Huge Multi Part Prob + Stats question from past paper

In summary, the probability that a randomly sampled page typed by secretary B contains more than 1 misprint is 0.25. The overall proportion of pages produced by the department that contain no misprints is approximately 75%. If a randomly sampled page is found to contain 2 misprints, the probability that this page was typed by secretary A is 0.75.
  • #1
hb2325
20
0
A university department has 2 secretaries (labelled A and B) who do all the word processing
required by the department. The number of misprints on a randomly sampled
page typed by secretary i (i=A,B) has a Poisson distribution with Mean Ui independent
of the number of misprints on any other page, where Mean Ua = 0.3 and Mean Ub = 1.2.


Assume that each page is typed entirely by a single secretary. Of the typing required
by the department, 75% is done by secretary A and 25% by secretary B.


(a) Find the probability that a randomly sampled page typed by secretary B contains
more than 1 misprint. [2 marks]


(b) Calculate the overall proportion of pages produced by the department that contain
no misprints. [3 marks]


(c) Suppose that a randomly sampled page produced by the department is found to
contain 2 misprints. Given this information, calculate the probability that this
page was typed by secretary A, Hence which secretary is most likely to have typed
the page concerned? [5 marks]


A book typed entirely by secretary A consists of 200 pages.
(i) Let X be the number of pages in the book that contain no misprints. Name
the (exact) distribution of X. Find approximately the probability that at
least 150 pages in the book are without misprints. [5 marks]


(ii) Find approximately the probability that the book contains at most 50 misprints
in total.


Attempt At soultions:

a) Using P(x=K) = (e^-u* u^k)/k! , I get the probability for X=0, X=1 add them and subtract from 1?

b) 0.25* Prob X=0 from part 1 + same thing for Sec A * 0. 75

c) Conditional prob. I am happy with this one.

d) i) I think this is a normal distribution.

Now if its a normal distribution, mean number of errors per page = 0.3 so number of errors in 200 page will be 60.

But how do I find The prob of 150+151...200 :S I am very confused and would love some help as my exam is on wednesday!

Thanks!
 
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  • #2
hb2325 said:
A university department has 2 secretaries (labelled A and B) who do all the word processing
required by the department. The number of misprints on a randomly sampled
page typed by secretary i (i=A,B) has a Poisson distribution with Mean Ui independent
of the number of misprints on any other page, where Mean Ua = 0.3 and Mean Ub = 1.2.


Assume that each page is typed entirely by a single secretary. Of the typing required
by the department, 75% is done by secretary A and 25% by secretary B.


(a) Find the probability that a randomly sampled page typed by secretary B contains
more than 1 misprint. [2 marks]


(b) Calculate the overall proportion of pages produced by the department that contain
no misprints. [3 marks]


(c) Suppose that a randomly sampled page produced by the department is found to
contain 2 misprints. Given this information, calculate the probability that this
page was typed by secretary A, Hence which secretary is most likely to have typed
the page concerned? [5 marks]


A book typed entirely by secretary A consists of 200 pages.
(i) Let X be the number of pages in the book that contain no misprints. Name
the (exact) distribution of X. Find approximately the probability that at
least 150 pages in the book are without misprints. [5 marks]


(ii) Find approximately the probability that the book contains at most 50 misprints
in total.


Attempt At soultions:

a) Using P(x=K) = (e^-u* u^k)/k! , I get the probability for X=0, X=1 add them and subtract from 1?

b) 0.25* Prob X=0 from part 1 + same thing for Sec A * 0. 75

c) Conditional prob. I am happy with this one.

d) i) I think this is a normal distribution.

Now if its a normal distribution, mean number of errors per page = 0.3 so number of errors in 200 page will be 60.

But how do I find The prob of 150+151...200 :S I am very confused and would love some help as my exam is on wednesday!

Thanks!

For (d): let p = probability that a page has no misprints (same p for all pages). The number of non-misprint pages in an n-page book is the number of "successes" in n independent trials with success probability p per trial. Can you name the corresponding distribution? In (d) you need to compute p (how?), and n = 200, so using the exact distribution might be difficult (especially on an exam without computer access), so using a good approximation would be the way to go. So much for part (i). For part (ii), can you identify the distribution of the total number of misprints in a 200-page book? The total number of misprints equals the number on page 1 plus the number on page 2 plus ... plus the number on page 200, and these numbers are all independent random variables with the same distribution Poisson(0.3). Again, the exact distribution might be hard to use in an exam situation, so a sensible approximation should be applied.

RGV
 
  • #3
Wait would it be a bernoulli distribution?
 
  • #4
Ok so I use the probability of X=0 for sec A that I found in part a of the question, that's p of a success, and I use the normal approximation of a bernoulli distribution?

But I am unsure of how to use the normal dist. to find these values? Please help me my exam is tomorrow :(
 
  • #5
hb2325 said:
Ok so I use the probability of X=0 for sec A that I found in part a of the question, that's p of a success, and I use the normal approximation of a bernoulli distribution?

But I am unsure of how to use the normal dist. to find these values? Please help me my exam is tomorrow :(

If you are talking about question (i), then yes, it would be a binomial distribution. So, what are the mean and variance of the number of error-free pages? (There are standard formulas---just use them.) Of course, you would use a normal approximation with exactly the same mean and variance. Then just use the standard methods that you have learned (or were supposed to have learned) how to apply.

RGV
 

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