## Howard-Wolowitz one year ago mean question

1. Howard-Wolowitz

2. Michele_Laino

using your data, we can write this: $\Large \frac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}}}{5} = 14$ where $$\large x_1,x_2,x_3,x_4,x_5$$ are the requested positive numbers Now, if we choose $$\large x_3= 10$$, then we have: $\Large {x_1},{x_2} < 10,\quad {x_4},{x_5} > 10$ by definition of median

3. Howard-Wolowitz

I see, but how would you simplify that

4. Michele_Laino

we can write this: $\Large \begin{gathered} {x_1} + {x_2} + 10 + {x_4} + {x_5} = 70 \hfill \\ \hfill \\ \left( {{x_1} + {x_2}} \right) + \left( {{x_4} + {x_5}} \right) = 60 \hfill \\ \end{gathered}$ now I can choose $$\large x_1=5,x_2=9$$ so, what is $$\large x_3+x_4 =...?$$

5. Howard-Wolowitz

I dont know

6. Howard-Wolowitz

I would tihnk that you couldnt combine

7. Michele_Laino

if I substitute, I get this: $\Large {x_4} + {x_5} = 60 - 9 - 5 = 46$

8. Michele_Laino

so I can choose these values: $\Large {x_4} = 22,\quad {x_5} = 24$

9. Howard-Wolowitz

so is that what you would put to answer the first question?

10. Michele_Laino

yes!

11. Michele_Laino

here are the five positive numbers: $\Large {x_1} = 5,\quad {x_2} = 9,\quad {x_3} = 10,\quad {x_4} = 22,\quad {x_5} = 24$

12. Howard-Wolowitz

ok so i put those and thats it?

13. Michele_Laino

yes!

14. Howard-Wolowitz

alright cool

15. Howard-Wolowitz

16. Michele_Laino

I have used the definition of $$\large median$$ and the definition of $$\large mean$$ of a distribution of data

17. Howard-Wolowitz

I see well ok that makes snese

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