/*! 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 14 The following data represent the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The following data represent the amount of time (in minutes) a random sample of eight students took to complete the online portion of an exam in Sullivan's Statistics course. Compute the mean, median, and mode time. $$ 60.5,128.0,84.6,122.3,78.9,94.7,85.9,89.9 $$

Short Answer

Expert verified
Mean: 93.225, Median: 85.25, No mode.

Step by step solution

01

- Arrange the Data

Sort the given data in ascending order: \[60.5, 78.9, 84.6, 85.9, 89.9, 94.7, 122.3, 128.0\]
02

- Calculate the Mean

To calculate the mean, sum all the data points and divide by the number of data points. \[\text{Mean} = \frac{60.5 + 78.9 + 84.6 + 85.9 + 89.9 + 94.7 + 122.3 + 128.0}{8} = \frac{745.8}{8} = 93.225\]
03

- Find the Median

To find the median, identify the middle value(s). Since there is an even number of data points, take the average of the two middlemost values. \[84.6, 85.9\]\[\text{Median} = \frac{84.6 + 85.9}{2} = \frac{170.5}{2} = 85.25\]
04

- Determine the Mode

The mode is the value that appears most frequently. Since all data points are unique, there is no mode.

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.

Understanding the Mean
The mean is one of the most commonly used measures of central tendency. It is also known as the average. To calculate it, you sum all the data points and then divide by the number of data points.
In our exercise, the data points are: 60.5, 128.0, 84.6, 122.3, 78.9, 94.7, 85.9, and 89.9.
First, we add all the numbers:
\[60.5 + 128.0 + 84.6 + 122.3 + 78.9 + 94.7 + 85.9 + 89.9 = 745.8\]
Next, we divide by the number of data points, which is 8:
\[ \text{Mean} = \frac{745.8}{8} = 93.225 \]
This means that on average, students took about 93.225 minutes to complete the online portion of the exam. It's a useful summary because it gives an overall sense of the dataset.
Grasping the Median
The median is the middle value in a data set when it is arranged in ascending order. If there is an even number of data points, the median is the average of the two middle numbers.
First, let's sort the data in ascending order: \([60.5, 78.9, 84.6, 85.9, 89.9, 94.7, 122.3, 128.0] \)
Since we have eight values, we need to find the average of the fourth and fifth values. These values are: \(84.6\) and \(85.9\).
So, the median calculation is:
\[\text{Median} = \frac{84.6 + 85.9}{2} = \frac{170.5}{2} = 85.25\]
This tells us that half of the students took less than 85.25 minutes, and half took more. The median provides a good indication of the central point of the dataset, especially when the data is skewed.
Learning About the Mode
The mode is the value that appears most frequently in a data set. It is possible to have more than one mode or no mode at all.
In our case, the data points are: \([60.5, 78.9, 84.6, 85.9, 89.9, 94.7, 122.3, 128.0] \).
Each value in the dataset is unique, meaning none of the values repeat.
Thus, there is no mode for this dataset.
The mode is particularly useful when you have data with many repeating values, and you want to understand which is the most common or popular.

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

In January \(2015,\) the mean amount of money lost per visitor to a local riverboat casino was \(\$ 135 .\) Do you think the median was more than, less than, or equal to this amount? Why?

Find the population variance and standard deviation or the sample variance and standard deviation as indicated. $$ \text { Population: } 3,6,10,12,14 $$

For Super Bowl XLVII, there were 82,566 tickets sold for a total value of \(\$ 218,469,636 .\) What was the mean price per ticket?

27\. Rates of Returns of Stocks Stocks may be categorized by sectors. Go to www.pearsonhighered.com/sullivanstats and download \(3_{-} 2_{-} 27\) using the file format of your choice. The data represent the one-year rate of return (in percent) for a sample of consumer cyclical stocks and industrial stocks for the period December 2013 through November 2014 . Note: Consumer cyclical stocks include names such as Starbucks and Home Depot. Industrial stocks include names such as \(3 \mathrm{M}\) and FedEx. (a) Draw a relative frequency histogram for each sector using a lower class limit for the first class of -50 and a class width of \(10 .\) Which sector appears to have more dispersion? (b) Determine the mean and median rate of return for each sector. Which sector has the higher mean rate of return? Which sector has the higher median rate of return? (c) Determine the standard deviation rate of return for each sector. In finance, the standard deviation rate of return is called risk. Typically, an investor "pays" for a higher return by accepting more risk. Is the investor paying for higher returns for these sectors? Do you think the higher returns are worth the cost? Explain.

The standard deviation of batting averages of all teams in the American League is 0.008 . The standard deviation of all players in the American League is \(0.02154 .\) Why is there less variability in team batting averages?

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.