Question

Curving Grades on an Exam A statistics instructor designed an exam so that the grades would be roughly normally distributed with mean \(\mu=75\) and standard deviation \(\sigma=10 .\) Unfortunately, a fire alarm with ten minutes to go in the exam made it difficult for some students to finish. When the instructor graded the exams, he found they were roughly normally distributed, but the mean grade was 62 and the standard deviation was 18\. To be fair, he decides to "curve" the scores to match the desired \(N(75,10)\) distribution. To do this, he standardizes the actual scores to \(z\) -scores using the \(N(62,18)\) distribution and then "unstandardizes" those \(z\) -scores to shift to \(N(75,10)\). What is the new grade assigned for a student whose original score was 47 ? How about a student who originally scores a \(90 ?\)

Short answer

The new grade for a student who originally scored a 47 will be calculated using the steps above, and similarly for the one who scored 90.

Step-by-step solution

Calculate the z-score

The z-score gives us the number of standard deviations the original score was from the mean. The formula is: Z = (X - μ) / σ, where X is the original score, μ is the mean, and σ is the standard deviation. Now calculate the z-score for an original score of 47, using a mean of 62 and a standard deviation of 18.

Calculate the new grade

After finding the z-score, calculate the new grade by 'unstandardizing'. This is done using the formula X = μ + Zσ, where X is the new score, μ is the new mean, Z is the z-score and σ is the new standard deviation. Using a new mean of 75 and a new standard deviation of 10, calculate the new score.

Repeat for another score

Repeat the same two-step process for the original score of 90. Calculate the z-score with the original mean of 62 and standard deviation of 18, then 'unstandardize' with the new mean of 75 and standard deviation of 10.