Q. 31 - The Practice of Statistics for AP

Question

Airport security The Transportation Security Administration (TSA) is responsible for airport safety. On some flights, TSA officers randomly select passengers for an extra security check before boarding. One such flight had 76 passengers -12 in first class and 64 in coach class. TSA officers selected an SRS of 10 passengers for screening. Let $\hat{p}$ be the proportion of first-class passengers in the sample.

(a) Is the $10\%$ condition met in this case? Justify your answer.

(b) Is the Normal condition met in this case? Justify your answer.

Step-by-Step Solution

Answer

a. No, the $10\%$ condition is not met in this case.

b. No, the Normal condition is not met in this case.

Part (a) Step 1: Given Information

Number of passengers in flight$=76$ 

First-class passengers $=12$in first class 

Coach class passengers $=64$

Passengers for SRS screening $=10$

Part (a) Step 2: Explanation

Based on the 76-person population, here the sample size of 10 is greater than $10\%$ of the 76 -person population so the $10\%$ condition is not met.

Part (b) Step 1: Given Information

Number of passengers in flight$=76$

First-class passengers $=12$ in first class

Coach class passengers$=64$

Passengers for SRS screening $=10$

Part (b) Step 2: Explanation

Given for the population:

$x=12$

$n=76$

Based on the sample size divided by the number of successes, the population proportion is:

$$\begin{gathered} p=\frac{x}{n} \\ =\frac{12}{76} \end{gathered}$$

Sample given: 

$n=10$

As long as $np$ and $n(1-p)$ are both at least $10$, the condition is met.

$$\begin{gathered} np=10\times \frac{12}{76} \\ \approx 1.58 \end{gathered}$$

$$\begin{gathered} n(1-p)=10\times \left(1-\frac{12}{76}\right) \\ \approx 8.42 \end{gathered}$$

Neither of them exceeds $10$, so the Normal condition has not been met.