To find the probability that a randomly selected test score is more than 90, we can apply the Empirical Rule (also known as the 68-95-99.7 rule).
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Identify the Mean and Standard Deviation:
- Mean (μ) = 79
- Standard Deviation (σ) = 5.5
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Calculate the Z-score for a score of 90: \[ Z = \frac{X - \mu}{\sigma} = \frac{90 - 79}{5.5} = \frac{11}{5.5} \approx 2 \]
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Using the Empirical Rule:
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According to the Empirical Rule:
- About 68% of the data falls within 1 standard deviation (μ ± σ).
- About 95% falls within 2 standard deviations (μ ± 2σ).
- About 99.7% falls within 3 standard deviations (μ ± 3σ).
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For a Z-score of 2 (which is 2 standard deviations above the mean), approximately 95% of the scores lie within 2 standard deviations (between 68 and 90).
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Calculate the Probability:
- This means that about 95% of test scores are less than 90, leaving a probability of: \[ 100% - 95% = 5% \]
Since we are looking for the probability that a score is more than 90, it is half of the remaining 5% (as it is a normal distribution), which results in approximately 2.5%.
Final Answer: 2.5%