Question

Defects. A consumer organization inspecting new cars found that many had appearance defects (dents, scratches, paint chips, etc.). While none had more than three of these defects, \(7 \%\) had three, \(11 \%\) two, and \(21 \%\) one defect. Find the expected number of appearance defects in a new car and the standard deviation.

Short answer

The expected number of defects is 0.64, and the standard deviation is approximately 0.933.

Step-by-step solution

Introduce the Random Variable

Define the random variable \(X\) as the number of defects in a new car. The variable \(X\) can take the values 0, 1, 2, or 3 defects.

State the Probabilities

List the probabilities of each number of defects:- \(P(X = 0) = 1 - (0.07 + 0.11 + 0.21) = 0.61\) (since missing probability must be 61%)- \(P(X = 1) = 0.21\)- \(P(X = 2) = 0.11\)- \(P(X = 3) = 0.07\)

Calculate the Expected Value

The expected number of defects, \(E(X)\), is calculated using the formula:\[E(X) = \sum_{i=0}^{3} x_i \, P(X = x_i) = 0 \times 0.61 + 1 \times 0.21 + 2 \times 0.11 + 3 \times 0.07\]Evaluate this to find:\[E(X) = 0 + 0.21 + 0.22 + 0.21 = 0.64\]

Calculate the Variance

The variance \(\text{Var}(X)\) is calculated using the formula:\[\text{Var}(X) = E(X^2) - (E(X))^2\]First, calculate \(E(X^2)\):\[E(X^2) = 0^2 \times 0.61 + 1^2 \times 0.21 + 2^2 \times 0.11 + 3^2 \times 0.07 = 0 + 0.21 + 0.44 + 0.63 = 1.28\]Then, \[\text{Var}(X) = 1.28 - (0.64)^2 = 1.28 - 0.4096 = 0.8704\]

Calculate the Standard Deviation

The standard deviation is the square root of the variance:\[\text{SD}(X) = \sqrt{0.8704} \approx 0.933\]

Key concepts

Random Variable

A random variable is a fundamental concept in probability and statistics that represents a set of possible outcomes in a random process. In this context, when analyzing defects in new cars, the random variable, denoted as \( X \), represents the number of appearance defects a car can have. Here, \( X \) can take on specific values: 0, 1, 2, or 3 defects.

Each value of \( X \) corresponds to a different possible state of the car regarding defects.

• 0 defects means the car is perfect in appearance.
• 1 defect indicates a minor issue like a single scratch.
• 2 defects suggest more noticeable problems.
• 3 defects represent the maximum number of issues observed.

Understanding the concept of a random variable is essential to tackle problems involving uncertainty and to compute other statistics like the expected value and standard deviation.

Probability Distribution

Probability distribution is a crucial statistical function that outlines the likelihood of different outcomes for a random variable. For the defects in our example, the distribution defines the probability of a car having 0, 1, 2, or 3 defects.

The probabilities provided are:

• \( P(X = 0) \) — Found by subtracting the sum of other probabilities from 1, resulting in 61% chance of zero defects.
• \( P(X = 1) = 0.21 \) — There is a 21% chance of exactly one defect.
• \( P(X = 2) = 0.11 \) — An 11% chance for two defects.
• \( P(X = 3) = 0.07 \) — A 7% likelihood of three defects.

Probability distribution helps us manage uncertainty by understanding and visualizing the possible outcomes and their associated likelihoods. This understanding serves as the basis for calculating more complex measures, such as the expected value.

Standard Deviation

Standard deviation is a measure of the dispersion or spread of a set of values for a random variable. It provides insight into how much the values deviate from the expected value, on average. In the context of the given exercise, the standard deviation of defects helps us understand the variability in the number of defects across different cars.

To find the standard deviation, we start by calculating the variance, which is the average of the squared differences from the mean expected value: \[\text{Var}(X) = E(X^2) - (E(X))^2 \]Here, \( E(X^2) \) is the expected value of the square of the number of defects. Once variance is computed, the standard deviation \( \text{SD}(X) \) is derived by taking the square root of the variance:\[ \text{SD}(X) = \sqrt{\text{Var}(X)} \approx 0.933 \] The standard deviation helps in assessing the risk or variability from the average, hence offering a clearer picture of how consistent the car defects are in distribution, whether most cars are indeed defect-free or if a significant number have multiple defects.