Question

scores on the gre (graduate record examination) are normally distributed with a mean of 536 and a standard deviation of 90. use the 68 - 95 - 99.7 rule to find the percentage of people taking the test who score between 266 and 806
the percentage of people taking the test who score between 266 and 806 is %

Explanation:

Step1: Calculate the number of standard deviations from the mean

For \(x = 266\):

$$z_1=\frac{266 - 536}{90}=\frac{- 270}{90}=-3$$

For \(x = 806\):

$$z_2=\frac{806 - 536}{90}=\frac{270}{90}=3$$

Step2: Apply the 68 - 95 - 99.7 Rule

The 68 - 95 - 99.7 Rule states that for a normal distribution, approximately \(99.7\%\) of the data lies within \(z=-3\) and \(z = 3\) (i.e., within \(3\) standard deviations of the mean).

Answer:

\(99.7\)