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I feel like there isn't much information, but the answer for the mean is 79. The median is said to not be able to be found. Can someone explain this?

Thank you!

Helpappreciated111
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  • What are you asking for an explanation of? Why the mean is 79? Why the median is not able to be found? Both? Also, what have you tried in either case? Answering these questions will make it easier for us to help you. – Silynn Jul 08 '14 at 16:18

2 Answers2

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Regarding the second part of your question, the median is a statistic that involves the individual values of a (multi)set $A \subseteq \mathbb R$, whereas the mean $\mu(A)$ only involves $\sum_{a \in A} a$ and the size $\lvert A \rvert$.

Consider a simplified example of your question. You have two sets $A$ and $B$ such that $\mu(A)=1$ and $\mu(B)=2$ and $\lvert A \rvert=2$ and $\lvert B \rvert =3$. We could have $A=\{1, 1\}$ and $B=\{2, 2, 2\}$ or $A=\{1, 1\}$ and $B=\{-2, 2, 6\}$. In the first case the median is 2 whereas in the second case the median is 1.

Shawn O'Hare
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The mean of $n$ numbers $\{x_1,\dots, x_n\}$ is, by definition, $$M_n=\frac{x_1+x_2+\dots+x_n}{n}.$$

Now, you have two sets of numbers, $\{x_1,\dots,x_n\}$ and $\{y_1,\dots,y|n\}$. The mean of all numbers together is $$\frac{x_1+\cdots+x_n + y_1+\cdots + y_m}{m+n}.$$

For you, $m$ and $n$ are known. You need to calculate $x_1+\cdots+x_n + y_1+\cdots + y_m$. How can you do that? Well, there may be a way for you to know what $x_1+\cdots+x_n$ is, just think about it!

5xum
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