Develop a Windows Form application to calculate and display the average, standar

Question

Develop a Windows Form application to calculate and display the average, standard deviation, minimum value, and maximum value 6 grades. All grades are to be stored in an Array

General Requirements – The following must be included in your project:

The Windows Form should be easy to understand using appropriate Label and ToolTip controls to help the user navigate through the application

Comments should be placed throughout your VB code. Variables and control objects should follow common naming conventions (see common naming conventions handout)

Array : Data entered into Textbox control objects should be stored in an Array using an appropriate data type

Use Textbox controls to enter the data and to display the output.

Use a ListBox control to display the solutions

Use Error Handling (Try...Catch…Finally) to account for common errors. For example, if a user enters incorrect data types into the data point textboxes

Display a message (using a MsgBox)to the user where necessary

Use a GroupBox control to contain the Checkbox controls

You can use a For-Next Loop to find the minimum and maximum values in the array

Use a Buton control to calculate the solutions and to clear all the data on the Form

Explanation / Answer

private double getStandardDeviation(List<double> doubleList) {

double average = doubleList.Average();

double sumOfDerivation = 0;

foreach (double value in doubleList) { sumOfDerivation += (value) * (value);

}

double sumOfDerivationAverage = sum():

private double getAverage() {

        Cursor cursor = db.getMarks(subject);

        List<Double> list = new ArrayList<Double>();

        cursor.moveToFirst();

        while(!cursor.isAfterLast()) {

            list.add(cursor.getDouble(cursor.getColumnIndex(DBAdapter.KEY_VALUE)));

        }

        if (list.isEmpty())

            return 0.0;

        long sum = 0;

        int n = list.size();

        for (int i = 0; i < n; i++)

            sum += list.get(i);

        return ((double) sum) / n;

    }

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.