Question

\(z\) -Scores For the data sets in Exercises \(1-3,\) find the mean the standard deviation, and the z-scores corresponding to the minimum and maximum in the data set. Do the z-scores indicate that there are possible outliers in these data sets? \(n=12\) measurements: 8,7,1,4,6,6,4,5,7,6,3,0

Short answer

Question: Calculate the mean, standard deviation, and z-scores of the minimum and maximum values of the data set {8, 7, 1, 4, 6, 6, 4, 5, 7, 6, 3, 0}, and determine if there are any possible outliers. Answer: The mean of the data set is 4.75, the standard deviation is approximately 2.496, the z-score for the minimum value (0) is approximately -1.900, and the z-score for the maximum value (8) is approximately 1.303. As neither z-score is greater than 2 or less than -2, there are no possible outliers in this data set.

Step-by-step solution

Calculate the mean

To calculate the mean of the data set, we sum up all the values and divide by the total number of measurements, i.e., \(n\). Mean, \(\mu = \frac{\sum x}{n} = \frac{8+7+1+4+6+6+4+5+7+6+3+0}{12} = \frac{57}{12} = 4.75\).

Calculate the standard deviation

To calculate the standard deviation, first, we compute the variance, which is the average of the squared differences from the mean. Variance, \(\sigma^2 = \frac{\sum(x-\mu)^2}{n} = \frac{(8-4.75)^2 + (7-4.75)^2 + \dots + (3-4.75)^2 + (0-4.75)^2}{12}\). \(\sigma^2 \approx 6.229\). Then, we find the standard deviation by taking the square root of the variance: Standard deviation, \(\sigma = \sqrt{\sigma^2} \approx \sqrt{6.229} \approx 2.496\).

Identify the minimum and maximum values

We identify the minimum and maximum values in the data set by inspection: Minimum value: 0 Maximum value: 8

Calculate the z-scores for the minimum and maximum values

The z-score for a data point \(x\) is given by the formula: \(z = \frac{x - \mu}{\sigma}\). For the minimum value (0): \(z_{min} = \frac{0 - 4.75}{2.496} \approx -1.900\). For the maximum value (8): \(z_{max} = \frac{8 - 4.75}{2.496} \approx 1.303\).

Determine if there are possible outliers based on the z-scores

One common rule of thumb for identifying potential outliers is to check if the z-score is greater than 2 standard deviations away from the mean, or in other words, if the z-score is greater than 2 or less than -2. In this case, neither \(z_{min}\) nor \(z_{max}\) is greater than 2 or less than -2: \(-1.900 > -2\) and \(1.303 < 2\). As a result, the z-scores do not indicate that there are possible outliers in this data set.

Key concepts

Mean

The mean is a fundamental concept in statistics, representing the average of a set of numbers. It's calculated by summing up all the values in a dataset and then dividing by the number of observations. This simple arithmetic operation provides a central value that can help reveal the overall trend among the numbers.
In our exercise, with 12 measurements: \(8, 7, 1, 4, 6, 6, 4, 5, 7, 6, 3, 0\), we find the mean by adding these up to get 57, and then dividing by 12.

• Mean (\(\mu\)): \(\frac{57}{12} = 4.75\).

So, 4.75 is the average value indicating the center of our dataset.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of data values. It's a critical concept used to understand how spread out the data is around the mean. A low standard deviation indicates that the values cluster closely around the mean, whereas a high standard deviation suggests a wider spread.
To calculate it, we first determine the variance, which is the mean of the squared differences between each data point and the mean. For our dataset, the variance \(\sigma^2\) is calculated as 6.229.
We then find the standard deviation \(\sigma\) by taking the square root of the variance:

• Standard deviation (\(\sigma\)): \(\sqrt{6.229} \approx 2.496\).

This value tells us how much the data deviates, on average, from the mean of 4.75.

Outliers

Outliers are data points that lie far away from other observations in a dataset. They can skew or misrepresent data analysis, leading to incorrect conclusions. A useful method of identifying potential outliers is by using z-scores.

• Z-score measures how many standard deviations an element is from the mean.
• Common practice marks a z-score above 2 or below -2 as a potential outlier.

For our data set:- Minimum value (0), \(z_{min} = \frac{0 - 4.75}{2.496} \approx -1.900\)- Maximum value (8), \(z_{max} = \frac{8 - 4.75}{2.496} \approx 1.303\)Neither z-value indicates an outlier, as both are well within the acceptable range of -2 to 2.

Variance

Variance provides a numerical value that represents the degree of spread in a dataset. It's essentially the average of the squared differences from the mean, offering an indication of how much the data points differ from the mean.

• Calculated as \(\text{Variance (\(\sigma^2\))} = \frac{(x_1-\mu)^2 + (x_2-\mu)^2 + \ldots + (x_n-\mu)^2}{n}\).

For our dataset, we found the variance to be approximately 6.229. By using variance, we can understand how homogeneous or heterogeneous the data is, which aids in further statistical analysis. Although not directly used to identify outliers, understanding variance is crucial as it leads us to the calculation of standard deviation.