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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

A mail-order computer software business has six telephone lines. Let \(x\) denote the number of lines in use at a specified time. The probability distribution of \(x\) is as follows: $$ \begin{array}{lrrrrrrr} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & .10 & .15 & .20 & .25 & .20 & .06 & .04 \end{array} $$ Write each of the following events in terms of \(x\), and then calculate the probability of each one: a. At most three lines are in use b. Fewer than three lines are in use c. At least three lines are in use d. Between two and five lines (inclusive) are in use e. Between two and four lines (inclusive) are not in use f. At least four lines are not in use

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuality, each one could arrive on Wednesday (W), Thursday (T), Friday (F), or Saturday (S). Suppose that the two magazines arrive independently of one another and that for each magazine \(P(\mathrm{~W})=.4, P(\mathrm{~T})=.3\), \(P(\mathrm{~F})=.2\), and \(P(\mathrm{~S})=.1\). Define a random variable \(y\) by \(y=\) the number of days beyond Wednesday that it takes for both magazines to arrive. For example, if the first magazine arrives on Friday and the second magazine arrives on Wednesday, then \(y=2\), whereas \(y=1\) if both magazines arrive on Thursday. Obtain the probability distribution of \(y\). (Hint: Draw a tree diagram with two generations of branches, the first labeled with arrival days for Magazine 1 and the second for Magazine 2.)

The lifetime of a certain brand of battery is normally distributed with a mean value of \(6 \mathrm{hr}\) and a standard deviation of \(0.8 \mathrm{hr}\) when it is used in a particular cassette player. Suppose that two new batteries are independently selected and put into the player. The player ceases to function as soon as one of the batteries fails. a. What is the probability that the player functions for at least \(4 \mathrm{hr}\) ? b. What is the probability that the cassette player works for at most \(7 \mathrm{hr}\) ? c. Find a number \(z^{*}\) such that only \(5 \%\) of all cassette players will function without battery replacement for more \(\operatorname{than} z^{*} \mathrm{hr}\)

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 value of \(120 \mathrm{sec}\) and a standard deviation of \(20 \mathrm{sec}\). The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?

Because \(P(z<.44)=.67,67 \%\) of all \(z\) values are less than \(.44\), and \(.44\) is the 67 th percentile of the standard normal distribution. Determine the value of each of the following percentiles for the standard normal distribution (Hint: If the cumulative area that you must look for does not appear in the \(z\) table, use the closest entry): a. The 91 st percentile (Hint: Look for area \(.9100 .\) ) b. The 77 th percentile c. The 50 th percentile d. The 9 th percentile e. What is the relationship between the 70 th \(z\) percentile and the 30 th \(z\) percentile?
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