if a test has a mean of 75, with a standard deviation of 11, what % of students scored an A (90 or higher)?
To find the percentage of students who scored an A (90 or higher), we need to calculate the Z-score corresponding to 90. The Z-score measures how many standard deviations an observation is from the mean.
First, let's calculate the Z-score using the formula:
Z = (X - μ) / σ
Where:
Z is the Z-score,
X is the observed value,
μ is the mean, and
σ is the standard deviation.
In this case, X is 90, μ is 75, and σ is 11.
Z = (90 - 75) / 11 = 1.36
Now, we look up the Z-score on a standard normal distribution table (also known as the Z-table) or use a calculator to find the corresponding percentile.
Using a Z-table, we find that 1.36 corresponds to a percentile of approximately 91.46%.
Therefore, approximately 91.46% of students scored an A (90 or higher) on the test.