The first quartile, denoted as \( Q_1 \), is the value that separates the lowest 25% of data from the remaining 75%. To find \( Q_1 \) for a normally distributed set of IQ scores with a mean of 100.6 and a standard deviation of 15.9, we can use the z-score formula.
Step 1: Convert the quartile position to a z-score. Since \( Q_1 \) corresponds to the 25th percentile, the z-score is -0.674 (look up in z-table or use a calculator).
Step 2: Use the formula \( x = \mu + z \cdot \sigma \), where \( x \) is the desired value, \( \mu \) is the mean, \( z \) is the z-score, and \( \sigma \) is the standard deviation.
\( Q_1 = 100.6 + (-0.674) \cdot 15.9 \)
Main answer: The first quartile, \( Q_1 \), is approximately 89.31.
Answer within 100 words: The first quartile, \( Q_1 \), is a value that separates the lowest 25% of the data from the remaining 75%. To find \( Q_1 \) for a set of IQ scores that are normally distributed with a mean of 100.6 and a standard deviation of 15.9, we can use the z-score formula. First, we need to convert the quartile position to a z-score. Since \( Q_1 \) corresponds to the 25th percentile, the z-score is -0.674. Next, we use the formula \( x = \mu + z \cdot \sigma \), where \( x \) is the desired value, \( \mu \) is the mean, \( z \) is the z-score, and \( \sigma \) is the standard deviation. Plugging in the values, we get \( Q_1 = 100.6 + (-0.674) \cdot 15.9 \). Therefore, the first quartile, \( Q_1 \), is approximately 89.31.
Conclusion: The first quartile, \( Q_1 \), for the IQ scores is approximately 89.31. This means that 25% of the adults have IQ scores below 89.31, while the remaining 75% have scores above this value.