Question

tough question stats Help !

Answer 1

Also could you please throw some light on what exactly the y-axis represent ?

Answer 2

If i am correct, the x-axis is binned in groups of 5 :
0-5
5-10
10-15
...

Answer 3

The y-axis represents the fraction of the numbers with a certain value, e.g. the largestbar has the y-value 0.4, so 40% of the values are between 5 and 10

Answer 4

oh just a sec please let me make sense of that

Answer 5

so, if i add all them up, il get 100% is it ?

Answer 6

Right!

Answer 7

Woah ! I see, awesome !!

Answer 8

could u help me wid options also...
i have eliminated option C, as the distribution is unimodal

Answer 9

after that i got stuck :x

Answer 10

OK, we could try to make sense of the bars. To make things easier, we take the middle of each bar to be the number that it counts.
So: say you've got a vase with 100 balls. The balls have numbers on them: 2.5, 7.5, 12.5 and so on.
40 balls have the number 7.5, so that is why the relative frequency is 0.4.
10 balls have the number 2.5, giving a rel. freq. of 10%.

Now what can we calculate?

1. The mean - it is (2.5*10 + 7.5*40 + 12.5*......)/100
2. The modus (modal value?), this is the most frequent value (7.5)
3. The median, that is the number in the middle, so the 50th number
4. The 1st and 3rd quartile (25th and 75th number.

Once you've got these numbers, you can easily see which of the other options are true or false...

Answer 11

Mean = 12.6875 ?

http://www.wolframalpha.com/input/?i=+%282.5*10+%2B+7.5*40+%2B+12.5*20+%2B+17.5*15+%2B+22.5*5+%2B+27.5*5+%2B+32.5*2.5+%2B+37.5*1.25+%2B+42.5*1.25%29%2F100

Answer 12

how do i find median ?

Answer 13

it can be 7.5 or 12.5 right ?

Answer 14

12.5

Answer 15

I used these numbers: (BTW, can you use a TI84 calculator?)

x f
2.5 10
7.5 40
12.5 20
17.5 15
22.5 5
27.5 5
32.5 3
37.5 1
42.5 1

Frequencies add up to 100 nicely.

With my TI84 (in L1 the x-values and in L2 the frequencies) I get:

mean 12.65
median 10
Q1 7.5
Q3 17.5

Answer 16

meadian cant be 10
u have interval
10_15 of 0.2
so take the mean of the interval

Answer 17

uhmm the value of 50th element falls in second bin.
so i thought median = 7.5 ?

Answer 18

Median is the middle number.
When there are 100 numbers, the middle one doesn't exist. In that case it is the mean of the 50th and 51th number. These would be 7.5 and 12.5 here, so that explains the median of 10.

Answer 19

ahh mean = average of 50 and 51 elements = (7.5 + 12.5)/2 = 10 !!!

Answer 20

Sweet xD

Answer 21

i dint thought of that :o

Answer 22

im trying to figure out how u got Q1 and Q3.. .

Answer 23

Yes! Although we don't know how many numbers we have here, the principle is the same: if we have 1000 numbers, we have to take the mean of the 500th and 501th number, which would give the same outcome.

Answer 24

yess i totally get it XD

Answer 25

I hope you can answer the question now!

Answer 26

Q1 = 7.5, cuz the 24, 25 elements fall in second bin !!!

Answer 27

+1

Answer 28

Q3 = 17.5 for the same reason

Answer 29

+1 again

Answer 30

IQR = 17.5 - 7.5 = 10
so striked off a, b, c options

Answer 31

mean = 12.65
median = 10

so strike off d also xD

Answer 32

guess i dont need to calculate outliers to tick the last option; from the graph itself it is evident that all values are trapped between 0 and 45 !!

thanks a lot everything makes complete sense now !! appreciate your help very much!!! (:

Answer 33

YW! Just to be sure:you agree answer 1 is true?

Answer 34

yess IQR = 17.5 - 7.5 = 10

ty !

Answer 35

i think i need to review more of it :|

Answer 36

@ikram002p: these stats questions are always tricky!

Answer 37

yes i think i made a mistake in my previous reply,
looks i need to calculate outliers lol :o

Answer 38

to eliminate that option..

Answer 39

but i can manage that... thanks everyone :))

Answer 40

i never been good at this im only good at qsb program

Answer 41

There is not much to calculate: an outlier is a value that lies at an abnormal distance from the other ones. Because all values in the distribution are in adjacent bars, you can be sure there are no outliers.

Answer 42

@mathx: if you know your maths, there is enough employment for high school math teachers over here :) Where do you live now?

Answer 43

yeah i thought the same, but that makes both option a and e true... so i suspect there must be an outlier or something..

Answer 44

@rational: if I read the question, I think more than one option can be true..

Answer 45

ohh yes you're right lol

Answer 46

wat about this one ?

Answer 47

Is it A ?

Answer 48

There are only a few very expensive houses in the city, because of the one expensive suburb.

Answer 49

This would mean most houses are below $650k...

Answer 50

yess that makes most of the house prices around $650

Answer 51

oh

Answer 52

Because the very expensive houses make the mean higher. Therefore, the other houses have a lower price.

Answer 53

Of course, because we don't know how many houses are cheap and how many are expensive, you could also argue that the few expensive houses have only very little impact on the mean of all the prices.
In that case, there would be about as many houses above and below 650k.

So we have not enough data.
The question said: what dio you think is most likely true.
IMO, most likely the expensive houses in the suburb increase the mean, so most other houses will be below 650k...

Answer 54

makes perfect sense ty :) you saved me.. i was about to submit a wrong answer lol xD

Answer 55

is it online exam ?

Answer 56

@mathx: I hope you enjoy your study! Sometimes it will be hard, but if you just go on, you will succeed!

Answer 57

online quiz

Answer 58

hmm can u gimme the link ?

Answer 59

if its available

Answer 60

you need to join the course, it started two weeks back

Answer 61

:|
my bad luck

Answer 62

ok good luck ^^

Answer 63

its free
https://class.coursera.org/statistics-001

Answer 64

u can still join if u want to, I am oly doing the first quiz..

Answer 65

for first quiz, the deadline is still there till March3..
so u can catchup easily

Answer 66

oh ic , i guess ill join it
need to improve in statistics

Answer 67

im enjoying the course... my stats concepts have been rusty for a long time...
and im getting confidence that i can finish the course smoothly with the kindof help im getting from OS :)

Answer 68

my concepts rusty too, i only had 2 course in
statics + i focuesd on analysis programs
so i need to review , i guess ill join the course ^^
good luck :)