Question

scores on the gre (graduate record examination) are normally distributed with a mean of 583 and a standard devia people taking the test who score between 481 and 685. the percentage of people taking the test who score between 481 and 685 is %

Explanation:

Step1: Calculate the number of standard deviations from the mean

Let \(\mu = 583\) (mean) and assume the standard deviation \(\sigma\). First, find \(z\)-scores. The formula for the \(z\)-score is \(z=\frac{x-\mu}{\sigma}\).
For \(x = 481\), \(z_1=\frac{481 - 583}{\sigma}\). For \(x = 685\), \(z_2=\frac{685 - 583}{\sigma}\).
If we assume \(\sigma=102\) (since \(583-481 = 102\) and \(685 - 583=102\))
\(z_1=\frac{481-583}{102}=\frac{- 102}{102}=-1\)
\(z_2=\frac{685 - 583}{102}=\frac{102}{102}=1\)

Step2: Use the empirical rule (68 - 95 - 99.7 rule)

The empirical rule for a normal distribution states that approximately \(P(\mu-\sigma<X<\mu + \sigma)\) gives the percentage of data within one - standard deviation of the mean.
The formula for the probability \(P(-1<Z<1)\) where \(Z\) is a standard normal random variable.
We know that \(P(-1<Z<1)=\Phi(1)-\Phi(-1)\)
From the standard normal table, \(\Phi(1)=0.8413\) and \(\Phi(-1)=0.1587\)
\(P(-1 < Z < 1)=0.8413-0.1587 = 0.6826\)

Answer:

\(68.26\)