Problem 15 - Basic Statistics for Business and Economics with Student

Question

Industry standards suggest that 10 percent of new vehicles require warranty service within the first year. Jones Nissan in Sumter, South Carolina, sold 12 Nissans yesterday. a. What is the probability that none of these vehicles requires warranty service? b. What is the probability exactly one of these vehicles requires warranty service? c. Determine the probability that exactly two of these vehicles require warranty service. d. Compute the mean and standard deviation of this probability distribution.

Step-by-Step Solution

Answer

a) 0.2824, b) 0.3766, c) 0.2301, d) Mean: 1.2, Std Dev: 1.039

Step 1: Define the distribution

The problem involves a fixed number of trials (12 vehicles), each with a success probability (needs service) of 0.10. This is a classic setup for a binomial distribution, which applies here: \[ n = 12, \ p = 0.10 \].

Step 2: Probability that none requires service

To find the probability that none of the 12 vehicles require service, we use the binomial probability formula: \[ P(X = 0) = \binom{n}{x} p^x (1-p)^{n-x} \].For \( x = 0 \), the formula becomes: \[ P(X = 0) = \binom{12}{0} (0.10)^0 (0.90)^{12} \].Calculate this to find: \[ P(X = 0) = 1 \times 1 \times 0.90^{12} \approx 0.2824 \].

Step 3: Probability that exactly one requires service

To find the probability that exactly one requires service, set \( x = 1 \) in the binomial probability formula: \[ P(X = 1) = \binom{12}{1} (0.10)^1 (0.90)^{11} \].Calculate this to find: \[ P(X = 1) = 12 \times 0.10 \times 0.90^{11} \approx 0.3766 \].

Step 4: Probability that exactly two vehicles require service

Set \( x = 2 \) in the binomial probability formula: \[ P(X = 2) = \binom{12}{2} (0.10)^2 (0.90)^{10} \].Perform the calculation: \[ P(X = 2) = 66 \times 0.01 \times 0.90^{10} \approx 0.2301 \].

Step 5: Compute the mean of the probability distribution

The mean of a binomial distribution is given by \( \mu = np \). Here, \( n = 12 \) and \( p = 0.10 \), thus: \[ \mu = 12 \times 0.10 = 1.2 \].

Step 6: Compute the standard deviation of the probability distribution

The standard deviation of a binomial distribution is given by \( \sigma = \sqrt{np(1-p)} \). Calculate this using the values: \[ \sigma = \sqrt{12 \times 0.10 \times 0.90} = \sqrt{1.08} \approx 1.039 \].

Key Concepts

Probability Computation

Understanding how to compute the probability of certain outcomes is crucial in the binomial distribution scenario. The binomial probability formula is used here for events with two possible outcomes, like success or failure.
Let's imagine you're asked whether a vehicle needs a warranty service.

• "Success" could mean the vehicle does need service.
• The probability of success, \( p \), is 0.10, as given in the problem

The formula is: \[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \] where

• \( n \) is the total number of trials (vehicles sold)
• \( x \) is the number of successes (vehicles needing service)
• \( p \) is the probability of success
• \( \binom{n}{x} \) represents the number of combinations

Calculating this formula allows us to find specific probabilities, such as: none, exactly one, or exactly two vehicles needing service. The calculations use the probability \( p \) for each vehicle, and adjust for those that do not require the service by using \( (1-p) \). This helps in determining the likelihood of each specific scenario happening.

Mean of Distribution

The mean of a distribution, often referred to as the average, provides insights into the expected outcome over numerous trials. In the context of a binomial distribution, the mean is easily found by multiplying the number of trials \( n \) by the probability of success \( p \). \[ \mu = np \] In our scenario of selling 12 vehicles, with a 10% probability that a single vehicle will need warranty service, we apply: \[ \mu = 12 \times 0.10 = 1.2 \] This result, 1.2, serves as an indicator. It signifies the average expected number of vehicles needing services based on the probability.
The mean provides a useful overview of what might be expected if this scenario occurs many times. It gives a quick snapshot of where the center of the distribution is likely to lie.

Standard Deviation Calculation

The standard deviation is an essential measure that expresses the variability or dispersion of a probability distribution. For a binomial distribution, the formula to compute the standard deviation \( \sigma \) is:\[ \sigma = \sqrt{np(1-p)} \] This takes into account both the probability of an event occurring and not occurring.

• To find this, we plug in the values: \( n = 12 \), \( p = 0.10 \)
• The calculation becomes: \( \sigma = \sqrt{12 \times 0.10 \times 0.90} = \sqrt{1.08} \)
• Upon solving this, we get: \( \sigma \approx 1.039 \)

In practical terms, this means that while we expect about 1.2 vehicles to require warranty service, the number could actually vary by about 1.039 vehicles. This variability helps us understand how spread out or tightly clustered our expectations might be.

Warranty Service Probability

When dealing with probabilities such as vehicles needing warranty service, understanding the context of these probabilities is key. We need to frame our computations in the context of real-life scenarios using likelihoods derived from the binomial distribution.

• The probability that no vehicles require service: Calculated using \( P(X = 0) \), which turns out to be about 0.2824 or 28.24%.
• The probability that exactly one vehicle requires service: Given by \( P(X = 1) \), resulting in approximately 0.3766, or 37.66%.
• For exactly two needing service: \( P(X = 2) \) results in roughly 0.2301, equating to 23.01%.

These values collectively hint at the overall chances you'd face when observing multiple vehicles over time. While exact numbers may vary, these calculated probabilities provide tangible insights into service likely required, empowering better preparation and planning in professional settings like vehicle sales.