Question

Which best describes the spread of a set of data that has an interquartile range of 14 and a mean absolute deviation of 6

A.
The average distance of all data values from the mean is 10 and the middle 50% of data values has a range of 8

B.
The average distance of all data values from the mean is 10 and the middle 50% of data values has a range of 20

C
The average distance of all data values from the mean is 6 and the middle 50% of data values has a range of 14

D
The average distance of all data values from the mean is 14 and the middle 50% of data values has a range of 6

Answer 1

please help i will medal!

Answer 2

@King.Void.

Answer 3

@k12andstudyislandhelp

Answer 4

@sammixboo

Answer 5

HELP

Answer 6

please

Answer 7

@godmod360 @Data_LG2 @Haseeb96 @welshfella @ZoeyLynne101 @jojo4eva @kevindragonlord @adajiamcneal @bohotness

Answer 8

ok

Answer 9

@Javk

Answer 10

thanks soooo much can you help with 2 more

Answer 11

wait thats not a answer

Answer 12

thats not a answer

Answer 13

@Javk do you know this

Answer 14

you were just going on open study and looking at the same questions!

Answer 15

@Javk do you know how to do this

Answer 16

just let me take a look

Answer 17

ok

Answer 18

were did you go

Answer 19

@bohotness

Answer 20

which one do yuo think isa your answer love:)

Answer 21

i dont know! they explained it horribly and i dont know what to do!

Answer 22

okay :D i can help you

Answer 23

thanks!

Answer 24

your welcome :D

Answer 25

Hints:
1. The IQR (Interquartile range) is between the 25th and 75th percentiles, so contains the middle half (50%) of data.
2. The mean absolute deviation is the average "distance" of data from the mean.
For example, dataset
{2,3,4,12,13,14} has a mean of 8 and a mean absolute deviation of 5 because
(|2-8|+|3-8|+|4-8|+|12-8|+|13-8|+|14-8|)/6=5

Answer 26

im still a little confused

Answer 27

form the hint what do you think yuor answer is?

Answer 28

6?

Answer 29

':D

Answer 30

is that correct? :)

Answer 31

?

Answer 32

The average distance of all data values from the mean is 14 and the middle 50% of data values has a range of 6. thats the one i think is correct is that right

Answer 33

@bohotness

Answer 34

oh my gosh someone help pleeaasee!!!

Answer 35

ok so first off we need to understand what absolute deviation is, absolute deviation is the absolute differance between that the element and the mean
This is where majority of your elements lie
So it would look something like this.

Answer 36

,

Answer 37

go on

Answer 38

ok

Answer 39

now tell me, when I am going from 25% to 50%, how many units is that?

Answer 40

25?

Answer 41

cant you just tell me the answer

Answer 42

@Javk

Answer 43

ok lets go back to the second pic I put up, we had 100%, we divided it into 4 parts: 0-25%, 25-50%, 50-75%, 75-100%, two of those parts is equal to 14 units.
True?
So that means any two parts will equal to 14 units.
True?
That means we can now say that from 0-50% (two parts), there are 14 units
True?

Answer 44

i already failed it was a time test i only had a hour to do it and my time ran out :(

Answer 45

thanks for helping though!

Answer 46

do you still want to carry on for next time though? its really simple from here

Answer 47

ok

Answer 48

ok so going back if 0-50% is 14 units, that means the furthest a piece of data can be is also 14 units. this is the spread of the data

majority of your values fall within the mean absolute deviation of 3 units either side of the mean. making the range 6.

that means your answer should be
`The average distance of all data values from the mean is 14 and the middle 50% of data values has a range of 6`