# Math Help - Finding number

1. ## Finding number

Hello everyone,

I'm curious to know:

A different set of natural numbers has median 20 and medium 17.

What will be the largest number in this series?

What I have done:

Because they are natural numbers then are positive;

also for the calculation of the median must be sorted in increasing.

Assuming that the number of terms is odd, then the formula for calculating the median is: $\displaystyle\frac{n + 1 }{2} = 20$, where $n = 39.$

And the formula for calculating the average we have:

$\displaystyle\frac{\displaystyle\sum_{i=1}^n{X_i}} {39} =17\longrightarrow{\displaystyle\sum_{i=1}^n{X_i = 663}}$

I guess as I now know how many terms and their sum, I calculate the nth term,

for if we assume that the numbers are ordered increasingly, the latter being the largest.

Any ideas?

Thank you very much.

Greetings

2. Originally Posted by Dogod11
Hello everyone,

I'm curious to know:

A different set of natural numbers has median 20 and medium 17.

What will be the largest number in this series?

What I have done:

Because they are natural numbers then are positive;

also for the calculation of the median must be sorted in increasing.

Assuming that the number of terms is odd, then the formula for calculating the median is: $\displaystyle\frac{n + 1 }{2} = 20$, where $n = 39.$

And the formula for calculating the average we have:

$\displaystyle\frac{\displaystyle\sum_{i=1}^n{X_i}} {39} =17\longrightarrow{\displaystyle\sum_{i=1}^n{X_i = 663}}$

I guess as I now know how many terms and their sum, I calculate the nth term,

for if we assume that the numbers are ordered increasingly, the latter being the largest.

Any ideas?

Thank you very much.

Greetings
This is not entirely clear, can you reword it or post the original question?

CB

3. Hi!

That's all it says the statement, but can not find a way to do it:

A different set of natural numbers has median 20 and medium 17.

What will be the largest number in this series?

A greeting

4. What is the medium? I google and all I can find is it is another name (somewhere) for the median. If so then that can't be right. Also, you don't need to assume the number of terms in your sequence is odd, since if it were even it wouldn't be an integer.

As for the problem, unless you know the numbers are the sequence of consecutive natural numbers or some other identifying information, I don't see how you are meant to be able to figure this out since the median doesn't offer information about the range of these numbers.

5. Hello, actually as you say:

I google and all I can find is it is another name (somewhere) for the median
The first value, 20 is the median.

The second (17) is the Media (average).

Then he says, what is the largest value of this series?

I do not see where they take it,

A greeting

6. Not sure if this is helpful, but, from Chebyshev's inequality, the maximum value of the standard deviation is the distance between the mean and the median, which, in your case, is 20 - 17 = 3.

Chebyshev's inequality - Wikipedia, the free encyclopedia

7. Ignoring a few things if the mean is $17$, the median $20$ and the sample size is $n$:

about half the numbers are less than $20$ and the maximum value is:

$x_{max} \le \sim 17n-\lfloor (n+1)/2\rfloor -1 \rfloor \times 20$

This is approximate as I have not checked that the corrections for everything being an integer etc and have allowed everything below the median to be $0$.

What I have tried to do is set everything below the median to zero, all but the largest value to $20$ and then found the corresponding largest value to give the total of $17n$. I have not been too fussy about the details so there may be a number of minor corrections needed to correct the bound.

Example: let there be $7$ values, then to get the largest possible $x_{max}$ we have $\{x_{max},20,20,20,0,0,0\}$ and so:

$x_{max} = 17\times7-3 \times 20= 59$

(the definition of median used here may be different from yours if so make the appropriate correction)

CB

Look at this from another forum:

The numbers might be, for example, $a, 20, b$ with $a <20 , so that the average would be:

$\dfrac{a+20+b}{3}=17$

thereby:

$a + b = 31$

$a <20 <31-a$

then $a <11.$

That is worth us $1\leq a\leq 10$. By this I mean that is not uniquely determined as many of that series.

Do you think of this?

A greeting