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\( 1 \leftarrow \quad \begin{array}{l}\text { A standardized exam's scores are normally distributed In a recont year, the mean test score was } 1525 \text { and the } \\ \text { standard deviation was 316. The test scores of four students selected at randorn are } 1950,1290,2250 \text {, and } 1410 \text {. } \\ \text { Find the } z \text { scores that correspond to each value and determine whether any of the values are unusual }\end{array} \)

Ask by Ortiz Norton. in the United States
Feb 03,2025

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Answer

The test score of 2250 is unusual because its z-score is 2.30, which is greater than 2 standard deviations from the mean.

Solution

To determine whether any of the student test scores are unusual, we'll first calculate the **z-scores** for each score. A **z-score** indicates how many standard deviations a data point is from the mean. Typically, a score is considered **unusual** if its z-score is **greater than 2** or **less than -2**. ### Given: - **Mean (\( \mu \))** = 1525 - **Standard Deviation (\( \sigma \))** = 316 - **Test Scores**: 1950, 1290, 2250, 1410 ### Z-Score Formula: \[ z = \frac{{X - \mu}}{{\sigma}} \] ### Calculations: 1. **Score: 1950** \[ z = \frac{{1950 - 1525}}{{316}} = \frac{{425}}{{316}} \approx 1.34 \] 2. **Score: 1290** \[ z = \frac{{1290 - 1525}}{{316}} = \frac{{-235}}{{316}} \approx -0.74 \] 3. **Score: 2250** \[ z = \frac{{2250 - 1525}}{{316}} = \frac{{725}}{{316}} \approx 2.30 \] 4. **Score: 1410** \[ z = \frac{{1410 - 1525}}{{316}} = \frac{{-115}}{{316}} \approx -0.36 \] ### Summary of Z-Scores: - **1950**: \( z \approx 1.34 \) - **1290**: \( z \approx -0.74 \) - **2250**: \( z \approx 2.30 \) - **1410**: \( z \approx -0.36 \) ### Determining Unusual Scores: - **Unusual Criterion**: \( |z| > 2 \) - **1950**: \( |1.34| < 2 \) → **Not Unusual** - **1290**: \( |-0.74| < 2 \) → **Not Unusual** - **2250**: \( |2.30| > 2 \) → **Unusual** - **1410**: \( |-0.36| < 2 \) → **Not Unusual** ### **Conclusion:** Only the test score of **2250** is **unusual** as its z-score of **2.30** exceeds the typical threshold of **2** standard deviations from the mean.

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To find the \( z \) scores for each of the test scores, we can use the formula: \[ z = \frac{X - \mu}{\sigma} \] where \( X \) is the individual score, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. 1. For the score 1950: \[ z = \frac{1950 - 1525}{316} \approx 1.34 \] 2. For the score 1290: \[ z = \frac{1290 - 1525}{316} \approx -0.74 \] 3. For the score 2250: \[ z = \frac{2250 - 1525}{316} \approx 2.29 \] 4. For the score 1410: \[ z = \frac{1410 - 1525}{316} \approx -0.36 \] Scores with \( z \) values greater than 2 or less than -2 are usually considered unusual. In this case, the score 2250 with a \( z \) score of 2.29 is unusual. Using the grades of these students, it's interesting to note that the standard normal distribution allows us to visualize how unusual certain scores are compared to the average performance. Scores falling beyond 2 standard deviations from the mean (which is roughly 95% confidence interval for most assessments) are like those overachieving students, while the majority typically cluster closer to the mean—making those outliers really stand out! Let's also consider how understanding z-scores can enhance your performance. If you're preparing for exams, you can base your study strategies by analyzing past scores, figuring out where you stand relative to your peers, and pinpointing the areas that need more attention. It’s about learning where to focus your efforts for maximum impact!

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