scores on the gre (graduate record examination) are normally distributed with a mean of 503 and a standard…

scores on the gre (graduate record examination) are normally distributed with a mean of 503 and a standard deviation of 130. use the 68 - 95 - 99.7 rule to find the percentage of people taking the test who score between 113 and 503.\nthe percentage of people taking the test who score between 113 and 503 is 49.85%

scores on the gre (graduate record examination) are normally distributed with a mean of 503 and a standard deviation of 130. use the 68 - 95 - 99.7 rule to find the percentage of people taking the test who score between 113 and 503.\nthe percentage of people taking the test who score between 113 and 503 is 49.85%

Answer

Explanation:

Step1: Calculate the number of standard - deviations.

First, find the difference between the mean ($\mu = 503$) and the given score ($x = 113$). The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\sigma = 130$. So, $z=\frac{113 - 503}{130}=\frac{- 390}{130}=-3$.

Step2: Apply the 68 - 95 - 99.7 Rule.

The 68 - 95 - 99.7 Rule states that about 99.7% of the data lies within 3 standard deviations of the mean in a normal distribution. That is, $P(-3<Z<3)\approx0.997$. The area between $z = - 3$ and $z = 0$ is half of the area between $z=-3$ and $z = 3$. So the percentage of data between $z=-3$ and $z = 0$ is $\frac{99.7%}{2}=49.85%$.

Answer:

$49.85%$