If the mean is 6 then the 5 numbers must total 6 * 5 = 30.
Median is 7, so that leaves 4 numbers totaling 23.
So we need 2 numbers less than (or equal to) 7 and 2 numbers greater than (or equal to) 7 which all add up to 23 and where the highest number minus the lowest number equals 10.
So how about,
1,2,7,9,11
or 1,1,7,10,11 etc
3,5,6,7,9
Yes. If the predominant data are higher than the median, the mean average will be higher than the median average. For example, the median average of the numbers one through ten is five. The mean average is five and one-half.
Because the mode is 5, we know at least two of the numbers are 5. Because the median is 12 and we have five numbers, we know one of them is 12. Because the range is 10 and the smallest number we know so far is 5, we'll try 15 for a fourth number. This gives us: 5, 5, 12, 15 The mean is 10, so we know the sum of all five numbers is 50. The sum of the four numbers we have so far is 5+5+12+15 = 37. Thus, the fifth number is 50-37 = 13. That gives us 5, 5, 12, 13, 15. These five numbers fit all the criteria.
1,1,2,3,43. Any set of 5 numbers such that their sum is 50 and when ordered, the middle (ie third) number is 2.
The median is the middle value of a list of numbers. In [1,6,34] the median value is 6.
What Five Numbers have a range of 5 a median of 16 and a mean of 15
44689
2 4 4 6 8
3, 4, 5, 6, 7, 8, and 9.
4 4 6 6 10
The mean and the median of the two numbers, 12107 and 1115 are 6611.There is no mode. The range is 10992.
5, 6, 9, 10, & 10
3, 3, 5, 9, 10
Prior to the introduction of 170, the mean, median, mode and range did not exist since there were no numbers at all. Once 170 is introduced, it becomes the mean, median and mode. The range is zero.
There are at least 7 such sets, among them, 1,2,6,6,6,7,7.
No.
in solving numbers