Question

Help me with 6 ii and iii pleasee! http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s06_qp_7.pdf

Answer 1

lol we just started s1

Answer 2

@LXelle

Answer 3

do you mean #5? Since #4 doesn't even have a part ii or iii, @LXelle

Answer 4

Oops, 6 ii and iii

Answer 5

Well you can model the number of faults in the dress by a Poisson random variable (from i). Say that is \(X\).

Now for the faulty belts, we can imagine this being a Bernoulli random variable \(Y\) where a faulty belt is \(Y=1\) with probability 0.03, and \(Y=0\) is a non-faulty belt with probability 1-0.03=0.97
So, ii) \(P(X=0 \cap Y=0)=P(X=0)P(Y=0)\) since we assume independence

iii) \(P(X\ge 1 \cap Y =1)=P(X\ge 1)P(Y=1)\) again by independence. But, it is much easier to use the complement to find \(P(X \ge 1)=1-P(X<1)=1-P(X=0)\) since a Poisson random variable starts its support at x=0

Answer 6

What?

Answer 7

which part is confusing you, and I can try to re-explain :)

Answer 8

no wait, tell me why is ii) e^-0.3 x 0.97?

Answer 9

Hm for the poisson probabilty part, I get \(e^{-0.6}\) , since I used \(\lambda=\dfrac{4.8}{20}, t=2.5\) So \(Poisson\left(\lambda t=\dfrac{4.8}{20}\cdot2.5=0.6\right)\)

Answer 10

i know where the 0.6 comes from or i wouldnt have done the first part. i wanna know why is it x 0.97

Answer 11

ah. Well the question says "neither the dress nor the belt is faulty" which would be translated as "Dress is not faulty, and Belt is not faulty", so I translated that as \(P(X=0 \cap Y=0)\) But this is equal to \(P(X=0)P(Y=0)\) by the independence stated in the problem. And, the probability of the belt not faulty is \(P(Y=0)=1-0.03=0.97\)

Answer 12

shouldn't it be e^-0.6 0.97^0/0!?

Answer 13

how can you divide by 0?!

Answer 14

I don't believe there is a Poisson assumption for the faulty belts. We don't know about its rate of occurrence. The best I can think of is modelling it with a Bernoulli random variable

Answer 15

its 0 factorial

Answer 16

but isn't 0! also 0?!

Answer 17

nevermind it's 1 XD

Answer 18

0! = 1

Answer 19

I thought it was written as whatever it is !? as in exclamation and question like HUH!

Answer 20

but how do we actually know whether to add in the poisson or not?

Answer 21

usukidoll or whatever, if youre trying to be a keyboard warrior just get out.

Answer 22

Well the faults in the dresses.. it specifies that you have a rate of occurrence for the faulty dresses. Usually you use the Poisson if an average rate of occurrence is given for a specified amount of time (or area, or volume...) and you are asked to find the number of events occurring in a certain time interval, or area, or volume.

Answer 23

so p(x=0) and (y=0) dont we have to times them?

Answer 24

yes indeed

Answer 25

huh?

Answer 26

youre not answering me

Answer 27

anyways moving on to iv

Answer 28

sorry had to restart my computer. You asked if we multiply \(P(X=0)\) and \(P(Y=0)\) correct? That is what I wrote above

Answer 29

yeah. what do we get when we multiply both?
can you help me out with iv please thanks

Answer 30

\[ \frac{e^{-0.6}\cdot 0.6^0}{0!}(0.97)\]

Answer 31

ohh got it. thanks.

Answer 32

iv?

Answer 33

You can view this [problem as a binomial type problem, in which the outcomes are Reject outfit/do not reject outfit... where the probability of rejecting an outfit is the same as what you found in iii). There are 300 trials.

So, if \(T\) is the number of rejected outfits, then \(T\) has a binomial distribution with \(n=300, p=0.97e^{-0.6}\). So \(P(T < 3)=P(T\le 2) = P(T=0)+P(T=1)+P(T=2).

But that is actually an exact value. The problem states to use a suitable approximation... In a lot of cases when n is large, the normal approximation to the binomial should be suitable. However, a Poisson approximation to the binomial is also suitable when n is large and p is small. It's not very clear which one the problem wants. But I suppose the normal approximation is the more common one to use.

Answer 34

So \(P(T < 3)=P(T\le 2) = P(T=0)+P(T=1)+P(T=2)\) **

Answer 35

okayy. ill get back to you again if i dont understand thanks

Answer 36

all right =]