Thousands of travelers pass through the airport in Guadalajara, Mexico, each day. Before leaving the airport, each passenger must pass through the Customs inspection area. Customs agents want to be sure that passengers do not bring illegal items into the country. But they do not have time to search every traveler’s luggage. Instead, they require each person to press a button. Either a red or a green bulb lights up. If the red light shows, the passenger will be searched by Customs agents. A green light means “go ahead.” Customs agents claim that the proportion of all travelers who will be stopped (red light) is$0.30$, because the light has probability $0.30$of showing red on any push of the button. To test this claim, a concerned citizen watches a random sample of $100$travelers push the button. Only $20$get a red light.

(a) Assume that the Customs agents’ claim is true. Find the probability that the proportion of travelers who get a red light is as small as or smaller than the result in this sample. Show your work.

(b) Based on your results in (a), do you believe the Customs agents’ claim? Explain.

Given in the question that, Thousands of travelers pass through the airport in Guadalajara, Mexico, each day. Before leaving the airport, each passenger must pass through the Customs inspection area. Customs agents want to be sure that passengers do not bring illegal items into the country. But they do not have time to search every traveler’s luggage. Instead, they require each person to press a button. Either a red or a green bulb lights up. If the red light shows, the passenger will be searched by Customs agents. A green light means “go ahead.” Customs agents claim that the proportion of all travelers who will be stopped (red light) is $0.30$, because the light has probability$0.30$of showing red on any push of the button. To test this claim, a concerned citizen watches a random sample of 100 travelers push the button. Only 20 get a red light

Given,

Probability of success$\left(p\right)=0.30$

Number of events$\left(x\right)=20$

Number of trials $\left(n\right)=100$

If $X$is a random variable represented as the number of people who get the red light.

Since, $np($mean) and $np(1-p)$(variance) are greater than $5$. Therefore, $X$will follow Normal approximation.

$X~N(np,np(1-p\left)\right)$

$X~N\left(100\right(0.30),100(0.30\left)\right(1-0.30\left)\right)$

role="math" $X~N(30,21)$

The probability that the proportion of travelers who get a red light less than $20$, can be calculated as

localid="1650254778514" $$\begin{gathered} P(X<20)=P\left(\frac{X-np}{\sqrt{np(1-p)}}<\frac{20-30}{\sqrt{21}}\right)4.5826 \\ =P(Z<-2.1822) \\ =0.0145 \end{gathered}$$ (From standard normal table)

Thus, the required probability is $0.0145$

Given in the question that, Thousands of travelers pass through the airport in Guadalajara, Mexico, each day. Before leaving the airport, each passenger must pass through the Customs inspection area. Customs agents want to be sure that passengers do not bring illegal items into the country. But they do not have time to search every traveler’s luggage. Instead, they require each person to press a button. Either a red or a green bulb lights up. If the red light shows, the passenger will be searched by Customs agents. A green light means “go ahead.” Customs agents claim that the proportion of all travelers who will be stopped (red light) is $0.30$, because the light has probability$0.30$ of showing red on any push of the button. To test this claim, a concerned citizen watches a random sample of $100$ travelers push the button. Only $20$ get a red light.

From the above part, the probability is$0.0146$. Since, the probability is low subsequently it seems, by all accounts, to be that the claim is false, Thus, the specialist doesn't completely accept that that the customs agent claim.