Question

Stats help needed

Answer 1

what's your question?

Answer 2

Good Question @chihiroasleaf

Answer 3

I have worked over this problem

Answer 4

Faizan :o

Answer 5

AM=4

x+3+2+0+8+y+3+4/8=4

x+y+20=32

x+y=12

Answer 6

going good so far

Answer 7

Now what i see is that 4 is given as the mode so x and y should also be 4 but that is not the case :/

Answer 8

Looks like the question is wrong

Answer 9

why x AND y ??
any one of them can be 4

Answer 10

No but as its the mode so it should be the most recurring.....

Answer 11

I mean to say that 3 occurs 2 times and it is not the mode.So 4 should occur 3 times to be the mode

Answer 12

And if 3 is 2 times 4 is 2 times and then 8 would be 2 times,We have 3 modes here

Answer 13

yes so what? we can select mode as any 1 of the 3, and we select here 4

Answer 14

Okay got it,What should my step be now?

Answer 15

What should i do after getting an eqn?

Answer 16

take any one of 'x' or 'y' as 4

Answer 17

Okay got it thanks a lot,but what should i state

Answer 18

Why am i taking x or y as 4?

Answer 19

I need to do some reasoning?

Answer 20

because mode =4
if none of x or y is 4, then mode cannot be 4

Answer 21

Got it bro thanks :D

Answer 22

welcome :)