Carrie attained a score of 650 on the verbal section of her SAT. The mean grade was a 500 with a standard deviation of 100. What percentage of test takers scored below and above Carrie?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z score. Multiply by 100.
k89
50
To find the percentage of test takers who scored below and above Carrie, we need to convert her score to a z-score, which will tell us how many standard deviations her score is from the mean.
The formula to calculate the z-score is:
z = (X - μ) / σ
Where:
X is Carrie's score,
μ is the mean, and
σ is the standard deviation.
In this case:
X = 650
μ = 500
σ = 100
Plugging these values into the formula, we get:
z = (650 - 500) / 100
z = 150 / 100
z = 1.5
Now, we can look up the z-score in a standard normal distribution table, which will give us the percentage of test takers who scored below and above Carrie.
From the table, we find that approximately 93.32% of test takers scored below Carrie, and 6.68% scored above Carrie.
Therefore, approximately 93.32% of test takers scored below Carrie, and approximately 6.68% scored above her.