3.3 - Normal Distributions Score: $0 / 33$ $0 / 11$ answered Question 10 The combined SAT scores for the students at a local high school are normally distributed with a mean of 1537 and a standard deviation of 303 . The local college includes a minimum score of 2234 in its admission requirements. What percentage of students from this school earn scores that satisfy the admission requirement? Give your answer as a percentage rounded to one decimal place. \[ P(X>2234)=\square \% \] Question Help: $\square$ Message instructor $D$ Post to forum Submit Question

Step 1 ：Calculate the z-score using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(X = 2234\) is the score we are interested in, \(\mu = 1537\) is the mean, and \(\sigma = 303\) is the standard deviation.

Step 2 ：Substitute the values into the formula to get \(Z = \frac{2234 - 1537}{303} = 2.3\).

Step 3 ：Find the percentage of students who have a z-score greater than 2.3. This is equivalent to finding the area under the normal distribution curve to the right of z = 2.3.

Step 4 ：Use a standard normal distribution table (z-table) to find this area. However, the z-table gives us the area to the left of the given z-score. So, we need to subtract the area to the left of z = 2.3 from 1 to get the area to the right of z = 2.3.

Step 5 ：The area to the left of z = 2.3 is approximately 0.9893. So, the area to the right of z = 2.3 is \(1 - 0.9893 = 0.0107\).

Step 6 ：This means that approximately 1.07% of students from this school earn scores that satisfy the admission requirement.

Step 7 ：So, the final answer is \(\boxed{P(X>2234) = 1.1\%}\) (rounded to one decimal place).