Question

Which statement holds true for a skewed histogram showing a distribution of the weights of students in a class?

Answer 1

A.The number of weights on the right side of the mode is equal to the number of weights on the left.
B.The number of weights on the left side of the mean is equal to the number of weights on the right.
C.The mean weight and the median weight will be the same for the distribution.
D.The nature of the skew can be verified by the position of the mean with respect to the median.
E.The nature of the skew can be verified by the position of the mean with respect to the mode.

Answer 2

@amistre64

Answer 3

what can you narrow this down to?

Answer 4

i think its D...but im not sure

Answer 5

A isnt an option i dont think

Answer 6

or E

Answer 7

The nature of the skew can be verified by
the position of the mean
with respect to the median.

this is correct, if the mean is to the left of the median ... its left skewed; if the mean is to the right of the median, it is right skewed

Answer 8

i thought so, i didnt wanna make any choices to soon and miss it, but thank you (:

Answer 9

C is the definiton for a normal distribution, no skew

Answer 10

truee

Answer 11

B and C state the same thing

Answer 12

can you help me out with another question, its pretty hard lol

Answer 13

Which data set is the farthest from a normal distribution?

Answer 14

im not sure i can do difficult ones today ..

Answer 15

lol, this one

Answer 16

A.2, 3, 3, 4, 4, 4, 5, 5, 6
B.3, 4, 5, 6, 6, 7, 7, 7, 8, 8, 9, 10
C.0.9, 1.0, 1.0, 1.1, 1.1, 1.1, 1.2, 1.2, 1.3
D.4, 4, 4, 5, 5, 6, 7, 7, 8, 8, 8
E.2, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 9, 10

Answer 17

have you already done this for me..

Answer 18

i forgot if you did...lol

Answer 19

ive already done this question for someone else

Answer 20

oohhh do you think its hard?

Answer 21

its simple enough

determine the mean and median of each set, then compare whcih set has the biggest difference between mean and median.

there should be 2 sets that differ in mean and median ....

Answer 22

A. 2, 3, 3, 4, 4, 4, 5, 5, 6 <-- normal (4,4)

B. 3, 4, 5, 6, 6, 7, 7, 7, 8, 8, 9, 10 <--- out of balance

C. 0.9, 1.0, 1.0, 1.1, 1.1, 1.1, 1.2, 1.2, 1.3 <-- normal (1.1,1.1)

D. 4, 4, 4, 5, 5, 6, 7, 7, 8, 8, 8 <-- normal (6,6)

E. 2, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 9, 10 <-- out of balamce

Answer 23

A. Mean = 4
B. Mean = 7
C. Mean = 1.1
D Mean= 6
E. Mean= 7

Answer 24

ohh i did it wrong lol

Answer 25

B and E are the main suspects

Answer 26

those are the medians..

Answer 27

but ill find the means right now for b and e

Answer 28

you did those correctly, those are the means

Answer 29

E= Mean = 6.6

Answer 30

B = Mean = 6.6

Answer 31

so both are 6.6?

Answer 32

i appear to have broken the wold this morning :/

Answer 33

what?

Answer 34

world*?

Answer 35

wolf, wolframalpha.com

Answer 36

ohh how did you break it?

Answer 37

i use that website for everything lol

Answer 38

its just sat there spinning while i tried to dbl chk

Answer 39

i can type it in if you want?

Answer 40

B = 6.667 mean and 7 median

E = 6.6429 mean 7 median

Answer 41

so is it 6.667 -7 and 6.6429 - 7

Answer 42

yes

Answer 43

it has to be E then

Answer 44

ignore the negative sign in the end

and yes; 6429 is fartherest from 7000; so E

Answer 45

sweet that problem wasnt even as hard as it seemed, thank you

Answer 46

my fingers havent woken up yet :)

Answer 47

hahah ill help you with that if you want

Answer 48

after a 2liter of MtDew ... ill be fine

Answer 49

hahahah