To find the percentage of students who scored between 82 and 86 on a normally distributed final exam with a mean score of 86 and a standard deviation of 2, we first need to calculate the z-scores for the scores of 82 and 86.
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Calculate the z-score for 82: \[ z = \frac{(X - \mu)}{\sigma} \] where \( X \) is the score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
Here, \( X = 82 \), \( \mu = 86 \), and \( \sigma = 2 \).
\[ z_{82} = \frac{(82 - 86)}{2} = \frac{-4}{2} = -2 \]
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Calculate the z-score for 86: \[ z_{86} = \frac{(86 - 86)}{2} = \frac{0}{2} = 0 \]
Now, we need to find the area under the normal distribution curve between these two z-scores, \( z_{82} = -2 \) and \( z_{86} = 0 \).
From the normal distribution properties, we know:
- The area from \( z = -2 \) to \( z = 0 \) can be found as:
- The percentage of area from \( z < 0 \) is 50% (since the normal distribution is symmetric).
- The area from \( z < -2 \) corresponds to the z-score area given in the problem.
In a standard normal distribution table, the cumulative area up to \( z = -2 \) is about 2.28%. So, the percentage of students who scored between 82 and 86 is:
\[ P(82 < X < 86) = P(Z < 0) - P(Z < -2) = 50% - 2.28% = 47.72% \]
Rounding this, we can say this is approximately 47.7%.
So, the answer is B. 47.7%.
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