# Determine the cut-off score (z-score) separating 5% of the highest scores from the lower scores. A set of normally distributed student test scores has a mean of 80 and a standard deviation of 6. Determine the cut-off score (z-score) separating 5% of the highest scores from the lower scores.

## Expert Answers We are given that the mean is 80 and the standard deviation is 6. We want to determine the score that is the cut score at 95%.

To find the `z` score that has 5% above and 95% below, we consult the standard normal table or technology to find `z~~1.6449` (Using a TI-83 graphing calculator)

The score that corresponds to this `z` score can be found as follows:

`z=(x-bar(x))/sigma`

`1.6449=(x-80)/6`

`9.8697=x-80`

`x=89.8697`

So a score of `x~~90` will be above 95% of the scores.

Approved by eNotes Editorial Team

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