It is estimated that 10% of the vehicles entering Canada from the United States carry undeclared goods.

Question

It is estimated that 10% of the vehicles entering Canada from the United States carry undeclared goods. Assuming the data can be modelled using a normal approximation, determine the probability that a search of 500 randomly selected vehicles will find fewer than 50 with undeclared goods.

Answer 1

Since we have a random sample of 500 vehicles from a significantly large population, it is reasonable to assume that 500 is less than 5% of the population size, so that the randomization and independence conditions are met. Moreover, we are given that p = 0.10, and so,

np(1-p) = 500*(0.10)*(0.90) = 45 >= 10, which implies that the sampling distribution of the sample proportion is approximately normal with mean = 0.10 and standard deviation = √(p*(1-p)/n) = √(0.10*0.90/500) = 0.0134

The sample proportion = 50/500 = 0.10. If we standardize 0.10, we get (0.10 - 0.10)/0.0134 = 0.

Hence, P(X < 50) = P(sample proportion < 0.10) = P(z < 0) = 0.50

That is, out of 100 samples of size 500 we can expect 50 samples having fewer than 50 vehicles with undeclared goods.