EXERCICE 1

On interroge 296 habitants sur leur age et la nuisance sonore.

Donnees globales
 Classe   d age   ni    fi    Fi
 (18,30)        56   18.9   18.9
 (30,40)        73   24.7   43.6
 (40,50)        82   27.7   71.3
 (50,65)        85   28.7  100.0
 Total         296  100.0    --

1) Calcul de la mediane Me
   - On cherche la classe ou la FCC depasse 50.
     FCC a 30 = 43.6 < 50
     FCC a 50 = 71.3 >= 50
     => classe mediane = (40,50)
   - Interpolation lineaire dans (40,50) :
     borne_inferieure = 40
     amplitude_classe  = 10
     Favant = 43.6
     Fe = 50
     Fapres = 71.3
     Me = 40 + 10*(50 - 43.6)/(71.3 - 43.6)
        = 40 + 10*(6.4)/(27.7)
        = 40 + 10*0.2310
        = 40 + 2.31
        = 42.31
     => Me approx 42.3

2) Moyenne et ecart type pour les 116 habitants de nuisance FORTE

   Table reduite :
    Classe   ni   fi    ci
    (18,30)  19  16.4   24
    (30,40)  28  24.1   35
    (40,50)  33  28.4   45
    (50,65)  36  31.0  57.5
    Total   116 100.0   --

   a) Moyenne xbar
      somme_ni_ci = 19*24 + 28*35 + 33*45 + 36*57.5
                  = 456   + 980   + 1485  + 2070
                  = 4991
      xbar = somme_ni_ci / 116
           = 4991 / 116
           = 43.0

   b) Ecart-type sigma
      somme_ni_ci2 = 19*24^2 + 28*35^2 + 33*45^2 + 36*57.5^2
                   = 19*576  + 28*1225 + 33*2025 + 36*3306.25
                   = 10944   + 34300   + 66825   + 119025
                   = 230094
      moment2 = somme_ni_ci2 / 116
              = 230094 / 116
              = 1983.2
      variance = moment2 - xbar^2
               = 1983.2 - 43.0^2
               = 1983.2 - 1849
               =   134.2
      sigma = sqrt(variance)
            = sqrt(134.2)
            = 11.58
      => sigma approx 11.6

3) Estimateurs sans biais (n = 116)
   - Moyenne m_hat = xbar = 43.0
   - Variance S2 = (1/(116-1))*sum((xi - xbar)^2)
     Or on peut poser S2 = variance calculee ci-dessus = 134.2

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EXERCICE 2

Xi ~ Bernoulli(p=0.02), xbar_n = (1/n)*sum(Xi)

1) Loi et moments
   E(xbar_n) = p = 0.02
   Var(xbar_n) = p*(1-p)/n = 0.0196/n

2) Taille minimale n par TCL pour
   P(0.015 <= xbar_n <= 0.025) >= 0.99

   On pose Z = (xbar_n - p)/sqrt(p*(1-p)/n) ~ N(0,1)

   Condition centrale :
     0.015 <= xbar_n <= 0.025
     <=> (0.015 - 0.02)/sqrt(0.0196/n) <= Z <= (0.025 - 0.02)/sqrt(0.0196/n)
     Les deux bornes sont symetriques en valeur absolue = 0.005/sqrt(0.0196/n)
     Il faut 0.005/sqrt(0.0196/n) >= z_{0.995} = 2.58

   Calcul :
     0.005 / sqrt(0.0196/n) >= 2.58
     <=> 0.005*sqrt(n)/sqrt(0.0196) >= 2.58
     <=> sqrt(n) >= 2.58 * sqrt(0.0196) / 0.005
               = 2.58 * 0.14 / 0.005
               = 0.3612 / 0.005
               = 72.24
     <=> n >= 72.24^2 = 5218.6
     => n_min = 5219

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EXERCICE 3

Echantillon x1,...,xn de loi exponentielle param T>0

1) Fonction de vraisemblance L(T)
   L(T) = product_{i=1..n}[1/T * exp(-xi/T)]
        = T^{-n} * exp(- sum(xi)/T)

   Log-vraisemblance lnL(T)
   lnL = - n*ln(T) - (1/T)*sum(xi)

2) Derivees
   d(lnL)/dT = -n*(1/T) + (sum(xi))*(1/T^2)
             = -n/T + sum(xi)/T^2

   d2(lnL)/dT2 = n/T^2 - 2*sum(xi)/T^3

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EXERCICE 4

3) EMV T_hat_n
   On cherche T qui annule d(lnL)/dT
     d(lnL)/dT = n/T^2 *(-T + xbar)
     Signe change a T = xbar
     => EMV T_hat = xbar_n

4) Propriete de T_hat_n
   - Sans biais : E(T_hat) = E(xbar_n) = T
   - Convergent : xbar_n -> T (loi forte ou TCL)
   - Efficace :
       Information de Fisher I(T) = n/T^2
       Var(T_hat) = Var(xbar_n) = T^2/n = 1/I(T)