Question

Task 6: Arrays: Statistics Calculations Write a C program which accepts a "large" mumber of real numbers and calculates their sample mean and sample variance. Revision: From statistics, the sample variance, ,of N samples, X1, X2,..XN, can be calculated as: N mcan(X) ) 2. (Xt (1) N-1 There are many algorithms for calculating a data set's variance. For this task you should implement the reasonably robust two-pass algorithm which calculates the sample mean and uses this value to calculate the variance as follows: integer n -0 Eloat suml-0 Eloat sum2-0 FOR each data point Xn n n + 1 suml-suml Xn ENDFOR mean-s uml /n FOR each data point Xn sum2-sum2 + (Xn mean) (Xn-mean) ENDFOR sum2 (n-1) variance Your program should input" the data via an array initialisation. This is to avoid having to manually repeatedly type test data into scanf ()while debugging You may use the following template: #include <stdio.h> int main ) The mean of x[ is 5.000000 /The variance of x[] is 7.500000 x has 9 elements Eloat x[9] - {1,2,3,4, 5,6,7,8, 9); Eloat variance, mean; Eloat sum1 0, sum2 0; int n / Counter variable Calculate the sample mean for (n ; n 9; n++ (NB: we are assuming N-9 ) Write your calculation here Calculate the sample variance Write your own for () loop printf ("Mean: f\nVariance: f\n", return 0 mean, variance):

Please Provide Code for above question , be sure that the C code compiles using GCC

Answer 1

//C program

#include<stdio.h>
int main(){
   float x[9] = {1,2,3,4,5,6,7,8,9};
   float variance , mean;
   float sum1=0, sum2=0;
  
   int n;
  
   for(n=0;n<9;n++)sum1=sum1+x[n];
   mean = sum1/n;
  
   for(n=0;n<9;n++)sum2 =sum2+(x[n]-mean)*(x[n]-mean);
  
   variance = sum2/(n-1);
  
   printf("Mean: %f\nVariance: %f\n",mean,variance);
   return 0;
  
}

//sample output

C\Users\IshuManish\Documents\variance.exe Mean 5.000000 Jariance: 7.500000 Process exited after 0.1391 seconds with return value 0 Press any key to continue