Scores on the GRE (Graduate Record Examination) are normally distributed with a mean of 528 and a standard deviation of 135. Use the 68-95-99.7 Rule to find the percentage of people taking the test who score below 123.
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To find the percentage of people taking the test who score below a certain value, we can use the 68-95-99.7 rule for a normal distribution. This rule states that:
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.
Given that the mean score on the GRE is 528 and the standard deviation is 135, we can calculate the Z-score for the value 123 as follows:
Z = (X - μ) / σ
Z = (123 - 528) / 135
Z = -405 / 135
Z = -3
Now we need to find the area to the left of the Z-score -3 on the standard normal distribution curve. This can be done by looking up the corresponding value in the z-table or by using a calculator.
From the z-table, we find that the area to the left of -3 is approximately 0.00135.
Therefore, the percentage of people taking the GRE who score below 123 is approximately 0.135% (0.00135 * 100).