Can anybody tell me how to choose 6 numbers that have a range of 4, a median of 9, a mean of 9 and a mode of 7?
so far, i have 7,7,9,11
i know the mean should be 54/6 but the last 2 number must = 20, so 2 10's would negate the mode of 7
thanks
[Math] finding 6 numbers that are in a range with mean, mode and median averages
average
Related Solutions
This is NOT a complete answer, but I try to sketch my idea.
Suppose we have a sorted set, say, in an increasing order. If we want to maximise the difference between mean and median, it is necessary (but not sufficient) to keep as many outliers as possible (because mean is very much affected by these outliers, whereas median is not). The problems here are: how many is "many"? and how to control the difference?
Suppose $x_1 < x_2 < ... < x_n$. We can calculate its mean and median. My alogorithm (which I've not yet been able to prove so far) is as follows:
- If mean $>$ meadian: we delete the median or the closest number which is BIGGER than median, that is $x_{(n+1)/2}$, if $n $ is odd or $x_{n/2+1}$, if $n$ is even. The effect of doing this is: after each step, the mean (not always) gets bigger and bigger but the median (not always) gets smaller and smaller (and of course, mean is always bigger than median in every step). But overall, the difference gets bigger.
- If mean $<$ median: we delete the median or the closest number which is SMALLER than median, that is $x_{(n+1)/2}$, if $n $ is odd or $x_{n/2-1}$, if $n$ is even. The effect of doing this is: after each step, the mean (not always) gets smaller and smaller but the median (not always) gets bigger and bigger (and of course, mean is always smaller than median in every step). But overall, the difference gets bigger.
It turns out that, finally we will only have $3$ numbers left (can't be $2$, otherwise mean $=$ median): the biggest, the smallest, and the second smallest (if mean $>$ median) or the second biggest (if mean $<$ median). These $3$ numbers will have the maximum difference between mean and median.
I did some testing using what you cal "MSR" but I call Square Mean Root (SMR). It looks useful for dealing with data where large measurements indicate the opposite of an effect. For example in reaction time measurements, large reaction times indicate the participant was not paying attention. Unfortunately the "tyranny of large numbers" means those large reaction times have an inaapropriate effect on the results.
OTHER AVERAGES 1. Normalized Averages. a) Convert each data set to relative values. b) find the mean relative value c) restore the mean to actual value by multiplying and adding back the means of the subtracted constants and mean fo the divisors. NB this method wcan also produce realistic standard deviations.
Kalman Averages These weight means inversely according to their standard deviations. Thus less reliable measurements have less influence. This type of average can be found on the net and in some statistics books.
Winsorized Means These set limits to outliers, but do not eliminate them. The 5% and95% values of a set of measurments are taken as maximum values. Any measurments above or below these values are set to these max and minimum values. A mean is calculated from these modified measurements in the usual way. This is less questionable than simply eliminating the "outliers".
When using any "average" suh as mean, median, RMS, .., it is important to keep in mind what has been measured and what the size of the measurement numbers mean in terms of the effect producing those numerical values.
Best Answer
We want to find $6$ integers with specific range, mean, median and mode. We call them $x_1,x_2,x_3,x_4,x_5,x_6$ such that $x_1 \leq x_2 \leq x_3 \leq x_4 \leq x_5 \leq x_6$ and then:
From all that you have that: \begin{align} &&\frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6}{6} &= 9 \\ &\Leftrightarrow& x_1 + x_2 + x_3 + x_4 + x_5 + x_6 &= 54 \\ \tag{Range constraint.} \\ &\Leftrightarrow& x_1 + x_2 + x_3 + x_4 + x_5 + x_1 + 4 &= 54 \\ &\Leftrightarrow& 2x_1 + x_2 + x_3 + x_4 + x_5 &= 50 \\ \tag{Median constraint.} \\ &\Leftrightarrow& 2x_1 + x_2 + 18 + x_5 &= 50 \\ &\Leftrightarrow& 2x_1 + x_2 + x_5 &= 32 \\ \tag{Mode constraint.} \\ &\Leftrightarrow& 2x_1 + 7 + x_5 &= 32 \\ &\Leftrightarrow& 2x_1 + x_5 &= 25 \\ &\Leftrightarrow& x_1 &= \frac{25 - x_5}{2} \\ \end{align} Now from the median constraint we have that $x_5 \geq 9$ and from the range constraint that $x_5 \leq 11$ and we have $3$ possibilities:
As mentioned in one of the comments if we relax the mode constraint to be non strict then $x_5 = 11$ becomes a viable solution and the following sets of numbers work: