Answer:
a)
And we can find this probability with this difference and with the normal standard table or excel:
b) 
And we can use the z score defined by:

And using the limits we got:


And we want to find this probability:

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".
Part a
Let X the random variable that represent the resitances of a population, and for this case we know the distribution for X is given by:
Where
and
We are interested on this probability
And the best way to solve this problem is using the normal standard distribution and the z score given by:
If we apply this formula to our probability we got this:
And we can find this probability with this difference and with the normal standard table or excel:
Part b
We select a sample size of n =4. And since the distribution for X is normal then we know that the distribution for the sample mean
is given by:
And we want this probability:

And we can use the z score defined by:

And using the limits we got:


And we want to find this probability:
