Test score are normally distributed with a mean of 500 and a standard deviation of 50. What percentage of students scored below a 575?
a
93.32%
b
50%
c
56.68%
d
6.68%
Z = (score-mean)/SD = (575-500)/50 = 75/50 = 1.5
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
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david lane normal distribution
To find the percentage of students who scored below 575, we need to determine the area under the normal distribution curve to the left of 575.
First, let's standardize the score of 575 using the formula:
Z = (X - μ) / σ
where X is the score, μ is the mean, and σ is the standard deviation.
Z = (575 - 500) / 50
Z = 75 / 50
Z = 1.5
Next, we need to find the cumulative probability associated with a Z-score of 1.5 using a standard normal distribution table or a statistical calculator.
Looking up the Z-score of 1.5 in the standard normal distribution table, we find that the cumulative probability is approximately 0.9332.
This means that approximately 93.32% of students scored below 575.
Therefore, the correct answer is (a) 93.32%.
To find the percentage of students who scored below a certain score in a normally distributed set of test scores, you can use the standard normal distribution table or a calculator.
In this case, with a mean (μ) of 500 and a standard deviation (σ) of 50, we need to calculate the z-score for the given score of 575. The z-score represents the number of standard deviations a particular value is from the mean.
The formula to calculate the z-score is:
z = (x - μ) / σ
Calculating the z-score:
z = (575 - 500) / 50
z = 1.5
Now, we can look up the z-score of 1.5 in the standard normal distribution table to find the percentage of students who scored below 575.
From the table or calculator, we find that the area to the left of the z-score of 1.5 is approximately 0.9332.
Therefore, approximately 93.32% of students scored below 575.
So the answer is option A: 93.32%.