Marbeya's z score when she scored a 90% on the last exam is 1.33.
Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Where x is the raw score, μ is the mean and σ is the standard deviation.
Given that μ = 70% = 0.7, σ = 15% = 0.15
For x = 90% = 0.9:
z = (0.9 - 0.7)/0.15 = 1.33
Marbeya's z score when she scored a 90% on the last exam is 1.33.
Find out more on Z score at: https://brainly.com/question/25638875