Stats help! URGENT EXAM - HotUKDeals
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Stats help! URGENT EXAM

jizzm Avatar
banned8y, 6m agoPosted 8 years, 6 months ago
Got my exam this afternoon, bit stuck.!
The question is:
"2% of a new model of mobile `phones are known to be defective. If a store orders a batch of 150 of these mobile 'phones what is the probability that

i) None are defective?

ii) 3 or more are defective?


I presume you have to use poisson, but if anyone can point me in the right direction, would be great.
jizzm Avatar
banned8y, 6m agoPosted 8 years, 6 months ago
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#1
it would be 3
banned#3
Yes i know 2% of 150 is 3.

Part i) is worth 10 marks.
Part ii) is worth 20 marks.
#4
Its not an either or question!

Its what is the probabilities that 0 and then 3 or more are available...
#5
the question is asking for the probability, not the amount of defective ones.

i think you right you would need to use a poisson distribution, havent got any tables here so i cant really help!

edit: yeah def poisson dist. http://upload.wikimedia.org/math/5/5/9/55978f02e2b22e9a93943595030ecf64.png

have fun!
#6
Agreed with the others!
1) Probability is UNLIKELY but not impossible!
2) Probability is LIKELY but not certain!
#7
big_jim83;2191810
the question is asking for the probability, not the amount of defective ones.

i think you right you would need to use a poisson distribution, havent got any tables here so i cant really help!

edit: yeah def poisson dist. http://upload.wikimedia.org/math/5/5/9/55978f02e2b22e9a93943595030ecf64.png

have fun!

yes I agree with this :w00t: ........... rather you than me !!
#8
The first part isnt poisson:

Probability of failure = 0.02 therefore probability of phone working = 0.98

The probability of no phones being broken in 150 (part a) is the same as saying the probability of 150 phones working...

Probability of 150 phones working = 0.98^150 = 0.0483..
#9
ok i think this is how you go about it am trying to get hold of my mate as shes all into maths

but from forumla above for part a

lamda = 147
k = 150
e = natural log

and for b you work out the changes of 0 defects, 1 defect, and 2 defects and subtract the answer from 1
#10
Is this A Level?

For poisson distribution:

X~Po(np)

λ (lambda) = np

Where np is the number multiplied by the probability of the event occurring.

np = 0.02 x 150 = 3

Now head over to the cumulative probabilities for poisson distribution your statistical tables and look for where λ (lambda, np) = 3.

For (i), you just look at:

λ (lambda) = 3
x = 0

Answer: 0.0498

For (ii), you just look at λ (lambda) = 3 once again.

BUT, because it is now THREE OR MORE you need to look at x=2 as this is the probability for (x=0 + x=1 + x=2)

When x is less than or queal to 2, the probability is 0.4232 (from tables).

But you want the probability that there are three or more, so you have take this answer away from 1 (the total of all of the probabilities).

1 - 0.4232 = 0.5768.

Answer: 0.5768


Good luck in your exam!!
#11
snowtiger
yes I agree with this :w00t: ........... rather you than me !!


its not too hard, you just need a set of probability tables and a calculator. then spend ages putting numbers into the equation which is very very dull :thumbsup:

done to much stats work over the years, really glad i dont have to do it anymore!!!!
#12
I would miss this question out and move on lol
#13
big_jim83
the question is asking for the probability, not the amount of defective ones.

i think you right you would need to use a poisson distribution, havent got any tables here so i cant really help!

edit: yeah def poisson dist. http://upload.wikimedia.org/math/5/5/9/55978f02e2b22e9a93943595030ecf64.png

have fun!


You could use the formula if it was "find the probability that P(X = 45)" or whatever, but if it's a "less than" or "more than" question you'd have to do the calculation separately for them all and add them up, which would mean doing the calculation 147 times.

Hence, tables are the best way in this question - but for others you may need the formula (which is usually at the top of the statistical tables if I recall correctly).
#14
iliko
Probability is 1/50 or 2% :-)


No, it's not. :p
#15
You sure about part i) ducky - I might well be wrong (its years since i did probability) but can't see the floor in my reasoning....

jah128;2191853
The first part isnt poisson:

Probability of failure = 0.02 therefore probability of phone working = 0.98

The probability of no phones being broken in 150 (part a) is the same as saying the probability of 150 phones working...

Probability of 150 phones working = 0.98^150 = 0.0483..
#16
3 or more ... 1% of 150 = 1.5 x 2 = 3 ?
#17
jah128
You sure about part i) ducky - I might well be wrong (its years since i did probability) but can't see the floor in my reasoning....


