Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

$\mu=600$ represent the population mean for the Graduate Record Exam (GRE)

$\sigma=25$ represent the population standard deviation for Graduate Record Exam (GRE)

2) Solution to the problem

Let X the random variable that represent the Graduate Record Exam (GRE) of a population, and for this case we know the distribution for X is given by:

$X \sim N(600,25)$

Where $\mu=600$ and $\sigma=25$

We want to find the z score for a score of 550. And in order to do this we need to apply the formula for the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$z=\frac{550-600}{25}=-2$

So the answer for our case would be Z=-2

This value means that the score of Rachel 550 it's 2 deviations below the mean of the population $\mu=600$

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3 years ago