What is the standard deviation after the bonus marks are applied?

unaffected
standard deviation=sqrt(variance)
variance(X)=E((X-E(X))^2)=E((X+4-E(X+4))^2)=variance(X+4)

Man I could be wrong but I think you just add 4 to the 13.65. Actually I am. IT WONT CHANGE.

The sum of the scores will be increased by the number of students times 4.

you get the average by dividing by the number of students so you get the original average + 4.

Then you take the difference between each score, like 4+Score1-original average + 4...so you get the same standard deviation.

The sum of all this will get the original sum, divide by the same number of people, and take the square root. It wont change.

The standard deviation remains the same, at 13.65 because adding 4 points to each score only changes the mean, not how the scores are spread out.

Because standard deviation is translation independent, i.e. adding/subtracting a constant value to all values will have no effect on the standard deviation, the standard deviation will remain constant at 13.65.

Basically, the difference between the marks and the average mark remain the same, as they've all been added to equally, and thus the standard deviation, which is the RMS (root mean square) of the average difference, remains the same.

If the teacher gives 4 points to everyone the standard deviation should remain the same. Everyone's grade goes up, the mean goes up, but each grade deviates from the mean the same amount.

A little bit of algebra should be all that's needed to solve this problem

Adding 4 points to everyone's score will shift the mean up by 4 points. (we won't use this fact)

Now I remember what this problem is all about:
The property of variance to be used is:
Let X be the RV of interest (assumed normal)
k a constant
Var(X+k) = Var(X) ; adding a constant to each score does not change the variance.

The standard deviation will be unchanged. The way to think about this is that the shape of the bell curve is the same it's just shifted to the right by 4 points.