Question

Emilia and Ferdinand took the same freshman chemistry course: Emilia in the fall, Ferdinand in the spring. Emilia made an 83 on the common final exam that she took, for which the mean was 76 and the standard deviation 8. Ferdinand made a 79 on the common final exam that he took, which was more difficult, since the mean was 65 and the standard deviation 12. The one who has a higher z-score did relatively better. Was it Emilia or Ferdinand?

Answer 1

Answer:

$z = \frac{83-76}{8}=0.875$

$z = \frac{79-65}{12}=1.167$

As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.

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

Emilia case

Let X the random variable that represent the scores of a test, and we know that

Where $\mu=76$ and $\sigma=8$

The z score formula is given by:

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

Since Emilia made 83 points we can find the z score like this:

$z = \frac{83-76}{8}=0.875$

Ferdinand case

Let X the random variable that represent the scores of a test, and we know that

Where $\mu=65$ and $\sigma=12$

The z score formula is given by:

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

Since Ferdinand made 79 points we can find the z score like this:

$z = \frac{79-65}{12}=1.167$

As we can see the z score for Ferdinad is higher than the z score for Emilia so on this case we can conclude that Ferdinand was better compared with his group of reference.