When you find the median you are trying to find the middle of a set of numbers. If you line the numbers up in numerical order and count one off on each side multiple times, you will get to the middle. However, if you have only two numbers or two numbers in the middle, you would add those two numbers together and then divide by two.
For example:
2, 4, 14, 27, 65, 83
First, you would count off 2 and 83 then 4 and 65. You would be left with 14 and 27 in the middle. You would add 14 and 27 to get 41. Then, you would divide 41 by two to get 20.5 which would be your median.
you find the two numbers in the middle and add them. divide the sum by two. that's the median.
To find the median with two numbers, do as you would to find the "average" or the "mean" by finding the sum between the two numbers and dividing the sum (or total) by two.
You find the median to find the middle number
The middle number (in a sorted list of numbers).To find the Median, place the numbers you are given in value order and find the middle number.Example: find the Median of {13, 23, 11, 16, 15, 10, 26}.Put them in order: {10, 11, 13, 15, 16, 23, 26}The middle number is 15, so the median is 15.(If there are two middle numbers, you average them.
How do you find missed frequency if median and mode are given
You add the two numbers in the middle of your range and then you divide them by two, or just find the number that is halfway between the two numbers, and you then have your median.
#include<stdio.h> #define MAXVAL 1000 void sort1(int a[],int n); void median(int a[],int n); void mode(int a[],int n); int main() { int n; int arr[MAXVAL]; int i; printf("Enter the number of elements:"); scanf("%d",&n); printf("Enter the values:"); for(i=0;i<n;i++) { printf("a[%d]=",i); scanf("%d",&arr[i]); } sort1(arr,n); median(arr,n); mode(arr,n); } void sort1(int a[],int n) { int i; int j; int temp; for(i=0;i<n;i++) { for(j=i;j<n;j++) { if(a[i]>a[j]) { temp=a[i]; a[i]=a[j]; a[j]=temp; } } } } void median(int a[],int n) { int median; int mid; if((n%2)==0) { mid=n/2; median=(a[mid-1]+a[mid])/2; } else { mid=(n+1)/2; median=a[mid-1]; } printf("The median is:%d\n",median); } void mode(int a[],int n) { int i; int count1[MAXVAL]; for(i=0;i<n;i++) { count1[i]=0; } for(i=0;i<n;i++) { count1[a[i]]++; } i--; int mode=count1[0]; int j; int k; int flag=0; for(j=0;j<=a[i];j++) { if(count1[j]>count1[mode]) mode=j; } for(j=0;j<=a[i];j++) { for(k=j+1;k<=a[i];k++) { if(count1[j]=count1[k] && count1[j]>count1[mode]) { flag=1; } } } if(flag==1) { printf("Mode cannot be calculated"); } else printf("the Mode is:%d",mode); }
import java.util.Arrays; class Solution { public int median(int[] A) { int[] b = new int[A.length]; System.arraycopy(A, 0, b, 0, b.length); Arrays.sort(b); if (A.length % 2 == 0) { return b[b.length / 2]; } else { return (int) ((b[(b.length / 2) - 1] + b[b.length / 2]) / 2.0); } } }
if you are trying to find the median and there is two middle numbers, you add the two Numbers together and divide is by two.
you find the two numbers in the middle and add them. divide the sum by two. that's the median.
when you have an even amount of numbers while trying to find the median, you first find the two numbers that are at the median and then take all the numbers between them and find the median of that. if that amount of digits is also even, then you must have a decimal median.
If you need to find a median between two numbers, pick one of them arbitrarily.
To find the median with two numbers, do as you would to find the "average" or the "mean" by finding the sum between the two numbers and dividing the sum (or total) by two.
If the two numbers are that same in the middle than the median is that number.However if the two numbers are different then you just find the average of those two numbers.
take the two middle numbers and find the sum. then take half and that's your median.
i think you divide the histogram in two, so there are two equal halves. The number in the middle is the median,
You would add the two numbers in the middle and divide them by two and the answer would be the median.