Question

Statistics question #2

Answer 1

How should I start?

Answer 2

the numbers are probably in ascending order

Answer 3

That's the thing I'm not sure of...

Answer 4

I mean ... the expressions should be ordered

Answer 5

ascending or descending

Answer 6

in this case x = 11

Answer 7

But it should be like this:
x+17> x-1 > x-3 > x-4 ?!
Oh... seems I got something from here..

Answer 8

Alright, I got it

Answer 9

:)

Answer 10

\[\langle x\rangle=\sum\limits_{n=0}^{\infty} xP(x)=\sum\limits_{n=0}^{\infty} x\frac{N(x)}{N}\]
\[8=\frac{1}{5}\sum\limits_{n=0}^{6} xN(x)\]\[8=\frac{1}{5} \left((15\times1)+(x-1) \times 1+(x-3)\times1 +(x-4)\times1+(x-17)\times 1 \right)\]

Answer 11

The mean is 19. Isn't it?

Answer 12

@thomas5267 Nope. That's not in the option

Answer 13

should be 12

Answer 14

x+17> x-1 > x-3 > x-4
15 could be the largest or the smallest

Case I: if 15 is the largest
15>x+17> x-1 > x-3 > x-4
Median = x-1 =8 => x=9
mean = 12

Case II: if 15 is the smallest
x+17> x-1 > x-3 > x-4 >15
median = x-3 = 8 => x=11
mean = 13.6 .... Oh fail....
I thought I got 12 as the only answer..

@dumbcow Time to explain~

Answer 15

If I assume x to be a positive integer, the descending order would be:
x+17,15,x-1,x-3,x-4

Answer 16

15 can't be smallest, if x=11 then x-4 < 15

Answer 17

actually 15 can't be largest either, if x=9 then x+17 > 15

order should be x+17 > 15> x-1 > x-3 > x-4

Answer 18

So that we are assuming the question only gives us positive integer. @dumbcow

Answer 19

no we are not assuming that but given the median is 8, it is impossible for x to be negative
x can only be 9

Answer 20

found this : Mode = Mean—3 (Mean—Median)
the mode in your sample is 0

Answer 21

0 =M -3 (M-8)

Answer 22

never seen that before...will that work in the general sense?
what if mean = median, there is no guarentee mode = mean in that case

Answer 23

http://www.transtutors.com/homework-help/statistics/central-tendency-examples/mean-median-mode.aspx

Answer 24

I haven't seen this also ... but I hated Stats.

Answer 25

does 12 work ?

Answer 26

1,2,3,4,5
mean = 3
median = 3
mode = 0

formula breaks down ? :(

Answer 27

12 is the answer

Answer 28

In a symmetrical distribution, mean, median and mode are identical and have the same value. However, in actual life most distributions are not symmetrical—they are skew. (These concepts of symmetry and skewness would be discussed in the chapter on Skewness). In cases of moderately skew (or moderately asymmetrical) distributions the value of the mean, median and mode have the following empirical relationship which is given by Karl Pearson :
(i) Mode = Mean—3 (Mean—Median)
= Mean—3 Mean + 3 Median
= 3 Median — 2 Mean
Thus the value of Mode — 3 Median—2 Mean. From this relationship,
Mode = Mean—3 (Mean—Median) we can derive
(ii) Mean—Mode = 3 (Mean—Median) or
(iii) Mean—Median = 1/3 (Mean—Mode)
Thus we see that the difference between Mean and Mode is three times, the difference between mean and median. In other words median is closer to the mean than mode. This relationship between Mean (X), Median (M) and Mode (Z) is shown in the following diagram.

Answer 29

read the link ... interesting !

Answer 30

ahh just did...thanks
yes formula only works in special case where mean does not equal median and distribution is skewed

Answer 31

yep

Answer 32

lucky us ... no idea how else it could be done

Answer 33

I understand dumbcow / Mani 's solution,
And the link provided by mathg8 is ... interesting...

Answer 34

I guess you can take each expression and make it median ... then check the answers .

Answer 35

That formula works in this case

Answer 36

Mani's solution was nice ...

Answer 37

But the way he derived this formula is... unknown :S

Answer 38

he = Karl Pearson

Answer 39

Thanks mathg8!

Answer 40

here's the proof : http://stats.stackexchange.com/questions/3787/empirical-relationship-between-mean-median-and-mode

Answer 41

@Mani_Jha welcome

Answer 42

http://www.amstat.org/publications/jse/v13n2/vonhippel.html

Answer 43

Thanks all :)

Answer 44

np