You're correct that your issues are rooted in not understanding static. There are many resources on this, but suffice to say here that something static belongs to a Class whereas something that isn't static belogns to a specific instance. That means that

public class A{
    public static int b;

    public int x;
    public int doStuff(){
        return x;
    }

    public static void main(String[] args){
        System.out.println(b); //Valid. Who's b? A (the class we are in)'s b.
        System.out.println(x); //Error. Who's x? no instance provided, so we don't know.
        doStuff(); //Error. Who are we calling doStuff() on? Which instance?

        A a = new A();
        System.out.println(a.x); //Valid. Who's x? a (an instance of A)'s x.
    }
}

So related to that your method histogram isn't static, so you need an instance to call it. You shouldn't need an instance though; just make the method static:

Change public String histogram(int randomNum) to public static String histogram(int randomNum).

With that done, the line histogram(randomNum); becomes valid. However, you'll still get an error on System.out.print(histogram + "\n");, because histogram as defined here is a function, not a variable. This is related to the return statement. When something says return x (for any value of x), it is saying to terminate the current method call and yield the value x to whoever called the method.

For example, consider the expression 2 + 3. If you were to say int x = 2 + 3, you would expect x to have value 5 afterwards. Now consider a method:

public static int plus(int a, int b){
    return a + b;
}

And the statement: int x = plus(2, 3);. Same here, we would expect x to have value 5 afterwards. The computation is done, and whoever is waiting on that value (of type int) receives and uses the value however a single value of that type would be used in place of it. For example:

int x = plus(plus(1,2),plus(3,plus(4,1)); -> x has value 11.

Back to your example: you need to assign a variable to the String value returned from histogram(randomNum);, as such:

Change histogram(randomNum) to String s = histogram(randomNum).

This will make it all compile, but you'll hit one final roadblock: The thing won't run! This is because a runnable main method needs to be static. So change your main method to have the signature:

public static void main(String[] args){...}

Then hit the green button!