Question

The mean score on a particular test is 79 and the standard deviation is 5.5. What is the probability that a randomly selected test score is more than 90? Use the Empirical Rule to solve.(1 point) Responses 34% 34% 68% 68% 2.5% 2.5% 95%

Answer 1

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).

1. **Identify the Mean and Standard Deviation**:
- Mean (μ) = 79
- Standard Deviation (σ) = 5.5

2. **Calculate the Z-score** for a score of 90:
\[
Z = \frac{X - \mu}{\sigma} = \frac{90 - 79}{5.5} = \frac{11}{5.5} \approx 2
\]

3. **Using the Empirical Rule**:
- 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).

- 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).

4. **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%**