Learning Goals
In this support activity you’ll become familiar with the following:
In the next section of the course material and in the following activity, you will need to be familiar with a method for measuring how far a given data value is from the mean. You’ve seen before how to calculate deviation from the mean for a given data value. This method standardizes the distance in units of standard deviation above or below the mean. The calculations will be performed by hand, and they can be a little tricky. In this corequisite support activity, you’ll get some practice with them and learn where and how they can go wrong.
Measuring Distance from Mean
We’ll use a hypothetical typical arm span for Americans in our exploration of this method for measuring distance from the mean. Arm span is the distance from the tip of the middle finger on one hand to the tip of the middle finger on the other hand when one’s arms are stretched out and opened wide.
Suppose that, based on many measurements, a statistician believes that the distribution of arm spans of Americans has a mean of [latex]173.40[/latex] centimeters (cm) and a standard deviation of [latex]12.21[/latex] cm. Let’s understand that these values to pertain to the population of all American arm spans (not just a sample).
For Questions 1, 3, and 5 below, you’ll calculate arm span values that are one, two, and one-and-a-half standard deviations above and below the mean of [latex]173.40[/latex] cm. Then, you’ll carefully follow the directions in Questions 2, 4, and 6, to make the specified calculations for each of your answers to the previous question. Note that in each of these, you’ll have found the difference between the given value and the mean, and divided that distance by the standard deviation. We call this standardizing the value, but we’ll cover that more deeply in the next section. For now, let’s just get some practice making the necessary calculations by hand.
Standard Deviation
interactive example
For Questions 1 and 2 below, you’ll be given the mean and the standard deviation for the population of arm spans. Here’s how to do that.
Question 1: Find a value that lies one standard deviation above and one below the mean.
To do so, add one standard deviation to the mean to find the the value that lies above the mean. Subtract to find the value that lies one standard deviation below the mean.
Ex. Let’s say a mean is given to be [latex]7.5[/latex] with a standard deviation of [latex]2[/latex],
Add [latex]7.5 + 2 = 9.5[/latex]. The value one standard deviation above the mean is [latex]9.5[/latex].
Now you try it. Find the value that lies one standard deviation below the mean.
Question 2: Standardize the value.
To do so, take the value you calculated in Question 1, subtract the mean, and divide by the standard deviation.
Ex. In the first part above, you found that the value of 9.5 was one standard deviation above the mean.
First, subtract the mean from the value, then divide by the standard deviation. Recall the mean was 7.5 with a standard deviation of 2.
[latex]\dfrac{9.5-7.5}{2}=1[/latex]. That is, the value of 9.5 lies one standard deviation from the mean. Since the calculation is positive, 9.5 is above the mean.
Now you try it. Find how far, in standard deviations, the value 5.5 is from the mean and whether is lies above or below. Hint: if it lies below the mean (and it does) your calculation result must be negative.
Now you try it. In Question 1, add or subtract the standard deviation to or from the mean to calculate the new value.
question 1
In Question 2, apply the formula [latex]\dfrac{\text{value}-\text{mean}}{\text{standard deviation}}[/latex] to each of the values you obtained in Question 1.
question 2
In Question 1 you calculated a value exactly one standard deviation above and one below the mean. In Question 2, you divided the difference between the value you calculated and the mean by the standard deviation. You should have obtained positive one and negative one for the values above and below, respectively. What does positive imply? How about negative?
