Math / Data & Statistics

QuestionScores on the GRE (Graduate Record Examination) are normally distributed with a mean of 558 and a standard deviation of 83. Use the 68-95-99.7 Rule to find the percentage of people taking the test who score between 309 and 807.
The percentage of people taking the test who score between 309 and 807 is $\square$ $\%$ .

Studdy Solution

STEP 1

1. GRE scores are normally distributed.
2. The mean ( $\mu$ ) of the scores is 558.
3. The standard deviation ( $\sigma$ ) is 83.
4. The 68-95-99.7 Rule (Empirical Rule) applies to normally distributed data.

STEP 2

1. Calculate the z-scores for 309 and 807.
2. Use the 68-95-99.7 Rule to determine the percentage of scores between these z-scores.

STEP 3

Calculate the z-score for 309 using the formula:
$z = \frac{X - \mu}{\sigma}$
For $X = 309$ :
$z = \frac{309 - 558}{83}$
$z \approx -3$

STEP 4

Calculate the z-score for 807 using the formula:
$z = \frac{X - \mu}{\sigma}$
For $X = 807$ :
$z = \frac{807 - 558}{83}$
$z \approx 3$

STEP 5

According to the 68-95-99.7 Rule, approximately 99.7% of the data falls within 3 standard deviations ( $z = -3$ to $z = 3$ ) of the mean in a normal distribution.
The percentage of people taking the test who score between 309 and 807 is:
$\boxed{99.7\%}$

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