How to solve for the mean in an array without the max and min?

Question

public class FunWithArrays {
    public static void main(String[] args) {
        int[] myFunArray = {7, 10, 3, 0, 615, -1000};
        int i;
 //minimum 
    int min;
    min = myFunArray[0];
    for (i=0; i < myFunArray. length; i++) {
    if (myFunArray[i] < min) {
        min = myFunArray[i];
    } 
 }
 //maximum 
 int max;
 max= myFunArray[0];
 for (i=1; i < myFunArray. length; i++) {
    if (myFunArray[i] > max) {
        max = myFunArray[i];
        }
    }
 System.out.println("minimum = " + min + " maximum = " + max);
        //mean without outliers
        double noOutliersMean = 0;
        for (i=0; i < myFunArray.length; i++){
            noOutliersMean = ((sum - (max + min)) /myFunArray.length );
            if ( myFunArray.length < 3){
                System.out.println("not enough numbers");
            }
        }
        System.out.println("Mean without Outliers: " +noOutliersMean);

        //Array reversed
        int[] reversed = myFunArray;
        for(int i1 = reversed.length - 1; i1 >= 0 ; i1--) {
            System.out.println( "Reversed " + reversed[i1]);
        }
    }
}

For the mean without the outliers (max and min) the answer should be 5 my line of code gives me 1.333. What am i doing wrong?

2nd question regarding the array reversed, when printed out it gives me the right answer but each number on a separate line. How do i get it to be on 1 line in the format of ; reversed: [-1000,615,0,3,10,7]

your code shows max and min incorporated.. what are its values.. and how are you calculating sum — Shreyas Sarvothama, Oct 19 '16 at 03:36
You need to remove the outliers from the count as well: `/(myFunArray.length-2)`. — shmosel, Oct 19 '16 at 03:39
And your whole printing question has already been asked and solved. http://stackoverflow.com/questions/409784/whats-the-simplest-way-to-print-a-java-array — OneCricketeer, Oct 19 '16 at 03:39
Why are you recalculating the mean for each element? You're not even using `i`. — shmosel, Oct 19 '16 at 03:40
The whole point is to exculde the biggest and smallest number from the array. The code is supposed to work even when the numbers in the array are changed as long as there are at least 3 numbers. I calculated the min and max in order to deduct it from the sum so that i could find the mean. — AlexSK, Oct 19 '16 at 03:56

Shreyas Sarvothama · Answer 1 · 2016-10-19T03:52:35.583

0

the answer to your second question is don't print using

println

, just use print

The code should be:

System.our.print("Reversed: [")
for(int i1 = reversed.length - 1; i1 >= 0 ; i1--) {
    System.out.print(reversed[i1]);
    if(i1>0){
       System.our.print(",");
    }
}
System.out.print("]");

To answer your first question you need to provide the following:

How you are calculating sum, max and min. Even though you have mentioned without max and min.. you still are using it in your code

edited Oct 19 '16 at 03:52

answered Oct 19 '16 at 03:39

Shreyas Sarvothama

486
2
10

you need to check the next element, if exists then comma should be appended – Saravana Oct 19 '16 at 03:44

score 0 · Answer 2 · edited Jun 20 '20 at 09:12

Think about the logic of the code when you manipulated the array... you will see that the mean with no min or max is the same as: the mean excluding the elements at index0 and index length-1 if the array were sorted, so that is what you have to do

sort the array
loop excluding 1st and last
calculate the mean

Example:

int[] myFunArray = { 7, 10, 3, 0, 615, -1000 };
// unsorted { 7, 10, 3, 0, 615, -1000 };
// sorted { -1000, 0, 3, 7, 10, 615 };

Arrays.sort(myFunArray);
System.out.println(Arrays.toString(myFunArray));
int acumm = 0;
double mean = 0.0;
for (int i = 1; i < myFunArray.length - 1; i++) {
    acumm += myFunArray[i];
}
mean = acumm / (myFunArray.length - 2);
System.out.println(mean);

How to solve for the mean in an array without the max and min?

2 Answers2

Example: