A class is given an exam. The distribution of the scores is normal. The main score is 74 and the standard deviation is 11. Determine the test score C such that the probability of a student having a school grader than she is 81% C rounded to one decimal place.

We’re dealing with a normal distribution of test scores. We know the average score (mean) and how spread out the scores are (standard deviation). We want to find a specific test score © where 81% of the students scored lower than that score.

Using the Z-Score

Convert to a Z-score: We need to convert our desired probability (81%) into a standard z-score. A z-score tells us how many standard deviations away from the mean a particular value is. You can use a Z-table or a calculator to find the z-score that corresponds to a cumulative probability of 0.81 (representing 81%). You’ll find that the z-score is approximately 0.88.

Apply the Z-score formula: The z-score formula is:

z = (X - μ) / σ

Where:

z = z-score

X = the raw score we’re trying to find (in this case, C)

μ = the mean (74)

σ = the standard deviation (11)

Solve for C: Plug in the values and solve for C:

0.88 = (C - 74) / 11

Multiply both sides by 11: 9.68 = C - 74

Add 74 to both sides: C ≈ 83.7

Answer

The test score C, rounded to one decimal place, is 83.7. This means that approximately 81% of the students scored lower than 83.7.