/*! 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 30 Following is a sample of the day... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Following is a sample of the day-byday change, rounded to the nearest 100 points, in the Dow Jones Industrial Average during 10 successive business days around the start of the financial crisis in October \(2008:^{20}\) $$ -100,400,-200,-500,200,-300,-200,900,-100,200 $$ Compute the mean and median of the given sample. Fill in the blank: There were as many days with a change in the Dow above \(\quad\) points as there were with changes below that.

Short Answer

Expert verified
The mean change in Dow Jones Industrial Average during the 10 successive business days is 10 points, and the median change is 50 points. There were as many days with a change in the Dow above 10 points as there were with changes below that.

Step by step solution

01

Calculate the mean

To find the mean, we need to sum all the data points and then divide it by the total number of data points. We have 10 data points, so: Mean = \(\frac{(-100+400-200-500+200-300-200+900-100+200)}{10}\)
02

Simplify and calculate mean

Now, we can simplify the expression and calculate the mean: Mean = \(\frac{(100)}{10}\) = \(10\) So, the mean change in Dow Jones Industrial Average during the 10 successive business days is 10 points.
03

Calculate the median

To find the median, we need to sort the data points and find the middle value. If there is an even number of data points, the median is the average of the two middle values. Let's arrange the data points in ascending order: $$ -500, -300, -200, -200, -100, 200, 200, 400, 900 $$ There are 10 data points, so the median is the average of the 5th and 6th data points: Median = \(\frac{-100 + 200}{2}\)
04

Simplify and calculate median

Now, we can simplify the expression and calculate the median: Median = \(\frac{(100)}{2}\) = \(50\) So, the median change in Dow Jones Industrial Average during the 10 successive business days is 50 points.
05

Determine the value for equal changes

We need to find the value for which there are the same number of changes above and below that value. Since the mean change is 10 points and the median change is 50 points, we can logically deduce that the number of changes above and below the mean is equal. So, the value is 10 points. There were as many days with a change in the Dow above 10 points as there were with changes below that.

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.

Statistical Analysis
In statistical analysis, calculating the mean and median provides key insights into a data set.
The mean, often referred to as the average, tells us the central point around which the data values tend to cluster.
In our exercise, the changes in the Dow Jones Industrial Average were anywhere from \(-500\) to \(900\) points. The mean was calculated by summing all these values and then dividing by 10, the number of data points, resulting in a mean of \(10\) points. This simple calculation gives us a central value from the entire data.
The median, on the other hand, provides insight into the middle point of a data set.
Unlike the mean, the median is less affected by extreme values or outliers because it focuses solely on the middle of the sorted data. By organizing our data in ascending order, we can find the median by identifying the middle value or, in the case of an even number of data points, the average of the two middle numbers. This resulted in a median of \(50\) points for our sample.
Both the mean and median can tell us different stories about the data; together, they offer a more robust view of the data's central tendencies.
Data Interpretation
Interpreting data is a skill that involves extracting meaningful insights from numerical information.
In the context of this exercise, understanding the Dow's day-to-day changes requires more than just knowing how to compute mean and median.
When interpreting these statistics, the mean of \(10\) points suggests that the average daily change across the 10 business days was low, despite the volatility shown in the individual entries like \(400\), and \(900\) points. Meanwhile, the median of \(50\) points pinpoints that half of the days had changes below this value and half above, again emphasizing the presence of outliers.
A critical aspect of data interpretation is recognizing these extreme values or outliers, such as the \(-500\) and \(900\) points.
These can skew the mean and provide a misleading sense of what is typical. The median offers a counterbalance by not being influenced significantly by such outliers, thereby giving a different picture of the central tendency of the data. This dual approach of mean and median provides a clearer understanding of the dataset, highlighting both general trends and specific variations.
Financial Mathematics
Financial mathematics utilizes statistical calculations, including mean and median, to interpret financial data patterns.
These concepts are crucial in analyzing market trends, assessing investment risks, and making strategic decisions.
For instance, the data examined in this exercise depicts market volatility during the financial crisis.
A low mean suggests overall slight changes, while the median of \(50\) points indicates the typical magnitude of daily market movements. To determine the balanced point where changes above and below were equal, we observed where about half of the changes lay above and below the mean of 10 points, affirming this point as a key marker in financial trend analysis.
Financial analysts and investors often leverage these statistical tools to evaluate risk.
For instance, understanding the dispersion of data with outliers like \(900\) points is crucial in risk assessment, as it highlights potential instability. By using the median and mean together, professionals can balance the gap between daily fluctuations and establish more informed strategies in response to market behaviors.

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

Compute the (sample) variance and standard deviation of the data samples given in Exercises \(1-8 .\) You calculated the means in the last exercise set. Round all answers to two decimal nlaces. $$ 4.2,-3.2,0,1.7,0 $$

Y o u r ~ p i c k l e ~ c o m p a n y ~ r a t e s ~ i t s ~ p i c k l e s ~ o n ~ a ~ s c a l e ~ o f spiciness from 1 to \(10 .\) Market research shows that customer preferences for spiciness are normally distributed, with a mean of \(7.5\) and a standard deviation of 1 . Assuming that you sell 100,000 jars of pickles, how many jars with a spiciness of 9 or above do you expect to sell?

The Acme Insurance Company is launching a drive to generate greater profits, and it decides to insure racetrack drivers against wrecking their cars. The company's research shows that, on average, a racetrack driver races four times a year and has a 1 in 10 chance of wrecking a vehicle, worth an average of \(\$ 100,000\), in every race. The annual premium is \(\$ 5,000\), and Acme automatically drops any driver who is involved in an accident (after paying for a new car), but does not refund the premium. How much profit (or loss) can the company expect to earn from a typical driver in a year? HINT [Use a tree diagram to compute the probabilities of the various outcomes.]

Your company, Sonic Video, Inc., has conducted research that shows the following probability distribution, where \(X\) is the number of video arcades in a randomly chosen city with more than 500,000 inhabitants. $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \boldsymbol{x} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) & .07 & .09 & .35 & .25 & .15 & .03 & .02 & .02 & .01 & .01 \\ \hline \end{array} $$ a. Compute \(\mu=E(X)\) and interpret the result. HINT [See Example 3.] b. Which is larger, \(P(X<\mu)\) or \(P(X>\mu) ?\) Interpret the result.

Slim tells you that the population mean is just the mean of a suitably large sample. Is he correct? Explain.

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.