/*! 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 491 Find the average deviation for t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the average deviation for the grouped data given below: $$ \begin{array}{|c|c|c|c|c|c|} \hline \text { Class } & \text { Frequency } \mathrm{f}_{\mathrm{i}} & \text { Class Mark } \mathrm{X}_{\mathrm{i}} & \overline{\mathrm{X}} & \left|\mathrm{X}_{\mathrm{i}}-\overline{\mathrm{X}}\right| & \mathrm{f}_{\mathrm{i}}\left|\mathrm{X}_{\mathrm{i}}-\overline{\mathrm{X}}\right| \\\ \hline 49-54 & 6 & 51.5 & 66.5 & 15 & 90 \\ \hline 55-60 & 15 & 57.5 & 66.75 & 9 & 135 \\ \hline 61-66 & 24 & 63.5 & 66.5 & 3 & 72 \\ \hline 67-72 & 33 & 69.5 & 66.5 & 3 & 99 \\ \hline 73-78 & 22 & 75.5 & 66.5 & 9 & 198 \\ \hline \end{array} $$

Short Answer

Expert verified
The average deviation for the given grouped data is \(5.94\).

Step by step solution

01

Identify and Group Data

The given data is already grouped, and all the required columns are filled out. The class mark \(X_{i}\) is already given along with the frequency \(f_{i}\) and the absolute values of their deviation from the mean |\(X_{i}-\overline{X}\)| for each class. The product \(f_{i}\cdot |X_{i}-\overline{X}|\) is also given.
02

Sum up the frequencies and the absolute deviations for each class

Identifying the column of \(f_{i}\cdot |X_{i}-\overline{X}|\), add up all these values. The sum \(\Sigma f_{i}\cdot |X_{i}-\overline{X}|\) equals 90+135+72+99+198 = 594. Similarly, add up the values of frequencies to get \(\Sigma f_{i}\) = 6+15+24+33+22 = 100.
03

Calculate the Average Deviation

Use the formula \(\text{Average deviation} = \frac{\Sigma\left(f_{i} \cdot \left|X_{i}-\overline{X}\right|\right)}{\Sigma f_{i}}\) to find out the average deviation. Substituting \(\Sigma\left(f_{i} \cdot \left|X_{i}-\overline{X}\right|\right) = 594\) and \(\Sigma f_{i} = 100\) into the formula, we get \(\text{Average deviation} = \frac{594}{100} = 5.94\). The average deviation for the given grouped data is 5.94.

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.

Frequency Distribution
When working with large datasets, organizing the values into a frequency distribution can be incredibly helpful for analysis. It essentially puts data into a table which groups values into classes or intervals, telling us how many times each class occurs, known as the frequency. This perspective allows us to quickly understand the distribution of values and identify patterns, such as which ranges are the most or least common.For instance, in the context of student grades, a frequency distribution might show us how many students received a grade between 90-100, 80-89, and so on. It is the first step in many statistical analyses and is an essential concept for understanding grouped data, which subsequently serves as the foundation for calculating summary statistics like the average deviation.
Class Mark
The class mark, often referred to as the class midpoint, plays a pivotal role in handling grouped data. It is the central value of each class interval and is calculated by averaging the upper and lower boundaries of the class. It serves as a representative value for all observations within that class.Consider a class interval of 40-49 in a dataset; the class mark would be (40 + 49) / 2 = 44.5. This is particularly useful for calculating measures of central tendency, like the mean or median for grouped data, as it simplifies the data while maintaining an approximation of the distribution. It's vital when seeking to measure the variability around the mean, as we do with average deviation.
Deviation from the Mean
Put simply, deviation from the mean measures the distance of a data point from the average of a dataset. It reveals the spread of the data values, and whether they are close to or far from the mean. In the case of grouped data, the deviation for each class mark from the dataset's mean provides insight into the dispersion of those grouped values.For example, If we have a mean of 50, a class mark of 45 would have a deviation of |45 - 50| = 5. Deviations can be positive or negative but are often represented in their absolute form to focus on the magnitude, not the direction, of the difference. This concept is fundamental when assessing the consistency or variability of a dataset, leading into the calculation of average deviation.
Sum of Frequencies
In any statistical analysis involving frequency distribution, the sum of frequencies (Σfi) is a key value which indicates the total number of observations in the dataset. It is the sum of all individual frequencies within the distribution and is often used as the denominator in calculations to determine various statistical measures for the entire dataset.For instance, in calculating the average, the sum of frequencies helps us understand how that average should be properly weighted. It is crucial when computing averages from grouped data, because it enables the translation of data from numerous disparate classes into a single, representative figure, such as the average deviation in our example.

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

A die was tossed 120 times and the results are listed below. $$ \begin{array}{|l|l|l|l|l|l|l|} \hline \text { Upturned face } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Frequency } & 18 & 23 & 16 & 21 & 18 & 24 \\ \hline \end{array} $$ Compute the \(\mathrm{X}^{2}\) statistic for this 1 by 6 contingency table under the hypothesis that the die was fair.

Consider a distribution \(\mathrm{N}\left(\mu, \sigma^{2}\right)\) where \(\mu\) is known but \(\sigma^{2}\) is not. Devise a method of producing a confidence interval for \(\sigma^{2}\)

A research worker wishes to estimate the mean of a population using a sample large enough that the probability will be \(.95\) that the sample mean will not differ from the population mean by more than 25 percent of the standard deviation. How large a sample should he take?

Show that if \((\mathrm{X}, \mathrm{Y})\) has a bivariate normal distribution, then the marginal distributions of \(\mathrm{X}\) and \(\mathrm{Y}\) are univariate normal distributions; that is, \(\mathrm{X}\) is normally distributed with mean \(\mu_{\mathrm{x}}\) and variance \(\sigma^{2} \mathrm{x}\) and \(\mathrm{Y}\) is normally distributed with mean \(\mu_{\mathrm{y}}\) and variance \(\sigma^{2}{ }_{\mathrm{y}}\).

A highly specialized industry builds one device each month. The total monthly demand is a random variable with the following distribution. $$ \begin{array}{|l|l|l|l|l|} \hline \text { Demand } & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{D}) & 1 / 9 & 6 / 9 & 1 / 9 & 1 / 9 \\ \hline \end{array} $$ When the inventory level reaches 3, production is stopped until the inventory drops to 2 . Let the states of the system be the inventory level. The transition matrix is found to be \(\begin{array}{rlllll} & & 0 & 1 & 2 & 3 \mid \\\ & 0 & 8 / 9 & 1 / 9 & 0 & 0 \\ & 11 & 2 / 9 & 6 / 9 & 1 / 9 & 0 \\\ \mathrm{P}= & 12 & 1 / 9 & 1 / 9 & 6 / 9 & 1 / 9 \\ & 13 & 1 / 9 & 1 / 9 & 6 / 9 & 1 / 9\end{array}\) Assuming the industry starts with zero inventory find the transition matrix as \(\mathrm{n} \rightarrow \infty\)

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.