Question

Write a program in which you create an array of random real numbers with a size of 10001. The values should fall between 900 and 1000. Using the provided statistics library, calculate all statistical values represented by the functions in the library and output the results. The median value can only be correctly found after the array is sorted. After sorting using the selection sort method, search for the median value using both the linear search and binary search methods. Return the index of the median value in both searches.

The language is C++

Answer 1

#include<iostream>
#include<math.h>
using namespace std;
//logic for selection sort
void selection(double a[])
{
    int i,j;
    double t;
    for(i=0;i<10001;i++)
    {
       for(j=i+1;j<10001;j++)
       {
           if(a[i]>a[j])
           {
               t=a[i];
               a[i]=a[j];
               a[j]=t;
           }
       }
   }
     
   }
   //method for binary search
   int binarysearch(double a[], double no)
   {
      int i,low=0,up=10000;
      for(i=(low+up)/2;low<=up;i=(low+up)/2)
           {
               if(a[i]==no)
                   {
                       return i;
                   }
                   else
                       if(a[i]>no)
                           up=i-1;
                       else
                           low = i+1;
           }
   }
   //method forlinear search
int linearsearch(double a[],double n)
{
   int i;
   for(i=0;i<10001;i++)
   if(a[i]==n)
   return i;
}
int main()
{
   int i;
   double arr[10001],mean,median,variance,sd,sum=0;
  
   for(i=0;i<10001;i++)
   arr[i]= 900 + rand()%(1000-900+1);
  
//   cout<<endl<<"Array elements before sort\n";
   for(i=0;i<10001;i++)
   {
       //cout<<" "<<arr[i];
       sum+=arr[i];
   }
   //computing mean
   mean=sum/11;
   //computing variance
   sum=0;
   for(i=0;i<10001;i++)
   {
   sum=sum+ (arr[i]-mean)   * (arr[i]-mean);
   }
   variance = sum/10000;
   //computing standard deviation
   sd=sqrt(variance);
   //display the mean
   cout<<endl<<"Mean value is "<<mean;
  
   //call to selection() to sort
   selection(arr);
   //display the elements after sorting
//   cout<<endl<<"Array elements after sort\n";
//   for(i=0;i<10001;i++)
//   cout<<" "<<arr[i];
   //find the median
   median=arr[(10001/2)];
   //display the median
   cout<<endl<<"Median value is "<<median;
   //display the variance
   cout<<endl<<"Variance is "<<variance;
   //display the standard variance
   cout<<endl<<"Standard Deviation is "<<sd;
   //call to linearsearch() to find the position of median
   i=linearsearch(arr,median);
   cout<<endl<<median<<" value is at "<<i<<" position";
   //call to binarysearch() to find the position of median
   i=binarysearch(arr,median);
   cout<<endl<<median<<" value is at "<<i<<" position";
  
  
}

OUTPUT

For the output i have taken the array size as 11

This output is taking with array size 10001. Here you will observe the variations of position of the median value in linearsearch and binary search because the array having duplicate elements.

NOTE: SINCE YOU HAVE NOT GIVEN THE STATISTICS LIBRARY SO I HAVE COMPUTED MEAN,MEDIAN VARIANCE AND STANDARD DEVIATION ONLY.