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| X | Frequency |
|---|---|
| 1 | 3 |
| 2 | 1 |
| 3 | 3 |
| 4 | 1 |
| 5 | 3 |
| 6 | 1 |
| 7 | 3 |
The variable \(\mathrm{x}\) takes on integer values between \(1\) and \(7\) inclusive, as shown above. What is the probability that the absolute value of the difference between the mean of the distribution which is \(4\) and a randomly chosen value of \(\mathrm{x}\) will be greater than \(\frac{3}{2}\)?
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
Now let's check each possible value from 1 to 7 to see which ones are more than 1.5 units away from 4:
So the qualifying values are \(\mathrm{x} = 1, 2, 6, \text{ and } 7\).
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.
Now we count how many times our qualifying values (1, 2, 6, 7) appear in the distribution:
Favorable outcomes = \(3 + 1 + 1 + 3 = 8\)
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.
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) \(\frac{8}{15}\).
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.
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.
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.