Question

SAT Scores The national average SAT score (for Verbal and Math) is

1028. If we assume a normal distribution with σ=92, what is the 90 the percentile score? What is the probability that a randomly selected score exceeds 1200?

Answer 1

To find the 90th percentile SAT score, find the z-score for 0.90 and use the formula to compute the SAT score. To calculate the probability of exceeding 1200, determine the z-score for 1200 and subtract the area to the left of this z-score from 1.

Step-by-step explanation:

The national average SAT score for Verbal and Math combined is 1028 with a standard deviation of 92. To find the 90th percentile score, we need to find the z-score that corresponds to 0.90 in the standard normal distribution table. Once we have the z-score, we calculate the corresponding SAT score using the formula: SAT score = (z-score × standard deviation) + mean.

To find the probability that a randomly selected score exceeds 1200, we first calculate the z-score for 1200 using the formula: z = (X - mean) / standard deviation. Then we look up the z-score in the standard normal distribution table to find the area to the left of this z-score. The probability of a score exceeding 1200 is equal to 1 minus this area. Let's go through these steps in detail:

1. Calculate the z-score corresponding to the 90th percentile. We use a standard normal distribution table or a statistical software to find this z-score. Suppose the z-score is around 1.28 for the 90th percentile (you would need to confirm this exact value with a z-table or software).
2. Compute the 90th percentile SAT score: SAT score = 1.28 × 92 + 1028 = approx. 1198.
3. Calculate the z-score for an SAT score of 1200: z = (1200 - 1028) / 92 = 1.87.
4. Find the area to the left of z = 1.87 using the z-table. Suppose this area is 0.9693. Then the probability that a score exceeds 1200 is 1 - 0.9693 = 0.0307, or 3.07%.