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The probability distribution of \(x\), the number of defective tires on a randomly selected automobile checked at a certain inspection station, is given in the following table: $$ \begin{array}{cccccc} x & 0 & 1 & 2 & 3 & 4 \\ p(x) & .54 & .16 & .06 & .04 & .20 \end{array} $$ a. Calculate the mean value of \(x\). b. What is the probability that \(x\) exceeds its mean value?

Short Answer

Expert verified
The mean value of \(x\) is 1.28, and the probability that \(x\) exceeds its mean value is 0.30 (or 30%).

Step by step solution

01

Calculate the Mean Value

The mean of a probability distribution, also known as the expected value, is calculated by multiplying each outcome by its probability and then summing those products. Therefore, the mean is given by: \[ \mu = \sum x*p(x) \] Substituting the given values into this equation results in: \[ \mu = 0*0.54 + 1*0.16 + 2*0.06 + 3*0.04 + 4*0.20 = 1.28 \]
02

Calculate the Probability of \(x\) Exceeding the Mean

Now it is required to determine the probability that \(x\) exceeds its mean value. This is done by finding all \(x\)’s that are greater than the mean and summing their probabilities. Given the mean as 1.28, the value of \(x\) that exceeds the mean will be 2, 3 and 4. So the probabilities of these values are added together: \[ P(x > \mu) = P(2) + P(3) + P(4) = 0.06 + 0.04 + 0.2 = 0.30 \]
03

Interpret the Results

In conclusion, the mean value of the number of defective tires per automobile is 1.28. Moreover, the probability that the number of defective tires will exceed this mean value is 0.30 (or 30%). These results allow us to understand the expected amount of defects as well as how likely it is to find an automobile with more defects than the average.

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Key Concepts

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

Expected Value
The expected value, also known as the mean of a probability distribution, is the long-term average we anticipate over numerous trials. In probability theory, it's essentially the "center" of the distribution of outcomes. To find the expected value, multiply each possible outcome by its probability, then sum up all these products. This provides a calculated "average" outcome. For example, in the original problem, each number of defective tires (\(x\)) is multiplied by its probability (\(p(x)\)), and these products are summed to give the expected value of 1.28. In the context of quality control, this value informs us about the average number of defects expected per automobile.
Mean Calculation
The mean calculation in a probability distribution involves computing an average that represents the overall data set. This measure helps in understanding the general trend or behavior of the dataset.When calculating the mean from a probability table, such as the one provided, each outcome is weighted by its probability. For example, to calculate the mean number of defective tires, multiply each number of defects (\(x = 0, 1, 2, 3, 4\)) by its respective probability (\(p(x)\)), then add these values together. The specific calculation with values is:- 0 defects at probability 0.54 contributes 0.- 1 defect at probability 0.16 contributes 0.16.- 2 defects at probability 0.06 contributes 0.12.- 3 defects at probability 0.04 contributes 0.12.- 4 defects at probability 0.20 contributes 0.80.Summing these gives the mean: \( \mu = 1.28 \). This analysis shows the typical scenario expected across many observations.
Probability Exceeding Mean
Determining the probability that a random variable exceeds its mean helps in evaluating how often events above average occur. This measure provides insight into the spread of the data and helps to predict extreme or uncommon events. To find this probability, identify outcomes greater than the calculated mean and their associated probabilities. For instance, in the given example, we need the probability of finding more than the mean value of 1.28 defective tires.Outcomes exceeding 1.28 defects are 2, 3, and 4. Each probability is added:- Probability of 2 defects: 0.06- Probability of 3 defects: 0.04- Probability of 4 defects: 0.20The total probability is the sum: \( P(x > 1.28) = 0.30 \). This means there is a 30% chance of having more than 1.28 defective tires, indicating scenarios where more defects occur than the average.
Defective Items Probability
The probability of defective items is essential in fields such as manufacturing, where quality assurance is crucial. Knowing the likelihood of defects helps in planning, minimizing waste, and improving processes.In the context of the problem, each value of \(x\) (defective tires) is assigned a probability, indicating the frequency of occurrence.Examples of these probabilities are:- No defective tires have a probability of 0.54, suggesting it's the most frequent event.- One or more defects have a combined probability of 0.46, pointing to less frequent occurrences but still significant.By understanding this distribution, businesses can focus on reducing occurrences, thereby enhancing product quality.
Probability Theory
Probability theory is the mathematical foundation that allows us to measure uncertainty and predict future outcomes. It's used across various disciplines, from finance to engineering, to make informed decisions based on data trends. The probability distribution, like the one shown, is a crucial concept, depicting how probabilities are spread across different potential outcomes. This helps in visualizing and analyzing data behavior. By applying the principles of probability theory, calculations such as expected value and probabilities of exceeding certain thresholds become possible, offering deeper insights into the data set. Overall, probability theory not only helps in understanding data but also in optimizing processes and mitigating risks related to uncertain future occurrences.

