Question

thousands of travelers pass through the airport in guadalajara, mexico, each day. before leaving the airport, each passenger must go 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 a red light flashes, the passenger will be searched by customs agents. a green light means it is ok for the passenger to \go ahead.\ customs agents claim that the light has probability 0.30 of showing red on any push of the button. assume for now that this claim is true. suppose we watch 20 passengers press the button. let r = the number who get a red light.

(a) explain why r is a binomial random variable.
this is a binomial setting because:
binary?
independent?
number?
same probability?

(b) find the probability that exactly 6 of the 20 passengers get a red light.
(round to 4 decimal places. leave your answer in decimal form.)

here is the histogram of the random variable r.
histogram with x - axis labeled number of passengers who get a red light and y - axis labeled probability
(c) describe the shape of the histogram.
the graph is with a single peak at

Explanation:

Part (a)

Step1: Binary?

Each passenger pressing the button results in a binary outcome: either a red light (success) or a green light (failure). So, there are two possible outcomes for each trial (passenger pressing the button).

Step2: Independent?

The outcome for one passenger (whether they get a red light) does not affect the outcome for another passenger. The probability of a red light for one passenger is not influenced by the results of other passengers pressing the button. So, the trials are independent.

Step3: Number?

We are observing a fixed number of trials, which is 20 passengers pressing the button. So, the number of trials \( n = 20 \) is fixed.

Step4: Same probability?

The probability of getting a red light for each passenger is the same, \( p = 0.30 \), for every press of the button. So, the probability of success (red light) is constant for each trial.

Since all four conditions for a binomial random variable are met (binary outcomes, independent trials, fixed number of trials, constant probability of success), \( R \) is a binomial random variable.

To find the probability that exactly 6 of the 20 passengers get a red light, we use the binomial probability formula:

$$ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} $$

where:

• \( n = 20 \) (number of trials),
• \( k = 6 \) (number of successes),
• \( p = 0.30 \) (probability of success on a single trial),
• \( \binom{n}{k} = \frac{n!}{k!(n - k)!} \) (combination of \( n \) things taken \( k \) at a time).

Step1: Calculate the combination \( \binom{20}{6} \)

$$ \binom{20}{6} = \frac{20!}{6!(20 - 6)!} = \frac{20!}{6!14!} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 38760 $$

Step2: Calculate \( p^k \) and \( (1 - p)^{n - k} \)

• \( p^k = (0.30)^6 \approx 0.000729 \)
• \( (1 - p)^{n - k} = (0.70)^{14} \approx 0.00678223 \)

Step3: Multiply the three components

$$ P(R = 6) = 38760 \times 0.000729 \times 0.00678223 \approx 0.1916 $$

Looking at the histogram, we can see that the distribution of the number of passengers who get a red light:

• The peak (highest bar) is around the middle - to - right of the histogram (since the mean of a binomial distribution \( \mu = np = 20\times0.3 = 6 \)? Wait, no, \( 20\times0.3 = 6 \)? Wait, no, \( 20\times0.3 = 6 \)? Wait, no, \( 20\times0.3 = 6 \)? Wait, no, \( 20\times0.3=6 \)? Wait, no, \( 20\times0.3 = 6 \)? Wait, maybe I miscalculated. Wait, \( n = 20 \), \( p = 0.3 \), so \( \mu=np = 6 \)? Wait, but the histogram's peak seems to be around a higher value? Wait, no, maybe the x - axis is labeled as "Number of passengers who get a red light". Wait, the histogram is symmetric or slightly skewed? Wait, for a binomial distribution, when \( p = 0.3 \) and \( n = 20 \), the distribution is skewed to the right? Wait, no, when \( p<0.5 \), the binomial distribution is skewed to the right. Wait, but looking at the histogram, the left side (lower number of red lights) has shorter bars, and the right side (higher number of red lights) also has shorter bars, with the peak in the middle. Wait, maybe the histogram is approximately symmetric? Wait, no, \( np = 6 \), \( n(1 - p)=14 \), since \( np \) and \( n(1 - p) \) are both greater than 5 (6 and 14), the binomial distribution can be approximated as symmetric. Wait, but the histogram here: the shape is approximately symmetric (or maybe slightly skewed to the right, but the main feature is that it has a single peak (unimodal) and is approximately symmetric around the mean (which is \( np = 6 \)? Wait, no, \( 20\times0.3 = 6 \), but the peak in the histogram looks like it's around, say, 6? Wait, the x - axis is "Number of passengers who get a red light". The graph is unimodal (has a single peak) and approximately symmetric (or maybe slightly skewed, but the key is that it's unimodal with the peak at around \( np = 6 \)).

Wait, re - examining: The histogram has a single peak (unimodal), and the left and right sides of the peak are roughly balanced (so approximately symmetric), with the peak at the mean of the binomial distribution, which is \( \mu=np = 20\times0.3 = 6 \). Wait, but maybe the peak is at a value like 6? Wait, the description: "The graph is unimodal (has a single peak) with a single peak at around \( np = 6 \) (or the value corresponding to the highest bar in the histogram). The shape is approximately symmetric (or slightly skewed, but the main characteristic is unimodal with the peak in the middle - range of the number of red lights)".

Answer:

(for part a explanation):

• Binary? Yes, each trial (passenger pressing button) has two outcomes: red (success) or green (failure).
• Independent? Yes, one passenger's outcome doesn't affect another's.
• Number? Yes, fixed number of trials (\( n = 20 \)).
• Same probability? Yes, probability of red light (\( p = 0.30 \)) is constant for each trial.

Part (b)