Question

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

Answer 1

Answer:

c. 0.16

Step-by-step explanation:

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:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

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

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

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 3000000, p = 0.00001$

So

$\mu = E(X) = np = 3000000*0.00001 = 30$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{3000000*0.00001*(1-0.00001)} = 5.48$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{35 - 30}{5.48}$

$Z = 0.91$

$Z = 0.91$ has a pvalue of 0.8186

1 - 0.8186 = 0.1814.

The closest answer is:

c. 0.16