Heh, I thought I was but you have me wondering. I'd have stuck with poisson throughout...
#18
what the......., I did A-level maths years ago and havent a clue!!! Sorry
#19
Yeah its got me confused, but I guess so has the marks awarded (ie part i=10, part ii=20). There probably is a good reason why you can't do one or the other but I can't remember it :giggle:
#20
MANJ_007
what the......., I did A-level maths years ago and havent a clue!!! Sorry


AFAIK you'd only do it if you did a module in statistics, and poisson isn't/hasn't been on on all syllabuses. Some only cover the Binomial distribution.
#21
jah128
Yeah its got me confused, but I guess so has the marks awarded (ie part i=10, part ii=20). There probably is a good reason why you can't do one or the other but I can't remember it :giggle:


Well tbh our answers weren't that far apart... :-D
#22
has your exam finished yet ?, make sure no one sees ya .. ;-)
#23
Just a quick question, if youve got the exam this afternoon, how come the have the question already?
#24
duckmagicuk2;2192004
Well tbh our answers weren't that far apart... :-D


True, but probably just about further than the margin of error on the tables? Surely we must have some A-level maths teachers on here to enlighten us...

Anyway, good luck in exam, follow duckys method and I'm sure you'll get the marks :thumbsup:
#25
MANJ_007
Just a quick question, if youve got the exam this afternoon, how come the have the question already?


It'll be from a past paper.
#26
duckmagicuk2;2192046
It'll be from a past paper.


Lol unless he's sitting it at the moment and using his phone :w00t:

"Youth arrested for browsing HUKD during an exam" :giggle:
#27
just to make it simple

There are 150 phones and 2% are defective

10% of 150 is 15.
1% of 150 is 1.5
so
2% of 150 is 3

3 phones are defective
#28
gemz700
just to make it simple

There are 150 phones and 2% are defective

10% of 150 is 15.
1% of 150 is 1.5
so
2% of 150 is 3

3 phones are defective


Don't think it is that simple by the sounds of it, also that is not worthy of an A-level question!
#29
gemz700
just to make it simple

There are 150 phones and 2% are defective

10% of 150 is 15.
1% of 150 is 1.5
so
2% of 150 is 3

3 phones are defective


No, no, no...

If it asked "how many phones would you expect to be defective?" then your are correct, but it doesn't.

It asks "what is the probability that three or more are defective?"...

Or, to make it more obvious that your answer is incorrect, a probability cannot be more than 1 (certain). You have just said that the probability is 3. :giggle:
#30
gemz700;2192173
just to make it simple

There are 150 phones and 2% are defective

10% of 150 is 15.
1% of 150 is 1.5
so
2% of 150 is 3

3 phones are defective


You missing the whole point of probability there though, its not 2% of phones are defective, its there is a 2% probability of any phone being defective. So you cant say '3 phones are defective', you can only say 'there is a n% probability that 3 phones will be defective)...
#31
Trouble is that this isn't a Poisson distribution. This is a binomial problem instead. There are two outcomes for each phone (working or not), so there is a predicted value for each number of phones not working.

I cheated and used Excel to do my calcs, but the values I get are
0 faulty = COMBIN(150,0) * (0.02)^0 * (0.98)^150 = 4.83%
1 faulty = COMBIN(150,1) * (0.02)^1 * (0.98)^149 = 14.78%
2 faulty = COMBIN(150,2) * (0.02)^2 * (0.98)^148 = 22.48%
3 faulty = COMBIN(150,3) * (0.02)^3 * (0.98)^147 = 22.63%
etc...

Probably too late for the exam now, sorry!!!

Good luck :thumbsup:
#32
just too late proberly if i rember right most start at 1:30..... though our nice teachers always snuck us some of the questions form the REAL paper :D
#33
Question: How did you know your question before the exam?:?
#34
jah128
You missing the whole point of probability there though, its not 2% of phones are defective, its there is a 2% probability of any phone being defective. So you cant say '3 phones are defective', you can only say 'there is a n% probability that 3 phones will be defective)...


Thats why it said "3 or more" If you're talking probability and they have said that a certain percent are defective then the answer would be "most likely some are faulty"
banned#35
Wheyy
Had my exam it went fairly well to be honest.
Thanks for all your replies, didnt get to see some of them before i went in.
But i had an idea. :)

Cheers
#36
AyrshireBacon;2192675
Trouble is that this isn't a Poisson distribution. This is a binomial problem instead. There are two outcomes for each phone (working or not), so there is a predicted value for each number of phones not working.

I cheated and used Excel to do my calcs, but the values I get are
0 faulty = COMBIN(150,0) * (0.02)^0 * (0.98)^150 = 4.83%
1 faulty = COMBIN(150,1) * (0.02)^1 * (0.98)^149 = 14.78%
2 faulty = COMBIN(150,2) * (0.02)^2 * (0.98)^148 = 22.48%
3 faulty = COMBIN(150,3) * (0.02)^3 * (0.98)^147 = 22.63%
etc...

gemz700;2193321
Thats why it said "3 or more" If you're talking probability and they have said that a certain percent are defective then the answer would be "most likely some are faulty"


No, its talking about a general 2 percent failure rate, and there is a random sample of 150 devices, its not why it said '3 or more'. "Most likely" is not a mathematical answer is it :roll:

Probably too late for the exam now, sorry!!!

Good luck :thumbsup:


:giggle: Good point, that makes sense to me. So probability that 3 or more are faulty = 1 - probabilty that 2 or less are faulty = 100 - 4.83 - 14.78 - 22.48 = 57.91%....
#37
I see what you're getting at now...silly me I didn't read the question right.

"schoolboy error!"

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