Problem 24 - Basic Statistics for Business & Economics

Question

It is asserted that \(80 \%\) of the cars approaching an individual toll booth in New Jersey are equipped with an E-ZPass transponder. Find the probability that in a sample of six cars: a. All six will have the transponder. b. At least three will have the transponder. c. None will have a transponder.

Step-by-Step Solution

Answer

a. 0.262144 b. 0.983040 c. 0.000064

Step 1: Identify the Problem Type

The problem is asking for probabilities concerning a binomial distribution scenario. You have a fixed number of trials (sampling 6 cars), two possible outcomes (having or not having an E-ZPass transponder), and a constant probability of success (car having a transponder).

Step 2: Define Key Variables

Let X be the number of cars with a transponder in a sample of six cars. The number of trials (n) is 6, probability of success (p) is 0.8, and number of successes (k) will vary based on the question part.

Step 3: Formula Utilization: Binomial Probability

Use the binomial probability formula for finding the probability of exactly k successes: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

Step 4: Part A: Probability All Have Transponder

To find the probability that all six cars have a transponder (k=6), apply the binomial formula:\[ P(X = 6) = \binom{6}{6} (0.8)^6 (0.2)^0 \]This simplifies to:\[ P(X = 6) = 1 \times (0.8)^6 = 0.262144 \]

Step 5: Part B: Probability of At Least Three

For at least three cars having a transponder, calculate:\[ P(X \geq 3) = 1 - P(X < 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) \]. Evaluate these probabilities using the binomial formula:- \( P(X = 0) = \binom{6}{0} (0.8)^0 (0.2)^6 = 0.000064 \)- \( P(X = 1) = \binom{6}{1} (0.8)^1 (0.2)^5 = 0.001536 \)- \( P(X = 2) = \binom{6}{2} (0.8)^2 (0.2)^4 = 0.015360 \)Add these probabilities:\( P(X < 3) = 0.000064 + 0.001536 + 0.015360 = 0.016960 \)Thus, \( P(X \geq 3) = 1 - 0.016960 = 0.983040 \)

Step 6: Part C: Probability None Have Transponder

Calculate the probability for none (k=0) having a transponder:\[ P(X = 0) = \binom{6}{0} (0.8)^0 (0.2)^6 = 0.000064 \]

Key Concepts

Probability Calculation

The essence of probability calculation is determining the likelihood of an event occurring. In the context of our exercise with toll booth cars, probability helps us predict outcomes based on given conditions.
Since only two outcomes exist—either a car has a transponder or it doesn't—the situation is ideal for binomial distribution.

When calculating probabilities, it is crucial to identify variables. Here, the number of trials is the total cars sampled, which is six. We also need the probability of success, or the probability that a car has a transponder, given as 80% or 0.8.
This groundwork allows us to apply specific formulas that predict the number of cars with transponders in different scenarios.

Binomial Probability Formula

The binomial probability formula is a powerful tool in statistical problem solving for cases with two possible outcomes. The formula used in our toll booth car scenario is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Let's break this down:

• \(n\) is the total number of trials, representing the six cars.
• \(k\) is the number of successes, which changes based on what we're looking for (e.g., all cars having transponders).
• \(p\) is the probability of a car having a transponder (0.8).
• \((1-p)\) is the probability of not having a transponder (0.2).

Using the formula, you can compute probabilities for different numbers of successes. For example, in Part A of our problem, we set \(k = 6\) to find the probability that all cars have transponders. Calculating the formula gives us a clear probability result that fits the scenario on hand.

Statistical Problem Solving

Statistical problem solving involves systematically working through challenges using math and logic. In exercises like this, start by identifying the type of statistical method to apply—in our case, binomial distribution.
Once identified, the next step is setting up the problem using key questions or parts outlined clearly, like probabilities for all cars with transponders, at least some, and none at all. It’s crucial to break things into small parts for each occurrence and understand how cumulative probabilities or individual calculations fit together.

Using our example, solving for \(P(X \geq 3)\) involved understanding the complementary probability.
Instead of calculating directly, we found \(P(X < 3)\) and subtracted it from 1. This approach is efficient, avoids repetitive calculations, and utilizes properties of probabilities effectively.
Statistical problem solving isn't just about reaching the answer but also understanding and interpreting each step in the context of real-world scenarios.