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Chapter 7: Problem 28

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: $$ \begin{array}{lrrrrr} x & 0 & 1 & 2 & 3 & 4 \\ p(x) & .54 & .16 & .06 & .04 & .20 \end{array} $$ 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 1.06 and the probability that \(x\) exceeds its mean value is 0.30.

Step by step solution

01

Calculate the mean value

The mean or expected value of a discrete random variable \(x\) can be calculated as the sum of the product of each outcome and its respective probability. This can be mathematically represented as \[E(x) = \sum x \cdot P(x)\] Using the given values, we have: \[E(x) = (0 * .54) + (1 * .16) + (2 * .06) + (3 * .04) + (4 * .20)\]
02

Calculate E(x)

On calculating the above equation, the mean or Expected value, \(E(x)\) is found to be 1.06.
03

Calculate P(x > E(x))

Now that we have the mean value, we can find the probability that \(x\) exceeds its mean value by summing up the probabilities of all outcomes greater than 1.06. Looking at the distribution, the outcomes greater than 1.06 are 2, 3 and 4. So, \[P(x > E(x)) = P(2) + P(3) + P(4) = .06 + .04 + .20\]
04

Calculate P(x > E(x))

Adding the respective probabilities, we find that the probability that \(x\) exceeds its mean value is 0.30.

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Key Concepts

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

Discrete Random Variable
In statistics, a discrete random variable is a type of variable that can take on a finite or countably infinite number of possible values. These values are distinct and separate, unlike continuous random variables that can take on any value within a range. For instance, in the problem we鈥檙e looking at, the variable \(x\) represents the number of defective tires on a vehicle, and it can take on values like 0, 1, 2, 3, or 4.
  • **Nature**: Discrete variables result from counting (e.g., number of defective tires).
  • **Probability**: Each outcome has a probability that can be calculated.
Understanding discrete random variables is crucial because it sets the foundation for probability calculations and understanding distributions.
Mean Value
In probability and statistics, the mean value of a discrete random variable is often referred to as its expected value. It is a measure of central tendency, which gives us an idea about the average value of the random variable in the long run.The formula for calculating the mean value (expected value \(E(x)\)) of a discrete random variable is:\[E(x) = \sum x \cdot P(x)\]
  • **Interpretation**: Represents the long-term average or "expected" result if the experiment were repeated many times.
  • **Calculation Example**: In our problem, the mean value was calculated as follows: - \((0 \times 0.54) + (1 \times 0.16) + (2 \times 0.06) + (3 \times 0.04) + (4 \times 0.20) = 1.06\)
The value 1.06 suggests that on average, slightly more than one tire per car is expected to be defective.
Expected Value
The expected value is an essential concept in the field of probability. It tells us what average outcome we can expect if we could repeat an experiment a large number of times.To calculate the expected value of a discrete random variable, use this simple process:1. Multiply each possible outcome by its probability.2. Sum all those products.For our scenario:
  • **Step 1:** Multiply each possible value of \(x\) (defective tires) by its probability \(P(x)\).
  • **Step 2:** Add these results to find \(E(x)\).
This result shows us the average number of defects, helping in decision-making about the inspection process or further quality control measures.
Probability Calculation
Probability calculation involves determining how likely it is for a particular event to occur. In our context, after we determined the mean value, we focused on finding probabilities related to how \(x\) exceeds this mean.For example, to find the probability that \(x\) (number of defects) is more than 1.06, we added the probabilities of all outcomes greater than the mean:\[P(x > 1.06) = P(2) + P(3) + P(4)\]Using our exercise:
  • **Outcome Probabilities**: \(P(2) = 0.06\), \(P(3) = 0.04\), and \(P(4) = 0.20\).
  • **Total Probability**: This sums up to 0.30, indicating a 30% chance that more than one tire per car will be defective.
Such probability calculations are critical for assessing risks and making informed decisions in business and engineering applications.

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Most popular questions from this chapter

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean \(45 \mathrm{~min}\) and standard deviation \(5 \mathrm{~min}\). a. If \(50 \mathrm{~min}\) is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if we wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all students to complete the exam?

Suppose that \(90 \%\) of all registered California voters favor banning the release of information from exit polls in presidential elections until after the polls in California close. A random sample of 25 California voters is to be selected. a. What is the probability that more than 20 voters favor the ban? b. What is the probability that at least 20 voters favor the ban? c. What are the mean value and standard deviation of the number of voters who favor the ban? d. If fewer than 20 voters in the sample favor the ban, is this at odds with the assertion that (at least) \(90 \%\) of the populace favors the ban? (Hint: Consider \(P(x<20)\) when \(\pi=.9 .)\)

In a study of warp breakage during the weaving of fabric (Technometrics [1982]: 63), 100 pieces of yarn were tested. The number of cycles of strain to breakage was recorded for each yarn sample. The resulting data are given in the following table: $$ \begin{array}{rrrrrrrrrr} 86 & 146 & 251 & 653 & 98 & 249 & 400 & 292 & 131 & 176 \\ 76 & 264 & 15 & 364 & 195 & 262 & 88 & 264 & 42 & 321 \\ 180 & 198 & 38 & 20 & 61 & 121 & 282 & 180 & 325 & 250 \\ 196 & 90 & 229 & 166 & 38 & 337 & 341 & 40 & 40 & 135 \\ 597 & 246 & 211 & 180 & 93 & 571 & 124 & 279 & 81 & 186 \\ 497 & 182 & 423 & 185 & 338 & 290 & 398 & 71 & 246 & 185 \\ 188 & 568 & 55 & 244 & 20 & 284 & 93 & 396 & 203 & 829 \\ 239 & 236 & 277 & 143 & 198 & 264 & 105 & 203 & 124 & 137 \\ 135 & 169 & 157 & 224 & 65 & 315 & 229 & 55 & 286 & 350 \\ 193 & 175 & 220 & 149 & 151 & 353 & 400 & 61 & 194 & 188 \end{array} $$ a. Construct a frequency distribution using the class intervals 0 to \(<100,100\) to \(<200\), and so on. b. Draw the histogram corresponding to the frequency distribution in Part (a). How would you describe the shape of this histogram? c. Find a transformation for these data that results in a more symmetric histogram than what you obtained in Part (b).

You are to take a multiple-choice exam consisting of 100 questions with 5 possible responses to each question. Suppose that you have not studied and so must guess (select one of the five answers in a completely random fashion) on each question. Let \(x\) represent the number of correct responses on the test. a. What kind of probability distribution does \(x\) have? b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the \(x\) distribution.) c. Compute the variance and standard deviation of \(x\). d. Based on your answers to Parts (b) and (c), is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

The Los Angeles Times (December 13,1992 ) reported that what airline passengers like to do most on long flights is rest or sleep; in a survey of 3697 passengers, almost \(80 \%\) did so. Suppose that for a particular route the actual percentage is exactly \(80 \%\), and consider randomly selecting six passengers. Then \(x\), the number among the selected six who rested or slept, is a binomial random variable with \(n=6\) and \(\pi=.8\). a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6)\), the probability that all six selected passengers rested or slept. c. Determine \(P(x \geq 4)\).
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