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.

1 answer

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 \( \boxed{0.03} \).