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The task is: Typist printed 1000 pages of text, and made 140 errors. What is the probability that a randomly chosen page contains zero errors? one? two? The error distribution is described with Poisson law.

Using Poisson's law, I got that $P(m=0)=0.86,$ while the correct answer is 0.79.
I tried to use ML-theorem, because $pnq=140*0.86 >> 20$ and $n >> 50.$ So under my $\exp$ function I get an argument like $-70$ and $e^{-70} \approx 4E-30,$ and so on. But its highly illogical, that the random page would contain mistake about for sure and its far enough from answer. Where am i wrong; What's the real way?

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  • $\begingroup$ Who says 0.79? $ $ $\endgroup$
    – Did
    Feb 24 '16 at 12:38
  • $\begingroup$ The book does(and so my teacher). I was sure its $exp(-0.14)~=1-0.14=0.86$. But where am i wrong with ML,way? $\endgroup$ Feb 24 '16 at 12:41
  • $\begingroup$ Which book? Do they explain why? (No, $e^{-0.14}$ is not exactly $1-0.14$, but close.) $\endgroup$
    – Did
    Feb 24 '16 at 12:51
  • $\begingroup$ Not, just an answer. I just never used TeX's ~ but nvm, exp(-0.14)>0.79 for sure. $\endgroup$ Feb 24 '16 at 12:57
  • $\begingroup$ There probability is at least $1-0.14$ - under any distribution law, incl. at most one error per page. $\endgroup$
    – A.S.
    Feb 24 '16 at 14:00
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From available data the Poisson mean is estimated as $\hat \lambda = 140/1000 = 0.14.$ The formula you should use for the probability of various numbers $X$ per randomly chosen page is $P(X = i) = e^{-\lambda}\lambda^i/i!.$

Here is a brief table of the PDF (to five places) of $Pois(.14)$ from R software, showing $P(X = 0) = e^{-0.14} = 0.86936,$ and so on.

 lam = 140/1000  # mean number of errors per page
 i = 0:6;  pdf = round(dpois(i, lam), 5)
 cbind(i, pdf)
   ## i     pdf
   ## 0 0.86936
   ## 1 0.12171
   ## 2 0.00852
   ## 3 0.00040
   ## 4 0.00001
   ## 5 0.00000
   ## 6 0.00000
 sum(pdf)     # sum of first few terms of infinite series
 ## 1       
 sum(i*pdf)   # sum of first few terms of infinite series
 ## 0.13999   # sum of entire series would be E(X) = 0.14
 exp(-0.14)
 ## 0.8693582 # check P(X = 0) with formula

enter image description here

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  • $\begingroup$ ok, but if i would use ML theore, what answeer should i get? $\endgroup$ Feb 25 '16 at 8:25
  • $\begingroup$ Norm approx to Pois: $P(X = 0) = P(X < .5) \approx P(Z <(.5-1.4)/\sqrt{1.4}) \approx ..83.$ But not an excellent approx., and why approximate when exact is so easy? $\endgroup$
    – BruceET
    Feb 25 '16 at 16:15
  • $\begingroup$ i cant get what does $\sqrt{1.4}$ comes from. Isn't the pois law exactly exp(-0.14) for $P(I=0)=0.8693..$ Im totally confused. $\endgroup$ Feb 27 '16 at 10:08
  • $\begingroup$ For $X \sim Pois(\lambda)$ we have $E(X) = V(X) = \lambda$ so $SD(X) = \sqrt{\lambda}.$ So I made a typing mistake. To standardize for normal approximation (ML), it would be $P(Z < (.5 - .14)/\sqrt{.14}) \approx .83.$ Sorry for the confusion. $\endgroup$
    – BruceET
    Feb 27 '16 at 11:35
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$Comment:$ The approximation $e^{-.14} \approx 1 - .14$ is mentioned in a previous comment. The idea that $e^x \approx 1 + x,$ for $x$ near $0,$ is based on the first two terms of the Taylor (Maclaurin) expansion of $e^x$:

$$e^x = \sum_{i=0}^\infty x^i/i!.$$

For several values of $x$ near $0,$ the following table compares $x$ with $S_2 = 1 + x,\;$ $S_3 = 1 + x + x^2/2,$ and the sum $S_5$ of the first five terms of the series.

 x = seq(-.2, .2, by=.06)
 s.2 = 1 + x;  s.3 = 1 + x + x^2/2
 s.5 = 1 + x + x^2/2 + x^3/6 + x^4/24
 cbind(x, s.2, s.3, s.5, exp(x))
 ##     x  s.2    s.3       s.5       exp
 ## -0.20 0.80 0.8200 0.8187333 0.8187308
 ## -0.14 0.86 0.8698 0.8693587 0.8693582  <- Your x
 ## -0.08 0.92 0.9232 0.9231164 0.9231163
 ## -0.02 0.98 0.9802 0.9801987 0.9801987
 ##  0.04 1.04 1.0408 1.0408108 1.0408108
 ##  0.10 1.10 1.1050 1.1051708 1.1051709
 ##  0.16 1.16 1.1728 1.1735100 1.1735109

The plot below shows the two-term linear approximation (s.2) in red and the three-term quadratic approximation (s.3) in dashed blue. Your value $x = -.14$ is emphasized as a vertical green line. The linear approximation may not be as good at $x = -.14$ as you might like. At the resolution of the graph, the quadratic approximations are not distinguishable from the exact values (thin black curve).

enter image description here

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  • $\begingroup$ thats clear for me. I do understand how Taylors works, i used linear approx because the linear is less for sure, than the real exp. But linear gives me .86 which is higher enough than .79. Thanks for graphics anyway. $\endgroup$ Feb 27 '16 at 10:16
  • $\begingroup$ (1) I think .79 must just be a mistake. Wouldn't be the first wrong answer given to a problem. (2) In reality $\hat \lambda = .14$ is only an estimate based on looking at 1000 pgs. Maybe it's 95% CI is something like $.14 \pm .02.$ Then $P(X=0)$ could be anywhere in $(.848,.888)$, but I don't think you were expected to go as far as looking at confidence intervals, so I didn't include them in my answer.(I've never seen CIs used in a problem at this elementary level.) $\endgroup$
    – BruceET
    Feb 27 '16 at 11:15
  • $\begingroup$ can u help with dealing De Moivre–Laplace Th? For this $p$, this aprox of binomial one is more suitable, isnt it? $\endgroup$ Feb 27 '16 at 11:32
  • $\begingroup$ I don't see what binomial distribution you're using. (What is $n$ and what is $p$?) In comment above I mentioned normal approx to Poisson (instead of normal approx to binomial). Here $\lambda$ is really too small for good Poisson approx to normal. $\endgroup$
    – BruceET
    Feb 27 '16 at 11:45
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    $\begingroup$ Thank you, im just trying to find any way to get the book's answers. Seems like there is just a mistake. Thanks for time, ill try to think it over. $\endgroup$ Feb 27 '16 at 12:19

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