SOLUTIONS FOR THE STATISTICS 3083 FINAL OF DECEMBER 1996
 QUESTION 1
            Arthritis  Arthritis    Total
                            Free                P[Arthritis | Positive]
   Positive      0.12      0.034    0.154            0.12 / 0.154
   Negative      0.03      0.816    0.846
   Total         0.15      0.850    1.000
 QUESTION 2

     t               1      2      3      4      5        SUM
     f(t)           0.15   0.18   0.27   0.22   0.18      1
     F(t)           0.15   0.33   0.6    0.82   1         ...   Cumulative
     f(t)*t         0.15   0.36   0.81   0.88   0.9       3.1  = E(T)
     D = t-3.1     -2.1   -1.1   -0.1    0.9    1.9       ...
     f(t)*D*D       0.6615 0.2178 0.0027 0.1782 0.6498    1.71 = V(T)

 (a)                                               E(T)=3.1    V(T)=1.71
 (b)  (0.27+0.22+0.18)       / (0.18+0.27+0.22+0.18)         =   67 / 85
 (c)  (2*0.18+3*0.27+4*0.22+5*0.18) / (0.18+0.27+0.22+0.18)  =  277 / 85
 (d)                 See the "Cumulative" row of the calculations
 QUESTION 3

  Presumably we are to assume
     (i)  The 50% rate applies every night.
     (ii) Customers are statistically independent.
           -- This may not, in fact, be true, for example, will there
           -- be a heavier demand for sets on nights of the Grey Cup?

    S = # of sets      0     1      2      3     4     5       SUM
    Probability       1/32  5/32  10/32  10/32  5/32  1/32      1
    Rental fee         0     2      4      6     6     6
    Prob*Rental        0   10/32  40/32  60/32  30/32 6/32     146/32

 (a) P[S  > 3 ]                     5/32 + 1/32 =  6 / 32  (evenings)
 (b) E( Rental )                         146/32 =   4.5625 (dollars)
 QUESTION 4

 (a)  c = 1/2

 (b)  F(y) =   0                      for y < 0
           =   1                      for u > 1
           =   y - (y*y/4)            otherwise

 (c)  F(1)/F(1.5) = 0.75/0.9375 =  0.8

 (d)  This is integral from 0 to 1.5 of  (0.5)(2-y) y dy  = 0.5625
      divided by                                  F(1.5)  = 0.9375
      to get the quotient                                   0.6

 (e)  Hazard rate:   f(y)     /    [1-F(y)]
                     (1-y/2)  /  [1 - y + y*y/4]  =  1 /[1 - y/2]
 QUESTION 5

 (a)  The cumulative function is        F(X) = 1-exp(-X/11)
      To find the half life, solve      F(X) = 1/2
                                        X    = 11 ln(2)

 (b)  The exponential distribution having no memory, this is the
      6 plus the unconditional expected life, i.e.    6 + 11 = 17

 (c)  Using again the memoryless property, this equals the proba-
      bility that a new component lasts 5 hours or more which is
                       1 - F(5) =  exp(-5/11)
 QUESTION 6

 (a)                                    z = (100-120)/30  = -2/3
                                        P[Z > -2/3]       =   0.252

 (b)  Inversely look up 0.57 in a table of the
      normal distribution to get                        z =   0.18
 QUESTION 7

 The hazard function is the derivative of the negative (natural) log
 complimentary cumulative function,  1-F.   Integrate  6t to get the
 the negative logarithm of 1-F(t)

           -ln (1-F(t)) =  3t2 + constant

            1-F(t)      =  [another const] exp(-3t2)

 The constant is chosen to make 1-F(0)=1 and thus is one.

            f(t)        =  6 t exp(-3t2)

This is a Weibull distribution, with
              alpha = 3,        beta = 2 .
Expectation and variance can be found in textbook table 4.8

                               E{t} = sqrt(1/3) G(3/2)
                               V{t} = (1/3)G(2) - (E{t})2
 Note:
     G is used in place of upper case Gamma
     which is not available in HTML.
 QUESTION 8

 (a) X and Y are not independent.  On inspection the body of the
     table is seen not to be the product of its marginal sums.

 (b) The sum of these numbers:        .   .   .    .
                                      .   .  0.12 0.05
                                      .   .  0.05 0.00
                                      .   .  0.08 0.00   which is 0.3
 (c)   E(X) = 4.1
       E(Y) = 2.3
       E(XY)= 9.24
                                  Cov  =  9.24 - 4.1 2.3   =   -0.19

 (d)  (5.76+3.2+5.4+0+15.36+0)/(0.12+0.05+0.05+0+0.08+0)   = 29.72 / 0.3

 (e)  The sum of these numbers:    0.12 0.08 0.15  .     which is 0.42
                                   0.07   .    .   .
                                     .    .    .   .
                                     .    .    .   .
 QUESTION 9
      Use the central limit theorem to approximate a probability:

      E{ W }     =  40*36 + 40*30 = 2640
      V{ W }     =  40*25 + 40*25 = 2000
                              std = 44.72
       z = (2600-2640) / 44.72    = -0.894
                                                 Probability  =  0.186
 QUESTION 10
   W is chi square with 90 degrees of freedom.  Near the middle
   the chi square distrtibution with many degrees of freedom is
   well approximated by a normal distribution with the same mean
   and variance.

   E{W} = 90      V{W}  = 180

   (a)  The normal distribution being symmetric, the probability is 1/2.

   (b)                    std{W}                 = 13.42
                          z = (100-90)/ 13.42    = -0.745
                                                   Probability  =  0.77