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On the first day of class last semester, 50 students were asked for the one- way distance from home to college (to the nearest mile). The resulting data follow: $$\begin{array}{rrrrrrrrrr} \hline 6 & 5 & 3 & 24 & 15 & 15 & 6 & 2 & 1 & 3 \\ 5 & 10 & 9 & 21 & 8 & 10 & 9 & 14 & 16 & 16 \\ 10 & 21 & 20 & 15 & 9 & 4 & 12 & 27 & 10 & 10 \\ 3 & 9 & 17 & 6 & 11 & 10 & 12 & 5 & 7 & 11 \\ 5 & 8 & 22 & 20 & 13 & 7 & 8 & 13 & 4 & 18 \\ \hline \end{array}$$ a. Construct a grouped frequency distribution of the data by using \(1-4\) as the first class. b. Calculate the mean and the standard deviation. c. Determine the values of \(\bar{x} \pm 2 s\), and determine the percentage of data within 2 standard deviations of the mean.

Short Answer

Expert verified
The grouped frequency distribution would have varying frequencies for the classes constructed. The mean and standard deviation would depend on the data calculated. Accordingly, the values for \(\bar{x} \pm 2s\) would also vary. The percentage of data within 2 standard deviations of the mean may not exactly be 95% since the data may not follow a perfect normal distribution.

Step by step solution

01

Construct a grouped frequency distribution

First, determine the classes for the distribution. Each class should be of size 4 starting from 1. So, the classes will be 1-4, 5-8, 9-12, 13-16, 17-20, 21-24, 25-28. After identifying the classes, count the number of data points in each class. The count for each class is the 'frequency' for that class.
02

Calculate the mean

The mean is calculated by summing all the values in the data set and dividing by the count of the data set. Add all the data points together and then divide by the total number of students, which is 50.
03

Calculate the standard deviation

The standard deviation is calculated by finding the square root of the variance. First calculate the variance by finding the average of the squared differences from the Mean, then take the square root of the calculated variance to get standard deviation.
04

Determine the value of \(\bar{x} \pm 2 s\)

The values of \(\bar{x} \pm 2 s\) indicates the range in which lies around 95% of the data if they follow a normal distribution, where \(\bar{x}\) is the calculated mean and \(s\) is the calculated standard deviation. Subtract and add twice the standard deviation to the mean to calculate this range.
05

Determine the percentage of data within 2 standard deviations

Count the number of data points that lie within the range \(\bar{x} - 2s \) to \(\bar{x} + 2s \). Divide the count by total data points and multiply by 100 to get the percentage.

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Key Concepts

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

Grouped Frequency Distribution
When working with large sets of data, it is helpful to organize them using a grouped frequency distribution. This method groups data points into classes or intervals, making it easier to interpret the distribution of values. In this particular exercise, data on the distances students travel to college is divided into classes such as 1-4 miles, 5-8 miles, and so on. Each class spans 4 units, providing a clear breakdown of how frequently distances fall into each category.
To achieve this, begin by defining the class intervals. Here, each class starts at an integer progressively 4 units apart. Count the data points within each interval and mark this as the class frequency. This distribution helps visualize where most travel distances cluster, offering insights at a glance.
Mean Calculation
The mean, or average, is a central value that gives us an idea of the typical data point in a set. Calculating the mean involves summing all data points and dividing by the total number of data points. In our example, you'd add all the miles students travel and then divide by 50, which is the number of students.
Finding the mean is essential because it represents the "center" of the data set, giving a single value that summarizes the entire data. This central tendency measurement simplifies comparison across data sets and provides a baseline for further statistical calculations.
Standard Deviation
The standard deviation measures how much the data varies from the mean, telling us about the spread or dispersion of the data points. A low standard deviation means data points are close to the mean, whereas a high standard deviation indicates more spread out data.
To compute the standard deviation, first calculate the variance, which is the average of the squared differences between each data point and the mean. Then, take the square root of the variance to get the standard deviation. This information allows us to assess the variability in students' travel distances, offering insights into consistency or variability in traveling patterns.
Data Variability
Data variability provides an understanding of the distribution's spread, indicating how much individual points deviate from the average. It's essential to comprehend this aspect as it affects interpretative decisions and predictions.
In a distribution, the range, which is determined using the mean and the standard deviation, can tell us a lot about where the bulk of data lies. In our exercise, calculating \(ar{x} \pm 2s\) defines a range where the majority of data points are expected to fall, assuming a normal distribution. By checking what proportion of data falls within this range, it is possible to gauge how typical this spread is under normal conditions. Such analysis is valuable in contexts needing an understanding of consistency or to pinpoint outliers.

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Most popular questions from this chapter

Construct a stem-and-leaf display of the number of points scored during each basketball game last season: $$\begin{array}{ccccccc} 56 & 54 & 61 & 71 & 46 & 61 & 55 & 68 \\ 60 & 66 & 54 & 61 & 52 & 36 & 64 & 51 \end{array}$$

Explain why it is possible to find the mean for the data of a quantitative variable, but not for a qualitative variable.

A study completed by International Communications Research for the Soap and Detergent Association (SDA) lists the item Americans say they would be most willing to give up in order to be able to hire someone to do their spring cleaning. The most popular response was \(\$ 100(29 \%),\) followed by dining out for a month \((26 \%),\) concert tickets \((19 \%),\) a weekend trip (9\%), and other ( \(17 \%\) ). a. Construct a Pareto diagram displaying this information. b. Because of the size of the "other" category, the Pareto diagram may not be the best graph to use. Explain why, and describe what additional information is needed to make the Pareto diagram more appropriate.

A constant objective in the manufacture of contact lenses is to improve those features that affect lens power and visual acuity. One such feature involves the tooling from which lenses are ultimately manufactured. The results of initial process development runs were examined for critical feature \(X .\) The resulting data are listed here: $$\begin{array}{lllllllll} \hline 0.026 & 0.027 & 0.024 & 0.023 & 0.034 & 0.035 & 0.035 & 0.033 & 0.034 \\\ 0.033 & 0.032 & 0.038 & 0.041 & 0.041 & 0.021 & 0.022 & 0.027 & 0.032 \\ 0.023 & 0.023 & 0.024 & 0.017 & 0.023 & 0.019 & 0.027 & \\ \hline \end{array}$$ a. Draw both a dotplot and a histogram of the critical feature \(X\) data. b. Find the mean for critical feature \(X\). c. Find the median for critical feature \(X\). d. Find the midrange for critical feature \(X .\) e. Find the mode, if one exists, for critical feature \(X .\) f. What feature of the distribution, as shown by the graphs found in part a, seems unusual? Where do the answers found in parts \(b, c,\) and \(d\) fall relative to the distribution? Explain. g. Identify at least one possible cause for this seemingly unusual situation.

What is the mean weekly pay if 5 employees earn \(\$ 425\) per week, 3 earn \(\$ 750\) per week, and 1 earns \(\$ 1340 ?\)

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