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The random variable \(x\) has the discrete probability distribution shown here: \begin{tabular}{l|lllll} \hline\(x\) & -2 & -1 & 0 & 1 & 2 \\ \(p(x)\) & .10 & .15 & .40 & .30 & .05 \\ \hline \end{tabular} a. Find \(P(x \leq 0)\). b. Find \(P(x>-1)\) c. Find \(P(-1 \leq x \leq 1)\). d. Find \(P(x<2)\). e. Find \(P(-1

Short Answer

Expert verified
a. 0.65, b. 0.75, c. 0.85, d. 0.95, e. 0.70, f. 0.65

Step by step solution

01

Understanding the Distribution

We have a discrete probability distribution for a random variable \( x \) with the possible values \(-2, -1, 0, 1,\) and \(2\). Each value has a corresponding probability \( p(x) \). To solve the problems, we'll look at the probability associated with these values as we make comparisons.
02

Calculate \( P(x \leq 0) \)

To find \( P(x \leq 0) \), add the probabilities of \( x = -2, -1, \) and \( 0 \). So, \( P(x \leq 0) = P(x = -2) + P(x = -1) + P(x = 0) = 0.10 + 0.15 + 0.40 = 0.65 \).
03

Calculate \( P(x > -1) \)

To find \( P(x > -1) \), consider the probabilities of \( x = 0, 1,\) and \(2\). So, \( P(x > -1) = P(x = 0) + P(x = 1) + P(x = 2) = 0.40 + 0.30 + 0.05 = 0.75 \).
04

Calculate \( P(-1 \leq x \leq 1) \)

For this range, add the probabilities of \( x = -1, 0,\) and \(1\). Therefore, \( P(-1 \leq x \leq 1) = P(x = -1) + P(x = 0) + P(x = 1) = 0.15 + 0.40 + 0.30 = 0.85 \).
05

Calculate \( P(x < 2) \)

To find \( P(x < 2) \), add the probabilities for \( x = -2, -1, 0,\) and \(1\). Thus, \( P(x < 2) = P(x = -2) + P(x = -1) + P(x = 0) + P(x = 1) = 0.10 + 0.15 + 0.40 + 0.30 = 0.95 \).
06

Calculate \( P(-1 < x < 2) \)

For \( -1 < x < 2 \), the relevant values are \( x = 0 \) and \( x = 1 \). Therefore, \( P(-1 < x < 2) = P(x = 0) + P(x = 1) = 0.40 + 0.30 = 0.70 \).
07

Calculate \( P(x < 1) \)

To find \( P(x < 1) \), consider \( x = -2, -1,\) and \(0\). Thus, \( P(x < 1) = P(x = -2) + P(x = -1) + P(x = 0) = 0.10 + 0.15 + 0.40 = 0.65 \).

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Key Concepts

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

Random Variable
A random variable is a fundamental concept in probability theory, representing a numerical outcome of a random process. In simple terms, it is a variable that can take on different values where each possible value is associated with a probability. This randomness makes it distinct from regular variables used in algebra, as its values are determined by the outcomes of an experiment or event.

In discrete probability distributions, the random variable is discrete, meaning it can only take on specific, separated values. In our example, the random variable \( x \) can take on any of the following values: \(-2, -1, 0, 1,\) or \( 2 \). Each of these values corresponds to a probability value \( p(x) \), which tells us how likely the random variable is to take on that particular value.
  • Discrete values like \(-2, -1, 0, 1,\) and \( 2 \) help us define the possible outcomes.
  • The probabilities must sum to 1 because one of these outcomes must occur.
This forms the backbone for computations in probability theory, helping us calculate the likelihood of different events happening.
Probability Calculation
Probability calculation in the context of discrete distributions involves adding up probabilities for specific outcomes or ranges of outcomes. This is done by summing the probabilities associated with each value of the random variable that satisfies the given condition.

For example, to calculate \( P(x \leq 0) \), we sum the probabilities of \( x = -2, -1, \) and \( 0 \):
  • \( P(x = -2) = 0.10 \)
  • \( P(x = -1) = 0.15 \)
  • \( P(x = 0) = 0.40 \)
Adding these up gives \( 0.10 + 0.15 + 0.40 = 0.65 \).
This example shows that by calculating probability, you determine how likely it is for the random variable to fall within a particular range or meet a certain condition, which in turn helps predict the occurrence of various outcomes.
Probability Theory
Probability theory involves the mathematical study of uncertainty and patterns. It is a scientific way to quantify the likelihood of events occurring. The theory encompasses fundamental principles like probability distributions, random variables, and event occurrences.

A key aspect of probability theory is understanding how probabilities are derived and interpreted. Taking our discrete probability distribution as an example:
  • All probabilities need to be non-negative values.
  • The total sum of probabilities for all possible values of a random variable equals 1, illustrating certainty that one of the values will occur.
Probability theory aids in making informed predictions about random events. This can apply to a wide array of fields including finance, engineering, and the sciences, where it helps to model data and make decisions under uncertainty.
Discrete Mathematics
Discrete mathematics is the branch of mathematics dealing with discrete elements that use distinct, separated values. It includes structures like graphs, logic, and of course, probabilities that are inherently non-continuous.

Probability distributions, such as the one in our example, are a part of discrete mathematics. Here, we deal with specific, individual outcomes that are clearly distinct (like \(-2, -1, 0, 1, \) and \( 2 \)), as opposed to continuous mathematics, which involves continuous datasets.

This field focuses heavily on:
  • Discrete set handling, often finite in nature.
  • Precise structures where concepts like logic and set theory help us define rules and solve problems.
In discrete mathematics, simplicity and structure make solving problems like those in probability distributions more manageable, clear, and logically feasible.

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Most popular questions from this chapter

Waiting for a car wash. A certain automatic car wash takes exactly 5 minutes to wash a car. On the average, 10 cars per hour arrive at the car wash. Suppose that, 30 minutes before closing time, five cars are in line. If the car wash is in continuous use until closing time, is it likely anyone will be in line at closing time?

Give the characteristics of a hypergeometric distribution.

Consider the probability distribution shown for the random variable \(x\) here: $$ \begin{array}{l|cccc} \hline x & 1 & 2 & 4 & 10 \\ p(x) & .2 & .4 & .2 & .2 \\ \hline \end{array} $$ a. Find \(\mu=E(x)\). b. Find \(\sigma^{2}=E\left[(x-\mu)^{2}\right]\). c. Find \(\sigma\). d. Interpret the value you obtained for \(\mu\). e. In this case, can the random variable \(x\) ever assume the value \(\mu\) ? Explain. f. In general, can a random variable ever assume a value equal to its expected value? Explain.

Explain the difference between sampling with replacement and sampling without replacement.

Space shuttle disaster. On January \(28,1986,\) the space shuttle Challenger exploded, killing all seven astronauts aboard. An investigation concluded that the explosion was caused by the failure of the \(\mathrm{O}\) ring seal in the joint between the two lower segments of the right solid rocket booster. In a report made one year prior to the catastrophe, the National Aeronautics and Space Administration (NASA) claimed that the probability of such a failure was about \(1 / 60,000\), or about once in every 60,000 flights. But a risk- assessment study conducted for the Air Force at about the same time assessed the probability to be \(1 / 35\), or about once in every 35 missions. (Note: The shuttle had flown 24 successful missions prior to the disaster.) Given the events of January \(28,1986,\) which risk assessment \(-\) NASA's or the Air Force's - appears to be more appropriate?

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