Chapter 23: Problem 1
Calculate (a) the mean and (b) the standard deviation of the data set $$ 17,26,31,19,25,20,19,29,11,27 $$
Short Answer
Expert verified
Answer: The mean of the data set is 22.4, and the standard deviation is approximately 5.89.
Step by step solution
01
Find the Total Number of Data Points
To find the mean and standard deviation of a set of numbers, we first need to determine the total number of data points (n) in the set. In our case, there are 10 data points:
$$
17, 26, 31, 19, 25, 20, 19, 29, 11, 27
$$
So, n = 10.
02
Calculate the Sum of the Data Points
To find the mean, we need to calculate the sum of all data points. Add together all the numbers in the data set:
$$
17 + 26 + 31 + 19 + 25 + 20 + 19 + 29 + 11 + 27 = 224
$$
03
Calculate the Mean
Now we can find the mean by dividing the sum of the data points by the total number of data points (n):
$$
\text{Mean} = \frac{\text{Sum of Data Points}}{\text{Total Number of Data Points}} = \frac{224}{10} = 22.4
$$
So, the mean of the data set is 22.4.
04
Calculate the Deviation of Each Data Point from the Mean
Next, we need to find the deviation of each data point from the mean. To do this, subtract the mean from each data point and square the result:
$$
(17 - 22.4)^2 = 29.16
$$
$$
(26 - 22.4)^2 = 12.96
$$
...
$$
(27 - 22.4)^2 = 21.16
$$
05
Calculate the Sum of the Squared Deviations
Now, add up all the squared deviations we calculated in Step 4:
$$
29.16 + 12.96 + 74.16 + 11.56 + 6.76 + 5.76 + 11.56 + 43.56 + 129.96 + 21.16 = 346.68
$$
06
Calculate the Variance
Now we will calculate the variance by dividing the sum of the squared deviations by the total number of data points (n):
$$
\text{Variance} = \frac{\text{Sum of Squared Deviations}}{\text{Total Number of Data Points}} = \frac{346.68}{10} = 34.668
$$
07
Calculate the Standard Deviation
Finally, we can find the standard deviation by calculating the square root of the variance:
$$
\text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{34.668} \approx 5.89
$$
So, the standard deviation of the data set is approximately 5.89.
To summarize, we found that:
(a) The mean of the data set is 22.4.
(b) The standard deviation of the data set is approximately 5.89.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mean Calculation
The mean, often referred to as the average, is a measure that represents the typical value of a set of numbers. It is calculated by adding together all the values in a data set and then dividing the result by the total number of points in the set.
For example, if we are given a data set such as \( 17, 26, 31, 19, 25, 20, 19, 29, 11, 27 \), the sum of these numbers is \( 224 \). With 10 numbers in this data set, we divide the sum by 10, resulting in a mean of \( 22.4 \). The mean gives us a central value which can be useful in understanding the overall distribution of data points.
For example, if we are given a data set such as \( 17, 26, 31, 19, 25, 20, 19, 29, 11, 27 \), the sum of these numbers is \( 224 \). With 10 numbers in this data set, we divide the sum by 10, resulting in a mean of \( 22.4 \). The mean gives us a central value which can be useful in understanding the overall distribution of data points.
Variance Calculation
Variance is a statistical measure that tells us how much the individual data points in a set vary from the mean. To calculate the variance, we first need to determine each data point's deviation from the mean by subtracting the mean from the data point value. Each of these deviations is then squared to remove negative values.
The formula for variance \( s^2 \) is given by the sum of squared deviations divided by the total number of data points minus one (in cases where we calculate sample variance) or just by the number of data points (for population variance). In our example, the variance is calculated as \( \frac{346.68}{10} = 34.668 \) by dividing the sum of squared deviations by the data points.
The formula for variance \( s^2 \) is given by the sum of squared deviations divided by the total number of data points minus one (in cases where we calculate sample variance) or just by the number of data points (for population variance). In our example, the variance is calculated as \( \frac{346.68}{10} = 34.668 \) by dividing the sum of squared deviations by the data points.
Squared Deviations
Squared deviations are used in the process of calculating variance and involve squaring the difference between each data point and the mean. This step is crucial because it achieves two things: it makes all the differences positive (as squaring a negative number results in a positive) and it gives more weight to larger deviations.
In our data set, for a data point like 17, the deviation from the mean (22.4) is \( -5.4 \), and its squared deviation is \( (-5.4)^2 = 29.16 \). Calculating the squared deviations for each data point and summing them up is vital for the next step in calculating variance and standard deviation.
In our data set, for a data point like 17, the deviation from the mean (22.4) is \( -5.4 \), and its squared deviation is \( (-5.4)^2 = 29.16 \). Calculating the squared deviations for each data point and summing them up is vital for the next step in calculating variance and standard deviation.
Data Set Analysis
Analyzing a data set involves using statistical measures like the mean and standard deviation to describe its characteristics. The mean provides a central value, while the standard deviation tells us about the spread of the data. In other words, standard deviation helps us understand how concentrated or spread out the data points are around the mean.
A small standard deviation suggests that the data points are closely clustered around the mean, whereas a high standard deviation indicates that the data points are spread out over a wider range of values. For the provided data set, with a mean of 22.4 and a standard deviation of approximately 5.89, one can conclude that while there are variations in the data, they are not extremely wide since the standard deviation is relatively small compared to the mean.
A small standard deviation suggests that the data points are closely clustered around the mean, whereas a high standard deviation indicates that the data points are spread out over a wider range of values. For the provided data set, with a mean of 22.4 and a standard deviation of approximately 5.89, one can conclude that while there are variations in the data, they are not extremely wide since the standard deviation is relatively small compared to the mean.