Maria handed in her final data management
project last week. The class mean was 83%
with a standard deviation of 8. If Maria’s
mark produced a z-score of 1.09, what was
her grade?
You can play around with Z table stuff at
http://davidmlane.com/hyperstat/z_table.html
I have tried but could not get through ,,,, any hint
Thx
well, yeah. Her score was 1.09 STD above the mean, so
83 + 1.09*8 = 91.72
Thx
Well, if Maria's mark produced a z-score of 1.09, it means she's one standard deviation above the mean. Since the class mean was 83%, we can calculate her grade by adding the standard deviation (8%) to the mean (83%). So, Maria's grade would be... *drumroll*... 91%! Good job, Maria! Now go celebrate with some statistical cake.
To determine Maria's grade, we first need to understand what a z-score represents. A z-score measures how many standard deviations an individual data point is from the mean. It helps us compare an individual's score to the mean of a dataset and understand how that score relates to the rest of the data.
To calculate Maria's grade, we'll use the formula for z-score:
z = (x - μ) / σ
Where:
- z is the z-score
- x is Maria's grade
- μ is the mean of the dataset
- σ is the standard deviation of the dataset
Given:
- z = 1.09 (Maria's z-score)
- μ = 83% (class mean)
- σ = 8 (standard deviation)
Rearranging the formula, we get:
x = z * σ + μ
Substituting the given values:
x = 1.09 * 8 + 83
x = 8.72 + 83
x = 91.72
Therefore, Maria's grade is approximately 91.72%.