Around De Moivre–Laplace theorem/Poisson law

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?

• Who says 0.79? 
– Did
Feb 24 '16 at 12:38
• 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? Feb 24 '16 at 12:41
• Which book? Do they explain why? (No, $e^{-0.14}$ is not exactly $1-0.14$, but close.)
– Did
Feb 24 '16 at 12:51
• Not, just an answer. I just never used TeX's ~ but nvm, exp(-0.14)>0.79 for sure. Feb 24 '16 at 12:57
• There probability is at least $1-0.14$ - under any distribution law, incl. at most one error per page.
– A.S.
Feb 24 '16 at 14:00

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 • ok, but if i would use ML theore, what answeer should i get? Feb 25 '16 at 8:25
• 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? Feb 25 '16 at 16:15
• 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. Feb 27 '16 at 10:08
• 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. Feb 27 '16 at 11:35

$Comment:$ 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). • 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. Feb 27 '16 at 10:16
• (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.) Feb 27 '16 at 11:15
• can u help with dealing De Moivre–Laplace Th? For this $p$, this aprox of binomial one is more suitable, isnt it? Feb 27 '16 at 11:32
• 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. Feb 27 '16 at 11:45
• 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. Feb 27 '16 at 12:19