/*
•½‹Ï<Xi> = ƒ°Xi/n
•ªŽUV = ƒ°(Xi-<Xi>)^2/n�@�E�E�E(1)
     = ƒ°(Xi^2-2<Xi>ƒ°Xi+nX^2)/n
     = ƒ°(Xi^2-2(ƒ°Xi)^2/n+(ƒ°Xi)^2/n)/n
     = ƒ°(Xi^2-(ƒ°Xi)^2/n)/n
     = ƒ°Xi^2/n-(ƒ°Xi)^2/n^2
     = <Xi^2> - <Xi>^2�@�E�E�E(2)
•W�€•Î�·s = �ãV
•Î�·’l = 50 + (’l-<Xi>)/s * 10
*/
#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()
{
	int a[5],i,n=5;
	int me;
	char str[10];
	float sum,sum2,av,V,hensachi;
	for(i=0;i<n;i++){
		printf("a[%d]-->",i);a[i] = atoi(gets(str));
	}
	/* •½‹Ï */
	sum = 0.0;
	for(i=0;i<n;i++){
		sum += a[i];
	}
	av = sum/n;
	printf("•½‹Ï=%.3f\n",av);

	/* •ªŽU */
	sum = sum2 = 0.0;
	for(i=0;i<n;i++){
		sum += (a[i] - av)*(a[i] - av);/* (1)‚̉ð–@ */
		sum2 += a[i]*a[i];/* (2)‚̉ð–@ */
	}
	V = sum/n;/* (1)‚̉ð–@ */
	V = sum2/n - av*av;/* (2)‚̉ð–@ */
	printf("•ªŽU=%.3f\n",V);
	printf("•W�€•Î�·=%.3lf\n",sqrt(V));

	printf("‚ ‚È‚½‚Ì“¾“_‚Í�H-->"); me = atoi(gets(str));
	hensachi = 50 + ((me-av)/sqrt(V))*10;
	printf("•Î�·’l‚Í%f‚Å‚·�B\n",hensachi);

	return 0;
}