CMIS 102 Hands-On Lab // Week 5 Overview: This hands-on lab allows you to follow and experiment with the critical steps of developing a program including the program description, analysis, test plan, design (using pseudocode visualization), and implementation with C code.  The example provided uses sequential, repetition statements and nested repetition statements. Program Description: This program will calculate the average of 3 exams for 5 students. The program will ask the user to enter 5 student names. For each of the students, the program will ask for 3 exam scores.  The average exam score for each student will be calculated and printed. Analysis: I will use sequential and repetition programming statements. I will define one String to store student name: StudentName. I will define three Float numbers: Examvalue, Sum, Avg to store exam values the sum of the exams and the average of the exams. The sum will be calculated by this formula: Sum = Sum + Examvalue For example, if the first value entered was 80.0 and second was 90.0 and the third exam was 100.0: sum = sum + Examvalue = 0.0 + 80.0 sum = 80.0 + 90.0 = 170.0 sum = 170.0 + 100.0 = 270.0 Avg is then calculated as: Avg = sum/3.0 For example 270.0/3.0 = 90.0 A nested repetition loop can be used to loop through each of the 5 students and each of the 3 exams: For (students=0; students <5; students++) For (exams=0;exams<3;exams++) End For End For Sum values will need to be reset for each student to ensure only one student data is used for calculations each time. Test Plan: To verify this program is working properly the input values could be used for testing: Test Case Input Expected Output 1 Studentname=Chris Examvalue1=80.0 Examvalue2=90.0 Examvalue3=100.0 Studentname=John Examvalue1=70.0 Examvalue2=90.0 Examvalue3=80.0 Studentname=Sally Examvalue1=100.0 Examvalue2=100.0 Examvalue3=100.0 Studentname=Pat Examvalue1=50.0 Eexamvalue2=70.0 Examvalue3=60.0 Studentname=Sam Examvalue1=90.0 Examvalue2=95.0 Examvalue3=100.0 Average for Chris is 90.0 Average for John is 80.0 Average for Sally is 100.0 Average for Pat is 60.0 Average for Sam is 95.0 Pseudocode: // This program will calculate the average of 3 exams for 5 students // Declare variables Declare StudentName as String Declare ExamValue, Sum, Avg as Float // Loop through 5 Students For (students=0; students <5 ; students++) // reset Sum to 0 Set Sum =0.0 Print “Enter Student Name” Input StudentName // Nested Loop for Exams For (exams=0; exams < 3; exams++) Print “Enter exam grade: n” Input ExamValue Set Sum = Sum + ExamValue End For Set Avg = Sum/3.0 Print “Average for “ + StudentName + “ is “ + Avg End For Flow Chart: C Code The following is the C Code that will compile in execute in the online compilers. // C code // This program will calculate the average of 3 exams for 5 students. // Developer: Faculty CMIS102 // Date: Jan 31, 2014 #include int main () { /* variable definition: */ char StudentName[100]; float ExamValue, Sum, Avg; int students,exams; // Loop through 5 Students for (students=0; students <5 ; students++) { // reset Sum to 0 Sum =0.0; printf("Enter Student Name n"); scanf("%s", StudentName); // Nested Loop for Exams for (exams=0; exams < 3; exams++) { printf ("Enter exam grade: n"); scanf("%f", &ExamValue); Sum = Sum + ExamValue; } Avg = Sum/3.0; printf( "Average for %s is %fn",StudentName,Avg); } return 0; } Setting up the code and the input parameters in ideone.com: Note the Student and ExamValues for this run were: John: 90.0 80.0 100.0 Jim: 80.0 70.0 90.0 Joe: 70.0 100.0 100.0 Sally: 100.0 95.0 91.0 Sam: 30.0 54.0 68.0 You can change these values to any valid integer values to match your test cases. Results from running the programming at ideone.com: Learning Exercises for you to try: What would you change in the design and the code if you wanted to input 10 students and 5 exams? What is the line of code doing? char StudentName[100]; (Hint: We haven’t covered arrays, but a Str...

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