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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 probabi… Read More

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.

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.

Options

## All Comments

(37) Jump to unreadPost a commentPart i) is worth 10 marks.

Part ii) is worth 20 marks.

Its what is the probabilities that 0 and then 3 or more are available...

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.

have fun!

1) Probability is UNLIKELY but not impossible!

2) Probability is LIKELY but not certain!

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.

have fun!

yes I agree with this :w00t: ........... rather you than me !!

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..

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

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 distributionyour statistical tables and look for where λ (lambda, np) = 3.For (i), you just look at:

λ (lambda) = 3

x = 0

Answer: 0.0498For (ii), you just look at

λ (lambda) = 3once again.BUT, because it is now THREE OR

MOREyou 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: 0.5768Good luck in your exam!!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!!!!

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.

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).

No, it's not. :p

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..

Heh, I thought I was but you have me wondering. I'd have stuck with poisson throughout...

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.

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:

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: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

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!

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 "

" then your are correct, but it doesn't.how manyphones would you expect to be defective?It asks "what is the

probabilitythat 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:

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)...

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:

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"

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

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...

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%....

"schoolboy error!"