Question

The top 5% of applicants on a test will receive a scholarship. If the test scores are normally distributed with a mean of 76 and a standard distribution of 12, what is the lowest test score that still qualifies for a scholarship? Use Excel, and round your answer to the nearest integer.

Answer 1

96

Explanation:

Z-Scores:

z-scores are defined as:
`z=(x-\mu )/(\sigma)`, and this may look a bit confusing at first but by looking at the numerator and then denominator we can get a more understandable definition.

The numerator is first finding the difference between the statistic and the mean. It then divides by the standard deviation, so essentially it's telling us how far the statistic is from the mean in terms of standard deviation.

We can actually rewrite the equation to express this:

`z=(x-\mu )/(\sigma)\\\\z\sigma = x-\mu\\\mu + z\sigma=x`

So in essence, how many standard deviations the statistic is away from the mean.

Now this may seem very off topic compared to what the problem is asking, but we want to convert the top 5% to a z-score. Now let's first convert this top 5% to a percentile. To be in the top 5% you just need to be in the 95th percentile and using technology we can convert this into a z-score which is approximately 1.645.

So this means the 95th percentile is 1.645 standard deviations away and in this case above (since it's positive) from the mean.

The good thing is we know the standard deviation and mean, now let's just apply it:
`76+1.645(12)=95.74`. We now want to round this to the nearest integer of 96, and now we have our answer!