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          Mon, 30 May 2005 04:47:07 PDT
Date: Mon, 30 May 2005 04:47:07 -0700 (PDT)
From: Vlado Had <vlado_had@yahoo.com>
Subject: Statistika
To: 2mi@st.fmph.uniba.sk
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 Tu je priklad zo skusky Statistika. Nather povedal,
ze jeden priklad (poistovne) bude na najblizsom
buducom termine, tak prikladam aj ten.

1)
 Pan X chodi do prace metrom. Metro chodi pravidelne
kazde 3 minuty. Aka je pravdepodobnost, ze za
23prac.dni bude cakat pan X menej nez 80 min.

R: f = 1/3 0<x<3 
      {
       0


E(x) = integral (x * f(x))dx = 3/2
D(xi) = E(x^2) - |E(X)|^2 = 3/4 

celkova doba cakania = X, potom X = suma (i=1, i<=46,
xi) = 46
E(xi) = 3/2 * 46 = 69
D(x) = suma(i=1, i<=46, D(xi)) = 3/4 * 46 = 34,5
P(x < 80) = P(n <  (80 - n*E(x))/sqrt(D(x))) = P(n <
(80-69)/sqrt(34,5)) = ... = P(n < 1,87) = theta(1,87)
= 0,97

2)
 Poistovna poistuje 1000 ludi rovnakeho veku. Poistne
je 120 korun. V pripade umrtia dostane rodina 5000
korun. Pravdepodobnost, ze clovek pocas roka umrie je
0,008. Aka je P, ze poistovna bude stratova?

R: P( |p-p^| >= epsilon) = 1 - sigma(a)

n = 1000
p = 0,008
a = (epsilon * sqrt(n)) / (sqrt(p * (1-p)))

kolko ludi musi umriet aby poistovna nebola stratova?
1000*120 - n*5000 > 0 => n < 24
p^ = 24/1000
|p - p^| = epsilon => epsilon = 0,016
uz mame vsetky hodnoty, aby sme vyratali 'a'
po vyratani 'a' vypocitame sigma(a) a nasledne
1-sigma(a)

ps mali by to byt spravne riesenia. 1. priklad mam od
jednej kolegyne z mf a 2. je od Rada.

Vlado Had


		
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