/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 82 A random variable \(x\) has this... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A random variable \(x\) has this probability distribution: $$\begin{array}{l|rrrrrr}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .1 & .3 & .4 & .1 & ? & .05\end{array}$$ a. Find \(p(4)\). b. Construct a probability histogram to describe \(p(x)\). c. Find \(\mu, \sigma^{2},\) and \(\sigma\). d. Locate the interval \(\mu \pm 2 \sigma\) on the \(x\) -axis of the histogram. What is the probability that \(x\) will fall into this interval? e. If you were to select a very large number of values of \(x\) from the population, would most fall into the interval \(\mu \pm 2 \sigma ?\) Explain.

Short Answer

Expert verified
What is the probability that x falls within this interval? Answer: In the given distribution, the interval μ ± 2σ does include all possible values of x. The probability that x falls within this interval is 1, meaning that all x values would fall into this interval in a very large number of selections from the population.

Step by step solution

01

Finding \(p(4)\)

The sum of the probabilities in a probability distribution must equal 1. In order to find the value of \(p(4)\), we can use the following equation: $$p(4) = 1 - (\sum_{i=0}^{3} p(i) + p(5))$$ Substitute known values and solve for \(p(4)\): $$p(4) = 1 - (0.1 + 0.3 + 0.4 + 0.1 + 0.05)$$ $$p(4) = 1 - 0.95$$ $$p(4) = 0.05$$
02

Construct a probability histogram for the distribution

To create a probability histogram, plot the discrete values of \(x\) on the horizontal axis and their corresponding probabilities \(p(x)\) on the vertical axis. Make sure to use the newly found value of \(p(4) = 0.05\). The probability histogram should have a bar for each value of \(x\), with a height equal to the probability.
03

Find \(\mu, \sigma^{2},\) and \(\sigma\)

The mean, variance, and standard deviation can be found using the following formulas: \(\mu = \sum_{i=0}^{5} x_i p(x_i)\) \(\sigma^2 = \sum_{i=0}^{5} (x_i - \mu)^2 p(x_i)\) \(\sigma = \sqrt{\sigma^2}\) Compute these using the given probabilities and the \(x\) values: Mean, \(\mu = (0)(0.1) + (1)(0.3) + (2)(0.4) + (3)(0.1) + (4)(0.05) + (5)(0.05) = 0 + 0.3 + 0.8 + 0.3 + 0.2 + 0.25 = 2.05\) Variance, \(\sigma^2 = (0-2.05)^2(0.1) + (1-2.05)^2(0.3) + (2-2.05)^2(0.4) + (3-2.05)^2(0.1) + (4-2.05)^2(0.05) + (5-2.05)^2(0.05) = 2.20575\) Standard Deviation, \(\sigma = \sqrt{2.20575} \approx 1.485\)
04

Locate the interval \(\mu \pm 2 \sigma\) on the histogram and find the probability

Compute the interval \(\mu \pm 2 \sigma = 2.05 \pm 2(1.485) = (-0.92, 5.02)\). This interval contains all the values of \(x\). The probability that \(x\) falls within this interval is the sum of all probabilities of \(x\), which is equal to 1
05

Discuss if most values of x fall into the interval \(\mu \pm 2 \sigma\)

In this specific distribution, \(\mu \pm 2 \sigma\) includes all possible values of \(x\). This means that in a very large number of selections from the population, all values of \(x\) would fall within this interval. In general, the interval \(\mu \pm 2 \sigma\) includes approximately 95% of the probability in a normal distribution, meaning that most values in a normal distribution would fall into this interval. Note that this might not always be the case for every distribution, as the shape of the distribution plays a significant role in the coverage of the interval.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Histogram
A probability histogram is a graphical representation that showcases the probability distribution of a discrete random variable. It resembles a bar graph, but with an important distinction: the height of each bar corresponds to the probability of each outcome rather than frequency or count.

