Loading...
Which of the following triples of numbers have the same standard deviation as the numbers \(\mathrm{r}, \mathrm{s}, \text{ and } \mathrm{t}\)?
Let's start by understanding what we're really being asked. We have three numbers: r, s, and t. We want to know which of the three given sets have the same "spread" or standard deviation as our original set.
Think of standard deviation as measuring how "spread out" numbers are from their average. If you have test scores of \(\mathrm{80}, \mathrm{85}, \mathrm{90}\), they're less spread out than scores of \(\mathrm{60}, \mathrm{85}, \mathrm{110}\), even though both sets have the same average of \(\mathrm{85}\).
Our job is to determine which transformations of \(\{\mathrm{r}, \mathrm{s}, \mathrm{t}\}\) keep this "spread" exactly the same.
Process Skill: TRANSLATE - Converting the abstract concept of "same standard deviation" into the concrete idea of "same spread from the average"
Let's think about what happens when we add or subtract the same number from every value in our set.
Imagine you have test scores: \(\mathrm{70}, \mathrm{80}, \mathrm{90}\). The average is \(\mathrm{80}\), and the individual scores are \(\mathrm{10}\) points below, right at, and \(\mathrm{10}\) points above the average.
Now, what if the teacher adds \(\mathrm{5}\) bonus points to everyone's score? We get: \(\mathrm{75}, \mathrm{85}, \mathrm{95}\). The new average becomes \(\mathrm{85}\), and notice that each score is still \(\mathrm{10}\) points below, right at, and \(\mathrm{10}\) points above this new average.
The key insight: When we shift all numbers by the same amount, the average shifts by that same amount, but the distances from the average stay exactly the same. Therefore, the standard deviation doesn't change.
Option I: \(\mathrm{r-2}, \mathrm{s-2}, \mathrm{t-2}\)
This is exactly what we just described - we're subtracting \(\mathrm{2}\) from every number. The spread stays the same, so this option preserves the standard deviation.
Now let's consider what happens when we don't shift every number by the same amount. This changes the relative positions of the numbers, which affects how spread out they are.
Option III: \(\mathrm{r-4}, \mathrm{s+5}, \mathrm{t-1}\)
Here we're subtracting \(\mathrm{4}\) from r, adding \(\mathrm{5}\) to s, and subtracting \(\mathrm{1}\) from t. These are different changes to each number.
Going back to our test score example: if we started with \(\mathrm{70}, \mathrm{80}, \mathrm{90}\) and applied changes of \(\mathrm{-4}, +5, -1\), we'd get \(\mathrm{66}, \mathrm{85}, \mathrm{89}\). The pattern of distances from the average has completely changed - this will have a different standard deviation from our original set.
Since we're applying different shifts to different numbers, this changes the relative spacing and therefore changes the standard deviation.
Let's examine the trickiest option:
Option II: \(\mathrm{0}, \mathrm{r-s}, \mathrm{t-s}\)
This looks different, but let's think carefully. We're taking our original values \(\{\mathrm{r}, \mathrm{s}, \mathrm{t}\}\) and subtracting s from each one to get \(\{\mathrm{r-s}, \mathrm{s-s}, \mathrm{t-s}\} = \{\mathrm{r-s}, \mathrm{0}, \mathrm{t-s}\}\).
Wait! This is actually the same as Option I - we're subtracting the same value (s) from every number in our original set. Just like subtracting \(\mathrm{2}\) from every number, subtracting s from every number shifts all values uniformly.
The fact that one of our new values is \(\mathrm{0}\) doesn't matter - what matters is that we applied the same shift to every original value. The relative distances between the numbers remain unchanged, so the standard deviation is preserved.
Process Skill: INFER - Recognizing that \(\{\mathrm{0}, \mathrm{r-s}, \mathrm{t-s}\}\) is actually \(\{\mathrm{r-s}, \mathrm{s-s}, \mathrm{t-s}\}\), which is a uniform shift of the original set
Summary:
Therefore, Options I and II both preserve the original standard deviation.
The answer is C: I and II only
1. Misunderstanding what standard deviation measures: Students may think standard deviation is about the actual values rather than the spread or variability. They might assume that if numbers change, the standard deviation must change, without recognizing that standard deviation specifically measures how spread out the data points are from their mean.
2. Not recognizing uniform transformations: Students may fail to identify that Option II (\(\mathrm{0}, \mathrm{r-s}, \mathrm{t-s}\)) is actually a uniform shift. They might see the presence of \(\mathrm{0}\) and the expression (\(\mathrm{r-s}\)) and think this is a completely different type of transformation, missing that it's equivalent to subtracting s from each original value.
3. Confusing different types of transformations: Students might not distinguish between uniform shifts (adding/subtracting the same value to all numbers) versus non-uniform changes (adding/subtracting different values). They may incorrectly assume all mathematical operations on the original set will preserve standard deviation.
1. Incorrect analysis of Option II: Even if students recognize that Option II involves subtracting s from each value, they might get confused by the reordering (putting \(\mathrm{0}\) first instead of \(\mathrm{r-s}, \mathrm{s-s}, \mathrm{t-s}\) in original order) and incorrectly conclude it changes the standard deviation.
2. Arithmetic errors in Option III: Students might attempt to use specific numerical examples to test Option III but make calculation errors when computing means and standard deviations, leading to incorrect conclusions about whether the standard deviation is preserved.
1. Partial credit selection: Students who correctly identify that Option I preserves standard deviation but incorrectly analyze Option II might select "A. I only" instead of the correct answer "C. I and II only".
2. Overinclusive selection: Students who incorrectly conclude that Option III also preserves standard deviation (perhaps by misunderstanding the effect of non-uniform transformations) might select "E. I, II, and III" instead of "C. I and II only".