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Chapter 6: Problem 108

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: \(\begin{array}{lccccc}x & 0 & 1 & 2 & 3 & 4 \\ p(x) & 0.54 & 0.16 & 0.06 & 0.04 & 0.20\end{array}\) a. Calculate the mean value of \(x\). b. Interpret the mean value of \(x\) in the context of a long sequence of observations of number of defective tires. c. What is the probability that \(x\) exceeds its mean value? d. Calculate the standard deviation of \(x\).

Short Answer

Expert verified
The mean value of \(x\) is 1.30, which represents the average number of defective tires on randomly selected automobiles checked at the inspection station. The probability that \(x\) exceeds its mean value is 0.30. The standard deviation of \(x\) is approximately 1.51.

Step by step solution

01

Calculate the mean value of \(x\)

To calculate the mean value of \(x\), we need to first find the expected value, given by: \(E(x) = \sum_{i}{x_i \cdot p(x_i)}\) Where \(x_i\) is each value of \(x\), and \(p(x_i)\) is the probability of that value. Using the given probabilities: \(E(x) = (0 \cdot 0.54) + (1 \cdot 0.16) + (2 \cdot 0.06) + (3 \cdot 0.04) + (4 \cdot 0.20)\) Now, calculate the expected value: \(E(x) = 0 + 0.16 + 0.12 + 0.12 + 0.80\) \(E(x) = 1.30\) The mean value of \(x\) is 1.30.
02

Interpret the mean value of \(x\)

The mean value calculated in Step 1 represents the average number of defective tires on randomly selected automobiles checked at the inspection station. In the context of a long sequence of observations of the number of defective tires, the mean value of 1.30 defective tires indicates that, on average, an automobile checked at the inspection station is expected to have 1.30 defective tires.
03

Calculate the probability that \(x\) exceeds its mean value

To find the probability that \(x\) exceeds its mean value, we need to find the probability of \(x\) being greater than 1.30. Since \(x\) can only take integer values, we need to find the probability of \(x = 2\), \(x = 3\), and \(x = 4\). We can find this by adding the probabilities of each of these values: \(P(x > 1.30) = P(x = 2) + P(x = 3) + P(x = 4)\) Using the given probabilities: \(P(x > 1.30) = 0.06 + 0.04 + 0.20\) \(P(x > 1.30) = 0.30\) The probability that \(x\) exceeds its mean value is 0.30.
04

Calculate the standard deviation of \(x\)

To calculate the standard deviation of \(x\), we first need to find the variance, given by: \(\operatorname{Var}(x) = E(x^2) - (E(x))^2\) We need to calculate \(E(x^2)\), which is the expected value of the square of \(x\), using the given probabilities: \(E(x^2) = (0^2 \cdot 0.54) + (1^2 \cdot 0.16) + (2^2 \cdot 0.06) + (3^2 \cdot 0.04) + (4^2 \cdot 0.20)\) \(E(x^2) = 0 + 0.16 + 0.24 + 0.36 + 3.20\) \(E(x^2) = 3.96\) Now, we can find the variance: \(\operatorname{Var}(x) = 3.96 - (1.30)^2\) \(\operatorname{Var}(x) = 3.96 - 1.69\) \(\operatorname{Var}(x) = 2.27\) Finally, we can find the standard deviation by taking the square root of the variance: \(\sigma_x = \sqrt{2.27}\) \(\sigma_x \approx 1.51\) The standard deviation of \(x\) is approximately 1.51.

