The College Board reported that SAT scores in the early 2000 s were normally distributed with mean 1026 and standard deviation 210 . Refer to the table of values $\Theta$ Area Under the Standard Normal Distribution as needed.
Part: $0 / 2$
Part 1 of 2
(a) What percent of students scored over 606 ?
$\%$ of students scored over 606 . (Round to the nearest tenth of a percent.)
Answer
\(\boxed{97.7\%}\) of students scored over 606
Step 1 ：Convert the score to a z-score using the formula: \(z = \frac{x - \mu}{\sigma}\) where \(x = 606\), \(\mu = 1026\), and \(\sigma = 210\)
Step 2 ：Calculate the z-score: \(z = \frac{606 - 1026}{210} = -2.0\)
Step 3 ：Find the area to the left of the z-score using the standard normal distribution table (Θ): \(0.02275\)
Step 4 ：Subtract the area to the left from 1 to find the area to the right of the z-score: \(1 - 0.02275 = 0.97725\)
Step 5 ：Convert the area to the right of the z-score to a percentage: \(0.97725 * 100 = 97.725\%\)
Step 6 ：\(\boxed{97.7\%}\) of students scored over 606
