- I would be grateful if the work can be shown, thank you. 2. In a city (outside

Question

- I would be grateful if the work can be shown, thank you.

2. In a city (outside North America) a traffic officer knows from experience that 30% of drivers do not have a driver's license or have an expired one. This officer stops 10 cars looking to obtain bribes to pay for his lunch. a) Given that the officer needs at least one bribe to have enough money for his lunch, what is the probability that the officer cannot get money for his lunch (i.e., all the drivers have a legal driver's license)? one? b) What is the probability that 4 or less than 4 of the drivers do not have a driver's license or have an expired c) What are the mean (i.e., expected value) and the standard deviation of the random variable? In one sentence, explain how to interpret the value of the mean. d) What is the probability that the number of drivers that do not have a driver's license (or have an expired one) is more than the mean?

Explanation / Answer

2.

This problem can be solved with the help of binomial distribution

p = probability of success = probability that driver do not have driver's license or have an expired = 0.30

n = number of trials = number of cars stopped by officer = 10

P(X=x) = nCx * px * (1-p)n-x

a)

Required probability = Probability that officer cannot get money for his lunch =

Probability that all drivers have a legal driver's license = P(X = 0) = 10C0 * p0 * (1-p)10-0

= (10! / 10! * 0!) * 0.300 * (1-0.30)10-0

= 0.7010 = 0.0282

b)

Required probability = Probability that 4 or less than 4 of the 10 drivers do not have a driver's licence or have an expired + P(X=1) + ......... + P(X=4)

= 10C0 * p0 * (1-p)10-0 + 10C1 * p1 * (1-p)10-1 + ..................+ 10C4 * p4 * (1-p)10-4

= (10! / 10! * 0!) * 0.300 * (1-0.30)10-0 + (10! / 9! * 1!) * 0.301 * (1-0.30)10-1 + ............................+

(10! / 6! * 4!) * 0.304 * (1-0.30)10-4

= 0.849

c)

Mean of binomial distribution = n * p = 10 * 0.30 = 3

Standard deviation of binomial distirbution = ( n * p * ( 1 - p ) )0.5 = (10 * 0.30 * ( 1 - 0.30 ))0.5 = 1.449

We can say that on an average 3 drivers out of 10 stopped car drivers by the officer do not have a driver's license or have an expired one.

d)

Required probability = P(X > 3) = P(X = 4) + P(X=5) + ....................+ P(X = 10)

= 10C4 * p4 * (1-p)10-4 + 10C5 * p5 * (1-p)10-5 + ..................+ 10C10 * p10 * (1-p)10-10

= (10! / 6! * 4!) * 0.304 * (1-0.30)10-4 + (10! / 5! * 5!) * 0.305 * (1-0.30)10-5 + ............................+

(10! / 0! * 10!) * 0.3010 * (1-0.30)10-10

= 0.3504

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