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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 is a fundamental concept in probability and statistics. It represents the long-term average or "mean" that we can expect from a random variable over many trials or experiments. To compute the expected value, you will multiply each possible outcome by its probability and then sum these products. This gives you a weighted average of the possible outcomes based on their likelihood of occurring. In the exercise above, we calculated the expected value of the number of defective tires on a randomly selected car as follows:
  • Multiply each number of defective tires (0, 1, 2, 3, and 4) by their respective probabilities.
  • Sum up these products, which results in the expected value.
The expected number of defective tires is 1.30 per car, meaning if you inspected many cars, on average, each car would have about 1.30 defective tires over the long term.
Standard Deviation
Standard deviation is a measure of how spread out the values in a data set are around the mean. It gives us an idea of the variability or dispersion of the values. To calculate the standard deviation, you first need to find the variance, which is the average of the squared differences from the mean. Through the exercise, here is how we approached it:
  • First, determine the expected value of the square of the random variable, denoted as \(E(x^2)\).
  • Then, find the variance using the formula \(\operatorname{Var}(x) = E(x^2) - (E(x))^2\).
  • The final step is to take the square root of the variance to find the standard deviation, \(\sigma\).
In the example given, our variance was calculated to be 2.27, and so the standard deviation ended up approximately being 1.51. This means that the number of defective tires typically varies by about 1.51 from the average of 1.30 tires per car.
Mean Interpretation
Understanding and interpreting the mean is crucial in statistics, especially when dealing with probability distributions. The mean provides insights into the central tendency of a random variable, which, in this case, is the average number of defective tires on a car. When interpreting the mean:
  • It helps to envision it over a long series of observations. For instance, if many cars are inspected, the average number of defective tires per car will hover around the mean value.
  • In this scenario, the mean of 1.30 defective tires suggests that, although individual cars may have different numbers of defective tires, the average across many cars is expected to be 1.30.
  • It does not imply that a car "can" have 1.30 defective tires, but rather that 1.30 is an average resulting from observing many cars.
Thus, the mean is a theoretical construct that helps us summarize the distribution of a random variable succinctly.

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Most popular questions from this chapter

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean of 120 seconds and a standard deviation of 20 seconds. The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

The Wall Street Journal (February 15,1972 ) reported that General Electric was sued in Texas for sex discrimination over a minimum height requirement of 5 feet, 7 inches. The suit claimed that this restriction eliminated more than \(94 \%\) of adult females from consideration. Let \(x\) represent the height of a randomly selected adult woman. Suppose that \(x\) is approximately normally distributed with mean 66 inches (5 ft. 6 in.) and standard deviation 2 inches. a. Is the claim that \(94 \%\) of all women are shorter than \(5 \mathrm{ft}\). 7 in. correct? b. What proportion of adult women would be excluded from employment as a result of the height restriction?

The paper referenced in the previous exercise also included data on left atrial diameter for children who were considered overweight. For these children, left atrial diameter was approximately normally distributed with a mean of \(28 \mathrm{~mm}\) and a standard deviation of \(4.7 \mathrm{~mm}\). a. Approximately what proportion of overweight children have left atrial diameters less than \(25 \mathrm{~mm}\) ? b. Approximately what proportion of overweight children have left atrial diameters greater than \(32 \mathrm{~mm} ?\) c. Approximately what proportion of overweight children have left atrial diameters between 25 and \(30 \mathrm{~mm}\) ? d. What proportion of overweight children have left atrial diameters greater than the mean for healthy children?

Consider the random variable \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose the probability distribution of \(y\) is as follows: \(\begin{array}{cccccc}y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ?\end{array}\) a. Only \(y\) values of \(0,1,2,3,\) and 4 have probabilities greater than \(0 .\) What is \(p(4) ?\) b. How would you interpret \(p(1)=0.20 ?\) c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. d. Calculate \(P(y<2),\) the probability that the carton contains fewer than two broken eggs. Why is this smaller than the probability in Part (c)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

Classify each of the following random variables as either discrete or continuous: a. The fuel efficiency (miles per gallon) of an automobile b. The amount of rainfall at a particular location during the next year c. The distance that a person throws a baseball d. The number of questions asked during a 1-hour lecture e. The tension (in pounds per square inch) at which a tennis racket is strung f. The amount of water used by a household during a given month g. The number of traffic citations issued by the highway patrol in a particular county on a given day
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