70, 75, 80, 85, 90, 105, 105, 130, 130, 130 The list shown consists of the times, in seconds, that...

1. Translate the problem requirements: We need to find how many running times are more than 1 standard deviation below the mean. This means finding values that are less than \(\mathrm{mean} - 1 \times \mathrm{standard\ deviation}\).
2. Calculate the mean of the running times: Add all 10 values and divide by 10 to find the average running time.
3. Determine the threshold value: Subtract 1 standard deviation (22.4 seconds) from the mean to find the cutoff point.
4. Count values below the threshold: Go through the list and count how many running times are less than this threshold value.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what this question is really asking. We have 10 running times and we need to find how many are "more than 1 standard deviation below the mean." In everyday terms, this means we're looking for the times that are significantly faster than average - specifically, times that are so much faster that they fall below a certain cutoff point.

Think of it this way: if the average running time is around 100 seconds, and the standard deviation is 22.4 seconds, then "1 standard deviation below the mean" would be around \(100 - 22.4 = 77.6\) seconds. We want times that are even faster than this cutoff - meaning times that are less than 77.6 seconds.

Process Skill: TRANSLATE - Converting "more than 1 standard deviation below the mean" to "less than \(\mathrm{mean} - 1 \times \mathrm{standard\ deviation}\)"

2. Calculate the mean of the running times

To find the average running time, we add up all 10 times and divide by 10.

Our times are: 70, 75, 80, 85, 90, 105, 105, 130, 130, 130

Let's add them up step by step to avoid errors:
\(70 + 75 = 145\)
\(145 + 80 = 225\)
\(225 + 85 = 310\)
\(310 + 90 = 400\)
\(400 + 105 = 505\)
\(505 + 105 = 610\)
\(610 + 130 = 740\)
\(740 + 130 = 870\)
\(870 + 130 = 1000\)

Total = 1000 seconds

Mean = \(1000 ÷ 10 = 100\) seconds

So the average running time is exactly 100 seconds - a nice round number that makes our calculations easier!

3. Determine the threshold value

Now we need to find the cutoff point. We know:
- Mean = 100 seconds
- Standard deviation = 22.4 seconds

The threshold for "1 standard deviation below the mean" is:
Threshold = Mean - 1 standard deviation
Threshold = \(100 - 22.4 = 77.6\) seconds

So we're looking for running times that are less than 77.6 seconds (these are the times that are "more than 1 standard deviation below the mean").

4. Count values below the threshold

Now let's go through our list of times and see which ones are less than 77.6 seconds:

Times: 70, 75, 80, 85, 90, 105, 105, 130, 130, 130

Checking each one:
- 70 seconds: \(70 < 77.6\) ✓ (This counts)
- 75 seconds: \(75 < 77.6\) ✓ (This counts)
- 80 seconds: \(80 < 77.6\) ✗ (This doesn't count - 80 is greater than 77.6)
- 85 seconds: \(85 < 77.6\) ✗ (This doesn't count)
- All remaining times (90, 105, 105, 130, 130, 130) are even larger, so they don't count

Therefore, exactly 2 running times (70 seconds and 75 seconds) are more than 1 standard deviation below the mean.

5. Final Answer

Two running times are more than 1 standard deviation below the mean.

The answer is B. two

Common Faltering Points

Errors while devising the approach

Students often confuse this phrase and think it means "more than 1 standard deviation above the mean" or "exactly 1 standard deviation below." The key insight is that "more than 1 standard deviation below" means we're looking for values that are LESS THAN \(\mathrm{mean} - 1 \times \mathrm{standard\ deviation}\). Since lower times indicate better performance, these are the exceptionally fast runners.

Even when students correctly identify the threshold as \(\mathrm{mean} - 1 \times \mathrm{standard\ deviation}\), they may set up the wrong inequality. They might look for values greater than 77.6 instead of less than 77.6. Remember: "below the mean" translates to "less than" the threshold value.

Errors while executing the approach

With 10 different values to add (70, 75, 80, 85, 90, 105, 105, 130, 130, 130), students often make addition mistakes. A systematic approach of adding step-by-step or grouping numbers strategically (like pairing 70+130=200, 75+85=160, etc.) can help avoid these errors.

Students might accidentally add instead of subtract when finding the threshold, calculating \(100 + 22.4 = 122.4\) instead of \(100 - 22.4 = 77.6\). This fundamental error would lead them to look for values above 122.4, finding only the three 130-second times.

When checking if \(80 < 77.6\), some students might incorrectly include 80 as being "below" the threshold, especially if they're working quickly. Since \(80 > 77.6\), the 80-second time should NOT be counted.

Errors while selecting the answer

Students might correctly identify the threshold as 77.6 but then mistakenly count values like 80 and 85 as being "close enough" to qualify. Only 70 and 75 are actually less than 77.6, giving us exactly 2 values, not 3 or 4.