Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Attendances at a high school's basketball games were recorded and found to have a sample mean and variance of 420 and \(25,\) respectively. Calculate \(\bar{x} \pm s, \bar{x} \pm 2 s,\) and \(\bar{x} \pm 3 s\) and then state the approximate fractions of measurements you would expect to fall into these intervals according to the Empirical Rule.

Short Answer

Expert verified
Answer: According to the Empirical Rule, the intervals of attendance are approximately (415, 425) for 68% of the data, (410, 430) for 95% of the data, and (405, 435) for 99.7% of the data.

Step by step solution

01

Calculate the standard deviation from variance

We are given the variance as \(25\). The standard deviation (s) is the square root of the variance: \(s = \sqrt{25}\) \(s = 5\)
02

Calculate the intervals \(\bar{x} \pm s, \bar{x} \pm 2 s,\) and \(\bar{x} \pm 3 s\)

The sample mean (\(\bar{x}\)) is given as 420. We will now calculate the three intervals using the standard deviation: 1. \(\bar{x} \pm s = 420 \pm 5 = (415, 425)\) 2. \(\bar{x} \pm 2s = 420 \pm 2(5) = (410, 430)\) 3. \(\bar{x} \pm 3s = 420 \pm 3(5) = (405, 435)\)
03

State the approximate fractions of measurements according to the Empirical Rule

Now that we have the three intervals, we can state the approximate fractions of measurements: 1. 68% of the data is expected to fall within the interval \(\bar{x} \pm s = (415, 425)\) 2. 95% of the data is expected to fall within the interval \(\bar{x} \pm 2s = (410, 430)\) 3. 99.7% of the data is expected to fall within the interval \(\bar{x} \pm 3s = (405, 435)\) In conclusion, according to the Empirical Rule, we can expect approximately 68% of the attendances to fall within the interval (415, 425), 95% within the interval (410, 430), and 99.7% within the interval (405, 435).

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 Vaia!

Key Concepts

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

Understanding Standard Deviation
Standard deviation is a key concept in statistics that helps us understand how much variation or dispersion exists in a set of data. It tells us how much individual data points deviate, on average, from the mean of the dataset. In simpler terms, it measures the extent to which numbers are spread out. - A small standard deviation means the data points tend to be close to the mean. - A large standard deviation means there is more spread among the data points. In our example, the standard deviation is calculated from the variance. Since the variance is given as 25, we take the square root of 25 to find the standard deviation, resulting in a value of 5. Simply put, the average deviation of attendance numbers from the mean (420) is 5.
Exploring Variance
Variance is a measure of how far each number in the dataset is from the mean and, consequently, from every other number in the dataset. It is essentially the average of the squared differences from the mean, providing a sense of how much the numbers in a dataset are spread out.In mathematical terms, variance is calculated as:\[ \sigma^2 = \frac{1}{N} \sum_{i=1}^{N}(x_i - \bar{x})^2 \]where \( \sigma^2 \) represents variance, \( x_i \) are individual data points, and \( \bar{x} \) is the mean.In our scenario with the basketball game attendances, the variance is provided as 25. This value helps us determine the standard deviation by taking its square root. Understanding variance is crucial because it underpins the calculation of standard deviation.
Normal Distribution
A normal distribution is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In data visualization, it is often represented as a bell curve. The normal distribution is fundamental because it describes how data is naturally distributed in many real-world situations.Key attributes of a normal distribution: - The mean, median, and mode are all equal. - The bell curve is symmetric at the center (around the mean, \( \bar{x} \)). - Approximately 68% of the data falls within one standard deviation from the mean. - 95% falls within two standard deviations, and 99.7% lies within three.Given our attendance data, it’s assumed to be normally distributed, which allows us to apply the Empirical Rule to determine the fraction of data within each interval.
Confidence Intervals and the Empirical Rule
Confidence intervals provide a range of values which is likely to contain a population parameter, like the mean, with a certain level of confidence. When applied to a normal distribution, these intervals help us understand the range in which most data points are expected to fall.The Empirical Rule, also known as the 68-95-99.7 rule, is closely related to confidence intervals. It gives us a way to estimate the probability of data falling within certain standard deviation ranges:- About 68% of data falls within one standard deviation (\( \bar{x} \pm s \))- Roughly 95% of data falls within two (\( \bar{x} \pm 2s \))- Nearly 99.7% of data falls within three standard deviations (\( \bar{x} \pm 3s \))In the context of our exercise, the intervals calculated using this rule help provide confidence about where most basketball game attendance figures are expected to fall in relation to the mean.

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

You are given \(n=8\) measurements: 3,1,5,6 4,4,3,5 a. Calculate the range. b. Calculate the sample mean. c. Calculate the sample variance and standard deviation. d. Compare the range and the standard deviation. The range is approximately how many standard deviations?

In the seasons that followed his 2001 record-breaking season, Barry Bonds hit \(46,45,45,5,26,\) and 28 homers, respectively, until he retired from major league baseball in 2007 (www.ESPN,com). \(^{16}\) Two box plots, one of Bond's homers through \(2001,\) and a second including the years \(2002-2007\) follow. The statistics used to construct these box plots are given in the table. \begin{tabular}{lccccccc} Years & Min & \(a_{1}\) & Median & \(a_{3}\) & IQR & Max & \(n\) \\ \hline 2001 & 16 & 25.00 & 34.00 & 41.50 & 16.5 & 73 & 16 \\ 2007 & 5 & 25.00 & a. Calculate the upper fences for both of these box plots. b. Can you explain why the record number of homers is an outlier in the 2001 box plot, but not in the 2007 box plot?34.00 & 45.00 & 20.0 & 73 & 22 \end{tabular}

Suppose you want to create a mental picture of the relative frequency histogram for a large data sct consisting of 1000 observations, and you know that the mean and standard deviation of the data set are 36 and 3 , respectively. a. If you are fairly certain that the relative frequency distribution of the data is mound-shaped, how might you picture the relative frequency distribution? (HiNr: Use the Ermpirical Rule.) b. If you have no prior information concerning the shape of the relative frequency distribution, what can you say about the relative frequency histogram? (Hivis: Construct intervals \(\bar{x} \pm k s\) for several choices of \(k\).

Find the five-number summary and the IQR for these data: $$ 19,12,16,0,14,9,6,1,12,13,10,19,7,5,8 $$.

Construct a box plot for these data and identify any outliers: $$ 25,22,26,23,27,26,28,18,25,24,12 $$.

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.

Sign-up for free