Oh, dude, it's like when you have a group of numbers and the mode, which is the most common one, is 100, but the mean, which is the average, is lower than the median, which is like the middle number when they're all lined up. So basically, it's just a funky little math situation where the numbers are playing musical chairs and the mean ends up being the odd one out below the median.
If the mode is 100, it means that 100 is the most frequently occurring value in the data set. If the mean is less than the median, it indicates that the data is negatively skewed, with a tail stretching towards lower values. This suggests that there are some low values dragging the mean down, while the median remains higher due to the presence of higher values on the other side.
(1, 5, 97, 99, 100, 100) The mode is 100. The median is 98. The mean is 67.
The mean is 1226.75. The median is 508. There is no mode as no number occurs more than any other.
The question is how do the mean and median affect the distribution shape. In a normal curve, the mean and median are both in the same point. ( as is the mode) If a distribution is skewed, its tail is either on the right or the left. If a distribution is skewed the median may be a better value to use than the mean since it has less effect on the shape. Also is there are large outliers, the median has less effect and is better to use. So the mean has a bigger effect on the shape many times than the median.
If the distribution is not symmetric, the mean will be different from the median. A negatively skewed distribution will have a mean hat is smaller than the median, provided it is unimodal.
Yes, the mode is the number that occurs most often. The median is the average.
IDN
(1, 5, 97, 99, 100, 100) The mode is 100. The median is 98. The mean is 67.
There cannot be such a set.If the mean is 10, the mode is 6 and the median is 9, the range must be at least 8.5.If the mode is 6, the median is 9 and the range is 8, the mean must be less than 9.8.
median
7,6,4,92,57,32
The mean is 1226.75. The median is 508. There is no mode as no number occurs more than any other.
I am guessing you are asking for an example of a set of numbers with these properties. Let's start with 5 numbers, so the median will be the middle number; say 1, 2, 3, 4, 5. The median is 3, but so is the mean. Now let's replace the 5 with 10. The median is still 3, but the mean is 4. To make the mode less than 3, let us change the 2 into a 1. Now the median is still 3, the mode is 1, and the mean is 3.8. So 1, 1, 3, 4, 10 will work.
The question is how do the mean and median affect the distribution shape. In a normal curve, the mean and median are both in the same point. ( as is the mode) If a distribution is skewed, its tail is either on the right or the left. If a distribution is skewed the median may be a better value to use than the mean since it has less effect on the shape. Also is there are large outliers, the median has less effect and is better to use. So the mean has a bigger effect on the shape many times than the median.
It depends on the definition of mode. If mode is simply the most frequently occurring outcome and more than one outcome in the sample space is allowed to be the mode, then all datasets do have a mean, median, and mode.
If the distribution is not symmetric, the mean will be different from the median. A negatively skewed distribution will have a mean hat is smaller than the median, provided it is unimodal.
Yes, the mode is the number that occurs most often. The median is the average.
If the distribution is positively skewed , then the mean will always be the highest estimate of central tendency and the mode will always be the lowest estimate of central tendency (If it is a uni-modal distribution). If the distribution is negatively skewed then mean will always be the lowest estimate of central tendency and the mode will be the highest estimate of central tendency. In both positive and negative skewed distribution the median will always be between the mean and the mode. If a distribution is less symmetrical and more skewed, you are better of using the median over the mean.