## anonymous one year ago The prices of some jewelry sets in a store are shown below: Store Price A $110,000 B$100,000 C $1,110,000 D$130,000 E \$120,000 Based on the data, should the mean or the median be used to make an inference about the price of the jewelry sets in the store? Mean, because it is in the center of the data Median, because it is in the center of the data Median, because there is an outlier that affects the mean Mean, because there are no outliers that affect the mean

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1. anonymous

@mathstudent55

2. anonymous

There's outlier in your data set

3. iYuko

How does the book tell you to do this?

4. anonymous

there is no book

5. iYuko

the lesson

6. anonymous

mean is skewed by the outlier in the data set

7. iYuko

Show me the work @Kim21

8. iYuko

I will tell you if it's right.

9. anonymous

but i dont know how to do this :(

10. anonymous

11. iYuko

You need to review

12. iYuko

What test is this?

13. anonymous

think back whats mean and median

14. anonymous

its a pretest so its just to see what i know so i haven't learned this yet but i want to still get the right answer cause it counts towards grade ;b

15. anonymous

median is the middle number of the data set and mean is when u add and divide the numbers in the data set

16. anonymous

The mean is the average of the numbers.

17. anonymous

yes so mean is$Mean = \frac{ x_1 + x_2 +...+ x_n }{ n }$

18. anonymous

what do you think happens if 1 of the x is a huge number

19. anonymous

not sure :?

20. anonymous

the means becomes skewed by that large x

21. anonymous

@mathstudent55

22. anonymous

you made any progress so far?

23. anonymous

no im gonna call @mathstudent55 for some help but thanks anways :)

24. anonymous
25. anonymous

26. anonymous

ok thanks! :)

27. mathstudent55

An outlier affects more the mean that the median. When you have outliers, using the median instead of the mean may give you a better picture of how the data is distributed.

28. mathstudent55

See the great explanation here: https://www.mathsisfun.com/data/outliers.html