FREQUENCY DISTRIBUTION OF THE VARIABLE X XFrequency 13 21 33 41 53 61 73 The variable x takes on integer...

1. Translate the problem requirements: We need to find the probability that \(|\mathrm{x} - 4| > \frac{3}{2}\), where x is randomly chosen from the given frequency distribution. This means we need values where the distance from 4 is greater than 1.5.
2. Identify qualifying values: Determine which x values satisfy \(|\mathrm{x} - 4| > \frac{3}{2}\) by checking each possible value from 1 to 7.
3. Calculate total sample space: Add up all frequencies to find the total number of outcomes possible.
4. Count favorable outcomes: Sum the frequencies of the qualifying x values to get favorable outcomes.
5. Compute probability ratio: Divide favorable outcomes by total outcomes to get the final probability.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking in plain English. We have a distribution of values from 1 to 7, each appearing with different frequencies. We're told the mean is 4, and we need to find the probability that a randomly chosen value will be more than \(\frac{3}{2}\) units away from this mean.

In everyday terms, we're looking for values that are "far enough" from 4. Since \(\frac{3}{2} = 1.5\), we want values where the distance from 4 is greater than 1.5.

Mathematically, this means we need \(|\mathrm{x} - 4| > \frac{3}{2}\), or \(|\mathrm{x} - 4| > 1.5\).

Process Skill: TRANSLATE - Converting the probability language into a concrete mathematical condition

2. Identify qualifying values

Now let's check each possible value from 1 to 7 to see which ones are more than 1.5 units away from 4:

• \(\mathrm{x} = 1\): Distance from 4 is \(|1 - 4| = 3\). Since \(3 > 1.5\), this qualifies.
• \(\mathrm{x} = 2\): Distance from 4 is \(|2 - 4| = 2\). Since \(2 > 1.5\), this qualifies.
• \(\mathrm{x} = 3\): Distance from 4 is \(|3 - 4| = 1\). Since \(1 < 1.5\), this does NOT qualify.
• \(\mathrm{x} = 4\): Distance from 4 is \(|4 - 4| = 0\). Since \(0 < 1.5\), this does NOT qualify.
• \(\mathrm{x} = 5\): Distance from 4 is \(|5 - 4| = 1\). Since \(1 < 1.5\), this does NOT qualify.
• \(\mathrm{x} = 6\): Distance from 4 is \(|6 - 4| = 2\). Since \(2 > 1.5\), this qualifies.
• \(\mathrm{x} = 7\): Distance from 4 is \(|7 - 4| = 3\). Since \(3 > 1.5\), this qualifies.

So the qualifying values are \(\mathrm{x} = 1, 2, 6, \text{ and } 7\).

3. Calculate total sample space

To find the total number of possible outcomes, we add up all the frequencies from the table:

Total outcomes = \(3 + 1 + 3 + 1 + 3 + 1 + 3 = 15\)

This means there are 15 total ways to randomly select a value from this distribution.

4. Count favorable outcomes

Now we count how many times our qualifying values (1, 2, 6, 7) appear in the distribution:

• \(\mathrm{x} = 1\) appears 3 times
• \(\mathrm{x} = 2\) appears 1 time
• \(\mathrm{x} = 6\) appears 1 time
• \(\mathrm{x} = 7\) appears 3 times

Favorable outcomes = \(3 + 1 + 1 + 3 = 8\)

5. Compute probability ratio

Probability is simply the ratio of favorable outcomes to total outcomes:

Probability = \(\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{8}{15}\)

This fraction is already in its simplest form since 8 and 15 share no common factors other than 1.

Final Answer

The probability that the absolute value of the difference between the mean (4) and a randomly chosen value of x will be greater than \(\frac{3}{2}\) is \(\frac{8}{15}\).

This matches answer choice (A) \(\frac{8}{15}\).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the absolute value condition
Students often struggle with translating "\(|\mathrm{x} - 4| > \frac{3}{2}\)" into practical terms. They may incorrectly think this means "x must be greater than \(4 + \frac{3}{2}\)" and miss that values less than \(4 - \frac{3}{2}\) also qualify. The key insight is that absolute value creates a "two-sided" condition - values can be either too far above OR too far below the mean.

2. Confusion about what constitutes a "randomly chosen value"
Students may think they need to calculate probability using just the unique values (1,2,3,4,5,6,7) rather than accounting for frequencies. They might incorrectly assume each value has equal probability of \(\frac{1}{7}\), forgetting that frequency affects the likelihood of selection.

3. Misunderstanding the boundary condition
Since we need \(|\mathrm{x} - 4| > \frac{3}{2}\) (strictly greater than), students may incorrectly include values where \(|\mathrm{x} - 4| = \frac{3}{2}\). With \(\frac{3}{2} = 1.5\), they might mistakenly include boundary cases if they existed, not recognizing the strict inequality requirement.

Errors while executing the approach

1. Arithmetic errors in distance calculations
When calculating \(|\mathrm{x} - 4|\) for each value, students commonly make sign errors or absolute value mistakes. For example, they might calculate \(|3 - 4|\) as +1 instead of 1, or more critically, forget to take the absolute value entirely, leading to incorrect qualification of values.

2. Incorrect frequency summation
Students may make addition errors when calculating either the total outcomes \((3+1+3+1+3+1+3)\) or favorable outcomes \((3+1+1+3)\). These seemingly simple arithmetic mistakes can lead to completely wrong probability calculations, especially under time pressure.

Errors while selecting the answer

1. Fraction simplification errors
After correctly calculating \(\frac{8}{15}\), students might attempt to "simplify" this fraction incorrectly, not recognizing that 8 and 15 share no common factors other than 1. They may force an incorrect simplification and match it to a wrong answer choice.

2. Calculating the complement probability
Students who correctly find that 7 values don't satisfy the condition might calculate \(\frac{7}{15}\) instead of \(\frac{8}{15}\), essentially finding \(\mathrm{P}(|\mathrm{x}-4| \leq \frac{3}{2})\) instead of \(\mathrm{P}(|\mathrm{x}-4| > \frac{3}{2})\). This is a classic error of solving for the opposite event.