A recent study shows that in a certain city 2, 477 cars out of 123, 850 are stol

Question

A recent study shows that in a certain city 2, 477 cars out of 123, 850 are stolen. The city policy are trying to find the stolen cars. Suppose that 100 randomly chosen cars are checked by the police. (a) Let X denote the number of stolen cars in this sample. Identify the type of distribution of X. Choose one of the followings: A. Binomial distribution B. Geometric distribution C. Hypergeometric distribution D. Negative binomial distribution E. Poisson distribution (b) Find the expression that gives the probability that exactly 3 of the chosen do not need to give the numerical value of this expression.

Explanation / Answer

(a) Distribution of X is most suitably defined by: A. Binomial distribution

Binomial distribution is the discrete probability distribution of number of successes in a sequence of trials, where for each trial we have a binary outcome such as Yes/No, Success/Failure and so on.

In this case, a car being a stolen car can be defined as a 'Success' event and otherwise a 'Failure'. Success and Failue are defined based on the case we are analyzing. Here, success is when we identify a stolen car.

(b) Total number of cars = 123,850

Number of stolen cars = 2,477

As, 100 randomly chosen cars are being inspected, we have: Number of trials, n = 100

Probability of success event = Probability of a car being stolen which is equal to Number of stolen cars/Total number of cars

Thus, probability of success event, p = 2477 / 123850 = 0.02

Probability of failure = Probability of car being genuine, q = 1 - p = 1 - 0.02 = 0.98

Thus, we have n=100, p=0.02, q=0.98

Given that X denotes the number of stolen cars in the sample. X follows Binomial distribution, i.e., X~B(n,p)

The probability mass function of Binomial distribution is given by:

f(k;n,p) = P(X=k) = nCk * pk * (1-p)(n-k)

where P(X=k) is the probability of getting exactly k number of successes in n trials.

Thus, we can calculate the probability of exactly 3 of the chosen cars to be stolen as follows:

k = 3

P(X=3) =100C3 * (0.02)3 * (1-0.02)(100-3)

P(X=3) = (100! / (3!*(100-3)!) ) * 0.000008 * 0.1409

P(X=3) = ((100*99*98*97!)/(6*97!))* 0.0000011272

P(X=3) = 161700 * 0.0000011272 = 0.1823

There is a 0.1823 probability that exactly 3 out of the 100 chosen cars are stolen.

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