Answer:
Where
and 
Since the distribution for X is normal then the distribution for the sample mean is also normal and given by:



So then is appropiate use the normal distribution to find the probabilities for 
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". The letter
is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: 
Solution to the problem
Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:
Where
and 
Since the distribution for X is normal then the distribution for the sample mean
is also normal and given by:



So then is appropiate use the normal distribution to find the probabilities for 