Question

Suppose that a new scoring system for universities was created which has a scale of 0-100. The scores on this system are normally distributed with a variance of 100 and a mean of 65. What score would need to have in order to be in the top 10% of universities?

Answer 1

Answer:

We need a score of 77.8 or higher in order to stay in the top 10% of universities

Step-by-step explanation:

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".  

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:  

$X \sim N(65,\sqrt{100}=10)$  

Where $\mu=65$ and $\sigma=10$ minutes.  

And we want the top 10% of universities, so we need a value a such that:

$P(X>a)=0.10$ or $P(X$

We need on the right tail of the distribution a value a that gives to us 90% of the area below and 10% of the area above. Both conditions are equivalent.

Let's use the condition $P(X$, the best way to solve this problem is using the z score with the following formula:

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

So we need a value from the normal standard distribution that accumulates 0.90 of the area on the left and 0.10 on the right. This value on this case is 1.28 and we can founded with the following code in excel:

"=NORM.INV(0.90,0,1)"

If we apply the z score formula to our case we have this:

$P(X$

So then based on the equalities we have this:

$\frac{a-65}{10}=1.28$

And if we solve for a we got:

$a=(10*1.28) +65=77.8$

So then we need a score of 77.8 or higher in order to stay in the top 10% of universities