/*! 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 41 Determine whether the statement ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. Both the variance and the standard deviation of a random variable measure the spread of a probability distribution.

Short Answer

Expert verified
The statement is true. Both the variance (\(Var(X) = \sum{(x_i -\mu)^2 * P(x_i)}\)) and the standard deviation (\(SD(X) = \sqrt{Var(X)}\)) of a random variable measure the spread of a probability distribution. They capture the average distance or deviation of each data point from the mean, revealing the degree of dispersion in the distribution. While the variance is calculated in squared units, the standard deviation is in the same unit as the mean, making it more interpretable for practical applications.

Step by step solution

01

Define Variance and Standard Deviation

The variance and standard deviation are two measures of dispersion (spread) of a random variable in a probability distribution. They help us understand how the data points are distributed around the mean. The variance of a discrete random variable X, denoted by \(Var(X) \), is calculated using the following formula: \[Var(X) = \sum{(x_i -\mu)^2 * P(x_i)}\] Where: - \(x_i\) are the possible values of the random variable X - \(\mu\) is the expected value (mean) of the random variable X - \(P(x_i)\) is the probability of each value \(x_i\) The standard deviation of the random variable X, denoted by \(SD(X)\), is the square root of the variance: \[SD(X) = \sqrt{Var(X)}\]
02

Analyze the Role of Both Variance and Standard Deviation in Measuring Spread

As seen from the formulas above, both the variance and standard deviation take into account the possible values of the random variable and their respective probabilities. By doing so, they measure the average distance or deviation of each data point from the mean. The main difference between these two measures of dispersion is that variance is calculated in squared units while the standard deviation is in the same unit as the mean. As a result, the interpretation of the standard deviation may be more intuitive, especially for practical applications, since it is in the same scale as the data. However, both measures capture the same underlying property of the distribution - the spread of the random variable. In this sense, we can say that they indeed measure the spread of a probability distribution.
03

Conclusion

The statement is true. Both the variance and the standard deviation of a random variable measure the spread of a probability distribution. The variance measures the average of the squared differences from the mean, while the standard deviation is the square root of the variance, making it more interpretable for practical applications. Nonetheless, both capture the degree of dispersion in the distribution, thus measuring the spread.

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!

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

Use the appropriate normal distributions to approximate the resulting binomial distributions. Preview Showcase, a research firm, screens pilots of new TV shows before a randomly selected audience and then solicits their opinions of the shows. Based on past experience, \(20 \%\) of those who receive complimentary tickets are "no-shows." The theater has a seating capacity of \(500 .\) Management has decided, for public relations reasons, that \(99 \%\) of all ticket holders will be seated. How many tickets should the company send out to prospective viewers for each screening?

A probability distribution has a mean of 50 and a standard deviation of \(1.4 .\) Use Chebychev's inequality to find the value of \(c\) that guarantees the probability is at least \(96 \%\) that an outcome of the experiment lies between \(50-c\) and \(50+c .\)

The weights, in ounces, of ten packages of potato chips are \(\begin{array}{llllllllll}16.1 & 16 & 15.8 & 16 & 15.9 & 16.1 & 15.9 & 16 & 16 & 16.2\end{array}\) Find the average and the median of these weights.

Let \(Z\) be the standard normal variable. Find the values of \(z\) if \(z\) satisfies a. \(P(Z>z)=.9678\) b. \(P(-z

Let the random variable \(X\) denote the number of girls in a five-child family. If the probability of a female birth is .5, a. Find the probability of \(0,1,2,3,4\), and 5 girls in a fivechild family. b. Construct the binomial distribution and draw the histogram associated with this experiment. c. Compute the mean and the standard deviation of the random variable \(X\).

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.