probability of passing the car...

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