Problem 15 - Basic Statistics for Business & Economics

Question

Industry standards suggest that \(10 \%\) 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, standard deviation ≈ 1.039.

Step 1: Identify Problem Type and Parameters

This problem can be modeled using a binomial distribution. We identify that the probability of a single car requiring warranty service, denoted by \( p \), is 0.10. The number of cars sold, denoted by \( n \), is 12. For each part of the problem, we are asked to find the probability of \( x \) vehicles requiring service.

Step 2: Calculate Probability for Zero Vehicles

We use the binomial probability formula: \( P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \). For part a, where \( x = 0 \), compute \( P(X = 0) = \binom{12}{0} (0.10)^0 (0.90)^{12} \). This simplifies to \( (0.90)^{12} \). Calculate to find the probability.

Step 3: Compute Probability for One Vehicle

For part b, where \( x = 1 \), use the formula \( P(X = 1) = \binom{12}{1} (0.10)^1 (0.90)^{11} \). This becomes \( 12 \times 0.10 \times (0.90)^{11} \). Calculate this result to find the probability.

Step 4: Determine Probability for Two Vehicles

For part c, where \( x = 2 \), use \( P(X = 2) = \binom{12}{2} (0.10)^2 (0.90)^{10} \). Calculate \( \binom{12}{2} = 66 \) and thus \( P(X = 2) = 66 \times (0.10)^2 \times (0.90)^{10} \). Compute this value.

Step 5: Calculate Mean and Standard Deviation

For a binomial distribution, the mean \( \mu \) is given by \( np \), and the standard deviation \( \sigma \) is given by \( \sqrt{np(1-p)} \). Calculate \( \mu = 12 \times 0.10 = 1.2 \) and \( \sigma = \sqrt{12 \times 0.10 \times 0.90} \). Compute \( \sigma \) to obtain the result.

Key Concepts

Probability

Probability is a measure of how likely an event is to occur. In the context of a binomial distribution, probability helps us understand the chance of a certain number of events happening out of a series of independent trials. Here, our event is a warranty service being needed for a vehicle.
For example, to find the probability that none of the cars require warranty service, we use the binomial probability formula: - \[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \] -
Where: - - \( n \) is the number of trials (vehicles sold) - \( x \) is the number of successes (vehicles needing service) - \( p \) is the probability of success in each trial (0.10 here)
To find \( P(X = 0) \), we plug in the values and calculate the probability of none of the vehicles requiring service, which simplifies to \( (0.90)^{12} \) because none of the cars needing service is the same as all running smoothly.

Mean

The mean, often referred to as the expected value, gives us the average outcome of a random variable over many trials.
For a binomial distribution, the mean \( \mu \) is calculated using the formula: - \[ \mu = np \]
In this problem, \( n \) (the number of trials) is 12, and \( p \) (probability of a vehicle needing service) is 0.10. Plugging these into our formula gives us: - \[ \mu = 12 \times 0.10 = 1.2 \]
This means that, on average, 1.2 cars out of every 12 sold are expected to need warranty service within the first year. It's a way of understanding the *typical* number of vehicles you might predict to require service.

Standard Deviation

Standard deviation is a measure of the spread or dispersion of a set of values. In a binomial distribution, it helps us understand how much variation exists from the mean.
The standard deviation \( \sigma \) in a binomial distribution is computed as: - \[ \sigma = \sqrt{np(1-p)} \]
With our given values, where \( n = 12 \) and \( p = 0.10 \), we find: - \[ \sigma = \sqrt{12 \times 0.10 \times 0.90} \]
This calculation tells us about the typical deviation from the average number of cars needing service. A smaller standard deviation would mean the number of cars requiring service is closer to the expected average, whereas a larger one shows more variability in the service numbers.

Warranty Service

The term "warranty service" refers to repairs or maintenance provided by the manufacturer at no additional charge to the owner within a specific time period. In this context, it represents a service needed because of a defect or issue encountered in the vehicle within the first year.
Understanding the probability of a car needing warranty service is crucial for dealerships and manufacturers. It helps them prepare for future service needs and allocate resources accordingly. When a new car is sold, knowing that 10% typically require attention within a year allows businesses to plan staffing and part supplies.
It also reassures potential buyers about the reliability of their purchase, showing a company's confidence in its product. Moreover, calculating how often service is needed can inform decisions on warranty terms and policies, ensuring they meet consumer expectations and maintain customer satisfaction.