0, 1, 4, 5, 6, 7, 8, 9, 9, 9, 11
mean = 6 and 3/11
median = 7
mode = 9
A single number, such has, 7152484654452062 can have only one mean, median, mode. And the range is 0.
The median of 5, 7, 11, 14, 8, 7, 11, 3, 0, and 7 is 7.
yes they are if you have 0 and 10 the mean is 5 and so is the median. The mean and the median can in fact be the same value. But basically to answer your question, One possible way is that if the values are ascending by 1 in the data set, then the number of values left to the median should be the same as the number of values right to the median. e.g. 6+7+8+9+10 6,7 = 2 terms 9,10 = 2 terms median =8 mode = 8
You are right. 1 is the median.The median of a set of data is found by arranging the data in increasing or decreasing order. So, we can arrange the items: 0, 1, 1, 1, 4, 4If the number of items of data is even, as it is in this case, then the median is the average (or mean) of the two middle numbers. So, the median is (1 + 1)/2 = 2/2 = 1If the number of items of data is odd, then the median is the middle number in the ordered set. For example the median of the data 1, 3, 3, 5, 6, 8, 9 is 5, because 5 is in the middle.It is also the mode. The mode of a set of data is the one item of data that occurs most often. 1 has the highest number of frequency in this sample. Thus 1 is the mode also.
If you have a list of positive and negative numbers then it is possible that the median would be 0.
Mean: 5.5 Median: 7 Mode: 9, 0, 8
The mean, median and mode of a single number is the number itself. The range is 0.
Mean = 2.125 Median = 1.5 Mode = 1 and 2.
Yes. In the set (10, 15, 20, 20, 25, 30) the mean, median and mode are all 20.
Median = 58 Mode = 58 Range = 0
A single number, such has, 7152484654452062 can have only one mean, median, mode. And the range is 0.
Mode = 2 = median
The range of a single number is 0.The mode, median, maximum and minimum of a single number is the number.
Mean is the average of all numbers. Median is the middle number of all: if you have an even number set, then there will be 2 left in the middle. then, you take the average of those 2 middle number. Mode is the most repeated number. Range is the Highest number minus the lowest number. Ex: 3,2,6,4,9,3,3,5,0 mean: 3+2+6+4+9+3+3+5+0=35/9= 3.89 median: 9 Mode: 3 Range: 9-0=9
#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); }
If a set of numbers is arranged in ascending or descending order the median is the value which lies halfway along the series. 9-6-5-0-6-11-10 when placed in ascending order becomes 0-5-6-6-9-10-11. The median is therefore 6. Note : When there are an even number of values the median is the mean of the middle two values.
The median of 5, 7, 11, 14, 8, 7, 11, 3, 0, and 7 is 7.