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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

Expert verified
Expected defects: 0.64; Standard deviation: 0.933.

Step by step solution

01

Define the Random Variable

Let the random variable \(X\) represent the number of appearance defects in a car. The possible values of \(X\) are 0, 1, 2, and 3, corresponding to the number of defects.
02

Provide the Probability Distribution

Based on the problem statement, the given probabilities are:\[ P(X = 1) = 0.21, \quad P(X = 2) = 0.11, \quad P(X = 3) = 0.07 \]The remaining probability is for \(P(X = 0)\), which can be calculated as: \[ P(X = 0) = 1 - (0.21 + 0.11 + 0.07) = 0.61 \].
03

Calculate the Expected Value

The expected value \(E(X)\) is calculated using the formula:\[ E(X) = \sum_{i=0}^{3} x_i P(X = x_i) = 0 \times 0.61 + 1 \times 0.21 + 2 \times 0.11 + 3 \times 0.07 \]Calculating this gives:\[ E(X) = 0 + 0.21 + 0.22 + 0.21 = 0.64 \]
04

Calculate the Variance

The variance \(Var(X)\) is calculated using:\[ Var(X) = E(X^2) - (E(X))^2 \]First, find \(E(X^2)\) using:\[ E(X^2) = 0^2 \times 0.61 + 1^2 \times 0.21 + 2^2 \times 0.11 + 3^2 \times 0.07 = 0 + 0.21 + 0.44 + 0.63 = 1.28 \]Then,\[ Var(X) = 1.28 - (0.64)^2 = 1.28 - 0.4096 = 0.8704 \]
05

Calculate the Standard Deviation

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variables
In probability and statistics, a random variable is a numerical description of the outcome of a random phenomenon. Think of it as a box where each possible outcome of an experiment or situation is stuffed with its respective value. Random variables can be discrete or continuous.

For the cars with appearance defects, we have a discrete random variable, denoted by \( X \). This variable stands for the number of defects a new car might have, with possible values of 0, 1, 2, or 3. Discrete means it can only take certain fixed values.
  • Discrete Random Variables: Take on a countable number of distinct values, like in our car defect example.
  • Continuous Random Variables: Can take on an infinite number of values within a range, such as heights or temperatures.
Understanding the type of random variable you're working with and the possible outcomes it can have is crucial for determining probabilities and other statistical measures.
Expected Value
Expected value is a fundamental concept in probability, describing the average outcome if you repeated an experiment numerous times. It's like the mean of a probability distribution and provides insight into the "center" or "typical" value that one would expect.

To find the expected number of defects in our situation, we use this formula:\[ E(X) = \sum x_i P(X = x_i) \]What we're doing here is simply summing up the products of each possible outcome (number of defects) and their probabilities. In this problem, we calculated the expected value as 0.64. This value tells us that, on average, a car will have about 0.64 defects, which helps in understanding and predicting manufacturing quality.
  • Intuition: Expected value shows us the mean or the 'center' of a probability distribution.
  • Application: Useful in a wide range of scenarios, from gambling to stock market predictions to everyday decision-making.
Standard Deviation
Standard deviation is a measure of spread or variability within a set of data, tidal in understanding how spread out the outcomes are around the mean or expected value.

In our exercise, after finding variance, we use standard deviation to quantify how much the number of defects deviates from the average number of defects. We calculate it as:\[ \sigma = \sqrt{Var(X)} \]The standard deviation of approximately 0.933 tells us the degree of variance one might witness in actual scenarios compared to the 0.64 average defects. If each car deviated minimally, it would mean most cars have the expected average defects. But, with a greater standard deviation, defect count differs more from the average. This helps in planning and quality control testing in production.
  • Key Insight: It gives a sense of what range the values typically fall within, which is crucial for statistical consistency.
  • Practical Uses: In assessing risks, quality control, finance, and various fields needing understanding of variability.
Variance
Variance measures the spread of a set of numbers. It tells us how far each number in the dataset is from the mean. While related to the standard deviation, variance offers different insights into data variability.

In the car defects scenario, variance is calculated first to help find the standard deviation. We determine it by calculating the expected value of the squared deviations from the expected value. The formula is:\[ Var(X) = E(X^2) - (E(X))^2 \]Here we found a variance of approximately 0.8704. This denotes how much the number of defects deviates from the mean on average squared. The square of these deviations gives us a clearer image about how data points deviate from the expected value.
  • Understanding Deviations: Variance is a depiction of the data's spread. Higher variance means data points are more spread out from the mean.
  • Context of Use: Helps in comparing datasets, risk assessment, and variance analysis in data science.

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Most popular questions from this chapter

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