In constructing a probability histogram, the horizontal axis represents the possible outcomes (in this case, the various values of the random variable x), while the vertical axis shows the probabilities associated with these outcomes. Completing such a histogram provides a visual understanding of how likely different results are. For our case, where the random variable x can take on the values 0 through 5, each bar's height is the probability of x taking each of those values.
Mean and Standard Deviation
In probability and statistics, the mean of a distribution, represented by the symbol \( \mu \) (Greek letter mu), is the average expected outcome and is computed by summing the product of each outcome and its probability. On the other hand, the standard deviation, represented by \( \sigma \) (Greek letter sigma), is a measure that indicates the amount of variation or dispersion present in a set of data values.

Standard deviation is the square root of variance, where variance \( \sigma^2 \) quantifies the spread of the values around the mean. A higher standard deviation signifies that the data points are spread out over a wider range of values. Understanding both mean and standard deviation is crucial as they together describe the shape and spread of the probability distribution.
Normal Distribution
The normal distribution, often known as the bell curve due to its shape, is a continuous probability distribution that is symmetric about the mean. It is characterized by two parameters: the mean \( \mu \) and the standard deviation \( \sigma \) of the dataset. The majority of values cluster around the mean, and the probabilities for values further away from the mean taper off symmetrically in both directions.

The Empirical Rule states that within a normal distribution, roughly 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three. This property is often used to predict the probability of a given outcome within these intervals. While not all distributions are normal, many real-world phenomena are modeled successfully by the normal distribution, making it an essential concept in statistics.
Probability Theory
Probability theory is the branch of mathematics concerned with analysis of random phenomena. It provides the mathematical foundation for the study of chance, governing the likelihood of events in processes from quantum mechanics to game theory. At its core, it involves assigning numbers between 0 and 1 to events with the intent of quantifying the randomness.

In the context of our exercise, probability theory principles were applied to compute the mean and standard deviation and then further to discuss the interval \( \mu \pm 2 \sigma \) within the probability histogram. The rules and theorems of probability theory help in understanding and predicting the behavior of this random variable \( x \) and form the underpinnings of statistical inference, enabling the transformation of data into actionable information.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Only \(40 \%\) of all people in a community favor the development of a mass transit system. If four citizens are selected at random from the community, what is the probability that all four favor the mass transit system? That none favors the mass transit system?

One professional golfer plays best on short-distance holes. Experience has shown that the numbers \(x\) of shots required for \(3-, 4-,\) and 5 -par holes have the probability distributions shown in the table: $$\begin{array}{lccccc}\text { Par-3 Holes } & \multicolumn{3}{c} {\text { Par-4 Holes }} & \multicolumn{2}{c} {\text { Par-5 Holes }} \\ \hline x & p(x) & x & p(x) & x & p(x) \\\\\hline 2 & .12 & 3 & .14 & 4 & .04 \\\3 & .80 & 4 & .80 & 5 & .80 \\\4 & .06 & 5 & .04 & 6 & .12 \\ 5 & .02 & 6 & .02 & 7 & .04\end{array}$$ What is the golfer's expected score on these holes? a. A par-3 hole b. A par- 4 hole c. A par-5 hole

An experiment is run as follows- the colors red, yellow, and blue are each flashed on a screen for a short period of time. A subject views the colors and is asked to choose the one he feels was flashed for the longest time. The experiment is repeated three times with the same subject. a. If all the colors were flashed for the same length of time, find the probability distribution for \(x\), the number of times that the subject chose the color red. Assume that his three choices are independent. b. Construct the probability histogram for the random variable \(x\)

Your family vacation involves a cross-country air flight, a rental car, and a hotel stay in Boston. If you can choose from four major air carriers, five car rental agencies, and three major hotel chains, how many options are available for your vacation accommodations?

Two tennis professionals, \(A\) and \(B\), are scheduled to play a match; the winner is the first player to win three sets in a total that cannot exceed five sets. The event that \(A\) wins any one set is independent of the event that \(A\) wins any other, and the probability that \(A\) wins any one set is equal to \(.6 .\) Let \(x\) equal the total number of sets in the match; that is, \(x=3,4,\) or \(5 .\) Find \(p(x)\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.