Answer:

So since we want the probability that "the sample mean would differ from the population mean by greater than 2.2 millimeters". Then we use the complement rule and we got
P=1-0.9558=0.0442
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".
Let X the random variable that represent the diameter of a population, and for this case we know the distribution for X is given by:
Where
and 
And let
represent the sample mean, the distribution for the sample mean is given by:

On this case 
Solution to the problem
We can begin the problem finding this probability

We can use 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 on this way:

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

So since we want the probability that "the sample mean would differ from the population mean by greater than 2.2 millimeters". Then we use the complement rule and we got:
P=1-0.9558=0.0442