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Most popular questions from this chapter

Components coming off an assembly line are either free of defects \((\mathrm{S}\), for success \()\) or defective \((\mathrm{F}\), for failure). Suppose that \(70 \%\) of all such components are defect-free. Components are independently selected and tested one by one. Let \(y\) denote the number of components that must be tested until a defect-free component is obtained. a. What is the smallest possible \(y\) value, and what experimental outcome gives this \(y\) value? What is the second smallest \(y\) value, and what outcome gives rise to it? b. What is the set of all possible \(y\) values? c. Determine the probability of each of the five smallest \(y\) values. You should see a pattern that leads to a simple formula for \(p(y)\), the probability distribution of \(y\).

The states of Ohio, Iowa, and Idaho are often confused, probably because the names sound so similar. Each year, the State Tourism Directors of these three states drive to a meeting in one of the state capitals to discuss strategies for attracting tourists to their states so that the states will become better known. The location of the meeting is selected at random from the three state capitals. The shortest highway distance from Boise, Idaho to Columbus, Ohio passes through Des Moines, Iowa. The highway distance from Boise to Des Moines is 1350 miles, and the distance from Des Moines to Columbus is 650 miles. Let \(d_{1}\) represent the driving distance from Columbus to the meeting, with \(d_{2}\) and \(d_{3}\) representing the distances from Des Moines and Boise, respectively. a. Find the probability distribution of \(d_{1}\) and display it in a table. b. What is the expected value of \(d_{1} ?\) c. What is the value of the standard deviation of \(d_{1}\) ? d. Consider the probability distributions of \(d_{2}\) and \(d_{3}\). Is either probability distribution the same as the probability distribution of \(d_{1}\) ? Justify your answer. e. Define a new random variable \(t=d_{1}+d_{2}\). Find the probability distribution of \(t\). f. For each of the following statements, indicate if the statement is true or false and provide statistical evidence to support your answer. i. \(E(t)=E\left(d_{1}\right)+E\left(d_{2}\right)\) (Hint: \(E(t)\) is the expected value of \(t\) and another way of denoting the mean of \(t\).) ii. \(\sigma_{t}^{2}=\sigma_{d_{1}}^{2}+\sigma_{d_{3}}^{2}\)

The article on polygraph testing of FBI agents referenced in Exercise \(7.51\) indicated that the probability of a false-positive (a trustworthy person who nonetheless fails the test) is 15 . Let \(x\) be the number of trustworthy FBI agents tested until someone fails the test. a. What is the probability distribution of \(x\) ? b. What is the probability that the first false-positive will occur when the third person is tested? c. What is the probability that fewer than four are tested before the first false-positive occurs? d. What is the probability that more than three agents are tested before the first false-positive occurs?

A breeder of show dogs is interested in the number of female puppies in a litter. If a birth is equally likely to result in a male or a female puppy, give the probability distribution of the variable \(x=\) number of female puppies in a litter of size 5 .

Suppose that a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of Boards 1 and 2 for inspection. a. List the 10 different possible outcomes. b. Suppose that Boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define \(x\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(x\).

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