Final answer:
To find the percentage of students scoring at least 600 on the SAT verbal test, we can use the z-score formula and the standard normal distribution table. Approximately 78.32% of students scored at least 600.
Step-by-step explanation:
The distribution of scores on the SAT verbal test in recent years follows a normal distribution with a mean of 515 and a standard deviation of 109 (approximately). To find the percentage of students scoring at least 600, we need to calculate the area under the normal curve to the right of 600. This can be done by using the z-score formula and the standard normal distribution table.
First, we need to find the z-score for 600 using the formula:
z = (x - μ) / σ
where x is the value of interest, μ is the mean, and σ is the standard deviation.
Plugging in the values:
z = (600 - 515) / 109
z ≈ 0.7798
Next, we can use the standard normal distribution table (also known as the z-score table) to find the percentage of students scoring at least 600. Looking up the z-score of 0.78 in the table, we find that the corresponding area is approximately 0.7832. This means that approximately 78.32% of students scored at least 600 on the SAT verbal test.