Problem 24
Question
In Exercises 17-26, find the standard deviation for each group of data items. Round answers to two decimal places \(6,6,6,6,7,7,7,4,8,3\)
Step-by-Step Solution
Verified Answer
The standard deviation is approximately \(1.41\) when rounded to two decimal places.
1Step 1: Calculate the mean
The mean - also known as the average - is calculated by adding up all of the numbers and dividing by the total count of numbers. In this case the total is \((6+6+6+6+7+7+7+4+8+3) = 60\), and there are 10 numbers. So, the mean is \(60/10 = 6\).
2Step 2: Calculate the variance
First, subtract the mean from each number to get a list of deviations. This results in \((0,0,0,0,1,1,1,-2,2,-3)\). Square all the deviations to get a list of squared deviations: \((0,0,0,0,1,1,1,4,4,9)\). Then, find the average of these squared deviations by adding them all up and dividing by the count. So, the variance is \((0+0+0+0+1+1+1+4+4+9)/10 = 2.0\).
3Step 3: Calculate the standard deviation
The standard deviation is the square root of the variance. So the standard deviation is \(\sqrt{2.0}\)
Key Concepts
Mean CalculationVariance CalculationStatistical Deviation
Mean Calculation
Understanding how to calculate the mean, which is the average of a set of numbers, is fundamental in statistics. To find the mean, you sum up all the values in your dataset and then divide by the number of values. For example, if we have a set of data points like in our exercise \(6,6,6,6,7,7,7,4,8,3\)\, we would add them all together to get a total of \(60\)\. This dataset contains \(10\)\ numbers, so to calculate the mean, we divide the total \(60\)\ by the number of items, \(10\)\, resulting in a mean value of \(6\)\.
Why is the mean important? It represents the center or the 'middle' of the data, and it's a measure of the central tendency of the dataset. However, it's also sensitive to extreme values or outliers. In sets with such outliers, the mean might not be the best representation of the central value, hence the importance of other measures like the median or the mode.
Why is the mean important? It represents the center or the 'middle' of the data, and it's a measure of the central tendency of the dataset. However, it's also sensitive to extreme values or outliers. In sets with such outliers, the mean might not be the best representation of the central value, hence the importance of other measures like the median or the mode.
Variance Calculation
Variance is a statistical measurement that describes the spread of numbers in a data set. It gives us an idea of how much the numbers in the dataset deviate from the mean. Following the mean calculation, to calculate the variance, we need to:
- Subtract the mean from each number to get their deviations. These represent the distance of each data point from the mean.
- Square each deviation to make sure we are only working with positive numbers, as the direction of the deviation is not as important as the magnitude when calculating variance.
- Sum all those squared deviations.
- Divide the total by the count of values in the dataset to get the average squared deviation, which is our variance.
Statistical Deviation
Statistical deviation is a measure of the difference between the observed value and some other value, often the mean of the data set. The term can refer to any of the several different measures of deviation within statistics, but in the context of standard deviation, it reflects how spread out the values in a data set are. High deviation means that the values are spread widely from the mean and low deviation means most of the numbers are very close to the mean. The standard deviation is one of the most common measures of statistical deviation. It gives a sense of how dispersed the data is in relation to the mean. As seen in our exercise, we calculate the standard deviation by taking the square root of the variance \(\sqrt{2.0}\)\. The result will give us a standard deviation that, when rounded to two decimal places as requested, will provide a clear and concise measure of the dispersion of our data set around the mean.
Other exercises in this chapter
Problem 23
In Exercises 17-26, find the standard deviation for each group of data items. Round answers to two decimal places \(1,1,1,4,7,7,7\)
View solution Problem 23
More college graduates had starting salaries in the \(\$ 51,000-\) \(\$ 55,000\) range than in the \(\$ 36,000-\$ 40,000\) range.
View solution Problem 24
If the sample is truly representative, then for a group of 400 college graduates, we can expect about 28 of them to have starting salaries in the \(\$ 31,000-\$
View solution Problem 25
In Exercises 17-26, find the standard deviation for each group of data items. Round answers to two decimal places \(9,5,9,5,9,5,9,5\)
View solution