Question

The national average SAT score (for Verbal and Math) is 1028. If we assume a normal distribution with standard deviation 92, what is the probability that a randomly selected score exceeds 1200? Express your answer as a decimal. Round your answer to the nearest hundredth.

Answer 1

To find the probability that a randomly selected SAT score exceeds 1200, we need to use the properties of the normal distribution.

1. **Define the parameters**:
- Mean (\(\mu\)) = 1028
- Standard Deviation (\(\sigma\)) = 92

2. **Standardize the score**: We will convert the score of 1200 into a z-score using the formula:
\[
z = \frac{X - \mu}{\sigma}
\]
where \(X\) is the score of interest.

3. **Calculate the z-score**:
\[
z = \frac{1200 - 1028}{92} = \frac{172}{92} \approx 1.8696
\]

4. **Use the z-score to find the probability**: We now need to find the probability that a z-score is greater than 1.8696. This can be found using the standard normal distribution table, or by using a calculator or statistical software.

5. **Find the probability for \(z\)**:
- We need to find \(P(Z > 1.8696)\).
- First, we find \(P(Z < 1.8696)\). Using the standard normal distribution table or a calculator, we find that:
\[
P(Z < 1.8696) \approx 0.9699
\]
- To find \(P(Z > 1.8696)\):
\[
P(Z > 1.8696) = 1 - P(Z < 1.8696) = 1 - 0.9699 = 0.0301
\]

6. **Round the result**:
To express the probability as a decimal rounded to the nearest hundredth:
\[
P(Z > 1.8696) \approx 0.0301 \approx 0.03
\]

Thus, the probability that a randomly selected score exceeds 1200 is \( \boxed{0.03} \).