Chapter 2: Problem 6
Use the data entry method in your scientific calculator to enter the measurements. Recall the proper memories to find the mean and standard deviation. Calculate the range. The range is approximately how many standard deviations? $$ n=10 \text { measurements: } 5,2,3,6,1,2,4,5,1,3 $$
Short Answer
Expert verified
Answer: Approximately 3.09 standard deviations.
Step by step solution
01
Enter the measurements into the scientific calculator
To begin, enter the 10 measurements into your scientific calculator using its data entry method. You may need to consult your calculator's user manual to find specific instructions for entering data.
02
Calculate the mean
After entering the measurements into the calculator, use its function to calculate the mean (average). Once again, refer to your calculator's user manual if you need assistance with this step. The mean can also be calculated by summing up all the measurements and dividing by the total number of measurements:
$$
\text{Mean} = \frac{5+2+3+6+1+2+4+5+1+3}{10} = \frac{32}{10} = 3.2
$$
03
Calculate the standard deviation
Using your calculator, calculate the standard deviation of the measurements. The standard deviation is a measure of how spread out the numbers are from the mean. In case you want to calculate it by hand, the formula for the standard deviation is:
$$
\text{Standard Deviation} = \sqrt{\frac{\sum(x_i-\text{Mean})^2}{n-1}}
$$
For this exercise, the standard deviation is approximately 1.62.
04
Calculate the range
To calculate the range, find the difference between the highest and lowest measurement in the dataset. In this case, the highest measurement is 6, while the lowest is 1. Thus, the range is:
$$
\text{Range} = 6 - 1 = 5
$$
05
Determine how many standard deviations the range is approximately
Finally, we will find how many standard deviations the range is approximately by dividing the range by the standard deviation:
$$
\text{Approximate Number of Standard Deviations} = \frac{\text{Range}}{\text{Standard Deviation}} = \frac{5}{1.62} \approx 3.09
$$
The range is approximately 3.09 standard deviations.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mean and Standard Deviation Calculation
Understanding how to calculate the mean (average) and standard deviation is crucial in statistical data analysis. The mean represents the average value of a dataset, showing the central tendency. To find the mean, sum up all the values and divide by the count of the values. For example, if you have 10 measurements: 5, 2, 3, 6, 1, 2, 4, 5, 1, and 3, the mean would be calculated as \( \text{Mean} = \frac{5+2+3+6+1+2+4+5+1+3}{10} = \frac{32}{10} = 3.2 \).
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. The formula for standard deviation \( \text{Standard Deviation} = \sqrt{\frac{\sum(x_i-\text{Mean})^2}{n-1}} \) accounts for each value's distance from the mean. When calculating by hand, you would square the difference between each measurement and the mean, sum all those squared differences, then divide by one less than the number of measurements (n-1), and finally take the square root of that result. The resulting standard deviation, in this case, is approximately 1.62.
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. The formula for standard deviation \( \text{Standard Deviation} = \sqrt{\frac{\sum(x_i-\text{Mean})^2}{n-1}} \) accounts for each value's distance from the mean. When calculating by hand, you would square the difference between each measurement and the mean, sum all those squared differences, then divide by one less than the number of measurements (n-1), and finally take the square root of that result. The resulting standard deviation, in this case, is approximately 1.62.
Range Calculation
In statistics, the range is a simple measure of variability in a dataset. It is calculated by simply subtracting the smallest value from the largest value. The importance of range lies in its ability to give you a quick sense of the spread of values in your data at a glance. For the given measurements (5, 2, 3, 6, 1, 2, 4, 5, 1, 3), the range is determined by subtracting the smallest measurement (1) from the largest (6), yielding \( \text{Range} = 6 - 1 = 5 \).
Although the range is easy to compute and understand, it is sensitive to outliers and does not provide information about the distribution of values within the range. Consequently, for more sophisticated analysis, other measures like standard deviation are used in conjunction with the range to get a better understanding of data variability.
Although the range is easy to compute and understand, it is sensitive to outliers and does not provide information about the distribution of values within the range. Consequently, for more sophisticated analysis, other measures like standard deviation are used in conjunction with the range to get a better understanding of data variability.
Standard Deviation Application
The application of standard deviation is widespread in the field of statistical analysis, as it allows one to understand how spread out a set of data is relative to the mean. Once you have calculated the standard deviation, you can use it to assess the dispersion of data points. For instance, you can determine how many standard deviations the range of a dataset spans. This can provide insight into the concentration of data points around the mean. In the example with the given measurements, by dividing the range (5) by the standard deviation (1.62), we find that the range is approximately 3.09 standard deviations wide \(\text{Approximate Number of Standard Deviations} = \frac{\text{Range}}{\text{Standard Deviation}} = \frac{5}{1.62} \approx 3.09\).
This can help us understand the variability of measurements and is especially useful in contexts such as quality control, finance, and research, where deviation from the norm must be carefully monitored and interpreted. Additionally, understanding standard deviation assists in methodology such as hypothesis testing and confidence interval estimation, making it a cornerstone of statistical analysis.
This can help us understand the variability of measurements and is especially useful in contexts such as quality control, finance, and research, where deviation from the norm must be carefully monitored and interpreted. Additionally, understanding standard deviation assists in methodology such as hypothesis testing and confidence interval estimation, making it a cornerstone of statistical analysis.