Question

You recently took a statistics exam in a large class. The instructor tells the class that the scores were Normally distributed, with a mean of 72 (out of 100 ) and a standard deviation of 12 . Your score was 90 . Your friend had a time conflict and took a course with another instructor. Your friend had a score of 75 on a test, with a mean of 60 and a standard deviation of 10 . What can you conclude?

Answer 1

Answer:

$z =\frac{90-72}{12}=1.5$  (you)

$z =\frac{75-60}{10}=1.5$ (friend)

So as we can see we got the same z score for both cases so then we can conclude that:

You and your friend are equally ranked.

Step-by-step explanation:

Previous concepts

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".  

Solution to the problem

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:

$X \sim N(72,12)$  

Where $\mu=72$ and $\sigma=12$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we have a z score greater we have a bettr result since the value would be on a higher percentile of the distribution. So if we find the two z scores for the scores given we got:

$z =\frac{90-72}{12}=1.5$

For the other instructor the distribution changes and would be:

$Y \sim N(60,10)$  

Where $\mu=60$ and $\sigma=10$

And the z score for the friend would be:

$z =\frac{75-60}{10}=1.5$

So as we can see we got the same z score for both cases so then we can conclude that:

You and your friend are equally ranked.