On a national standardized test, with normally distributed results, the middle 68% of students scored between 40 points and 70 points, and the mean score was 55 points.

Based on this information, which of the following represents a score in the 97.5th percentile?

According to normal distribution,

Mean = 55
55-40 = 70-55 = 15 = SD
and 97.5 is 2SD from mean.

So , 85 is 97.5th percentile.

score is 2

To find the score in the 97.5th percentile, we need to understand that a percentile represents the percentage of scores that fall below a given value.

Given that the middle 68% of students scored between 40 points and 70 points, we can deduce that the remaining 32% (100% - 68%) of students are distributed equally in the tails of the distribution.

Since the distribution is normal, we can use the standard deviation (σ) to determine the score at the 97.5th percentile.

First, we need to find the standard deviation. Since the mean is 55 points, we know that the middle 50% (68% / 2) of students, or the area between the mean and one standard deviation (σ) on both sides, is 15 points (70 - 55).

Therefore, σ = (70 - 55) / 1 = 15.

Next, we need to find the z-score corresponding to the desired percentile. In this case, the z-score at the 97.5th percentile is 1.96.

Finally, we can find the score at the 97.5th percentile using the formula:

Score = z-score * σ + mean
Score = 1.96 * 15 + 55
Score ≈ 84.4

Therefore, a score of approximately 84.4 represents the 97.5th percentile.