The average grade of second year Business Management tudents in a certain college is normally distributed with mean of 84 and a standard deviation of 5 . What is the mallest interval that contains the middle \( 90 \% \) of scores of the said students? Use whole numbers as the bounds of your interval. If our answer is \( 60-80 \), write: 60 to 80

To find the smallest interval that contains the middle **90%** of the scores for second-year Business Management students, we can use the properties of the normal distribution. ### Given: - **Mean (\(\mu\))**: 84 - **Standard Deviation (\(\sigma\))**: 5 - **Desired Confidence Level**: 90% ### Steps: 1. **Determine the Z-scores for 90% Confidence:** - For a 90% confidence interval, there is 5% in each tail of the distribution. - The Z-score corresponding to the 95th percentile (since 100% - 5% = 95%) is approximately **1.645**. 2. **Calculate the Interval:** - **Lower Bound**: \(\mu - Z \times \sigma = 84 - 1.645 \times 5 = 84 - 8.225 = 75.775\) - **Upper Bound**: \(\mu + Z \times \sigma = 84 + 1.645 \times 5 = 84 + 8.225 = 92.225\) 3. **Round to Whole Numbers:** - Since we need whole numbers, we round **75.775** up to **76** and **92.225** down to **92**. ### Final Interval: **76 to 92** This interval is the smallest whole-number range that captures the middle 90% of the students' scores. **Answer:** 76 to 92