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Chapter 2: Problem 70

Use the following information to answer the next two exercises: The following data are the distances between 20 retail stores and a large distribution center. The distances are in miles. 29; 37; 38; 40; 58; 67; 68; 69; 76; 86; 87; 95; 96; 96; 99; 106; 112; 127; 145; 150 Find the value that is one standard deviation below the mean.

Short Answer

Expert verified
One standard deviation below the mean is approximately 41.45 miles.

Step by step solution

01

Calculate the Mean

To find the mean, add up all the distances and divide by the number of stores. Sum of distances = 1509. Number of distances = 20. So, the mean \( \mu \) is \( \frac{1509}{20} = 75.45 \).
02

Calculate Squared Deviations

For each distance, calculate its deviation from the mean, square that deviation, and then sum all the squared deviations. For example, for the first value: \((29 - 75.45)^2 = 2168.9025\).Perform similar calculations for all values, and sum these squared deviations.
03

Calculate the Variance

Variance \( \sigma^2 \) is calculated by dividing the sum of squared deviations by the number of data points. Sum of squared deviations = 23128.55. Number of distances = 20. Therefore, variance \( \sigma^2 \) is \( \frac{23128.55}{20} = 1156.4275 \).
04

Calculate the Standard Deviation

The standard deviation \( \sigma \) is the square root of the variance. So, \( \sigma = \sqrt{1156.4275} = 34.001 \).
05

Find One Standard Deviation Below the Mean

To find one standard deviation below the mean, subtract the standard deviation from the mean: \( 75.45 - 34.001 = 41.449 \).

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Key Concepts

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

Understanding Mean Calculation
The mean, often referred to as the average, represents the central point of a data set, providing a single value that summarizes all numbers included. It's calculated by summing all observations in a set and dividing by the number of observations. For example, with the data set of distances between retail stores and a distribution center, you first add all distances: \( 29 + 37 + 38 + \ldots + 150 = 1509 \).
Then, divide this total by the number of observations, in our case 20 stores, yielding a mean of \( \mu = \frac{1509}{20} = 75.45 \).
  • Mean is helpful for understanding the typical value in a data set.
  • It acts as a balancing point for the data, equally dividing all values.
Mean calculation is foundational in statistics and acts as a starting point in finding other indices like the variance and standard deviation.
Explaining Standard Deviation
Standard deviation is a measure of how spread out numbers are in a data set. It provides insight into the data's dispersion relative to the mean, telling us how much data points differ from the average value. After calculating the variance, the standard deviation \( \sigma \) is determined by taking the square root of the variance.
For example, if the variance in our store distances is 1156.4275, the standard deviation is \( \sqrt{1156.4275} = 34.001 \).
  • A low standard deviation indicates data points are close to the mean.
  • A high standard deviation suggests greater variability, or spread, among the data points.
Understanding standard deviation helps assess the reliability and predictability of the data.
Decoding Variance
Variance quantifies the amount of variation or dispersion in a set of data values. It's calculated by finding the mean of the squared deviations of each data point from the mean. This method ensures that differences are uniformly positive and emphasizes larger distances.
To calculate variance \( \sigma^2 \), use the sum of squared deviations (23128.55) and divide it by the count of data points (20), yielding \( \sigma^2 = \frac{23128.55}{20} = 1156.4275 \).
  • Variance captures the degree of spread in data.
  • A larger variance indicates a wider spread of data points around the mean.
Although variance is useful, its units are squared, which is why the standard deviation, being in the same unit as original measurements, is preferable.
Calculating Squared Deviations
Squared deviations are essential for evaluating variance and further, standard deviation. They capture how each data point diverges from the mean. To find these, subtract the mean from each data point and square the result. These squared values collectively emphasize data points that fall far from the mean, hence contributing heavily to the overall variance.
For instance, take a distance value, like 29, and compute its squared deviation: \( (29 - 75.45)^2 = 2168.9025 \). Repeat for all data points, and then sum these values.
  • The summing of squared deviations protects against negatives canceling out positives, maintaining total deviation magnitude.
  • This calculation is a pivotal step preceding variance and standard deviation.
Squared deviations highlight the impact of outliers and drive our understanding of data dispersion.

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