One strange phenomenon that sometimes occurs at U.S. airport security gates is that an otherwise law-abiding passenger is caught with a gun in his/her carry-on bag. Usually the passenger claims he/she forgot to remove the handgun from a rarely-used bag before packing it for airline travel. It’s estimated that every day 3,000,000 gun owners fly on domestic U.S. flights.

Suppose the probability a gun owner will mistakenly take a gun to the airport is 0.00001. What is the probability that tomorrow more than 35 domestic passengers will accidentally get caught with a gun at the airport?

Choose the closest answer.

a. 0.02
b. 0.91
c. 0.16
d. 0.84
e. 0.28

For each passenger there are only two possible outcomes. Either they are caught with a gun, or they are not. The probability of a passenger being caught with a gun is independent from other passengers. So we use the binomial probability distribution to solve this question.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 3000000, p = 0.00001[/tex]

So

[tex]\mu = E(X) = np = 3000000*0.00001 = 30[/tex]

[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{3000000*0.00001*(1-0.00001)} = 5.48[/tex]

What is the probability that tomorrow more than 35 domestic passengers will accidentally get caught with a gun at the airport?

This probability is 1 subtracted by the pvalue of Z when X = 35. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{35 - 30}{5.48}[/tex]

[tex]Z = 0.91[/tex]

[tex]Z = 0.91[/tex] has a pvalue of 0.8186

1 - 0.8186 = 0.1814.

The closest answer is:

c. 0.16