Science scores for HS seniors in the US are normally distributed with a mean of 77 and a standard deviation of 7. Students scoring in the top 4% are eligible for a special prize. What is the approximate cutoff score a student must get in order to receive the prize?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion (.04) related to the Z score and solve the equation.
To find the approximate cutoff score that a student must get in order to receive the prize, we need to find the z-score corresponding to the top 4% of scores and then convert it back to the actual score using the z-score formula.
1. First, find the z-score corresponding to the top 4% of scores. The z-score represents the number of standard deviations a score is from the mean. We can use a standard normal distribution table (also known as a Z-table) to find the z-score.
The top 4% of scores would be the top 100% minus 4% = 96%. We need to find the z-score at the 96th percentile.
Using the Z-table, we find that the z-score corresponding to the 96th percentile is approximately 1.75.
2. Next, convert the z-score back to the actual score using the z-score formula:
z = (X - μ) / σ
Where:
- z = z-score
- X = actual score
- μ = mean
- σ = standard deviation
In this case, we're trying to find X. Rearranging the formula, we have:
X = z * σ + μ
Substituting the values we have:
X = 1.75 * 7 + 77
X = 3.5 + 77
X = 80.5
Therefore, the approximate cutoff score a student must get in order to receive the prize is 80.5.