Solutions SM230 Makeup Final exam Spring 1997 (8 May 97)

 

  1. Oil change
    2.

     

    No Oil

    Oil

     

    Brakes

    0.34

    0.06

    0.40

    Muffler

    0.36

    0.24

    0.60

     

    0.70

    0.30

     
    1. See above
    2. 0.16
    3. 0.06/0.40=0.15
    4. Prob(Oil|Muffler)=0.24/0.6=0.40 > Prob(Oil) = 0.30, so positive corr.
  2. Searching
    3.

     

    A

    B

    C

     

    Detect

    0.28

    0.20

    0.08

    0.28

    Not

    0.12

    0.20

    0.12

    0.72

     

    0.4

    0.4

    0.2

     
    1. Search A. 0.28 is highest probability of detection
      2.

       

      A

      B

      C

       

      Detect

      0.1167

      0.2778

      0.1111

       

      Not

       

       

       

       

       

      0.12/0.72

      0.4/0.72

      0.2/0.72

       
    2. Search B because 0.2778 is highest
  3. CSTS
    1. 1-BCDF(p=0.20,N=20,X=4)=0.37035
    2. BCDF(X=7)-BCDF(X=2)=0.7618
  4. O reps
    1. BCDF(p=0.25,N=20,X=3)=0.2252
    2. Solve BCDF(p=0.25,N,X=3)=0.05 -> N=29
  5. Watch bill
    1. 1-HCDF(G=35,B=25,S=30,X=29)=0
    2. HCDF(X=14)-HCDF(X=3)=0.0577
    3. HCDF(X=20)-HCDF(X=9)=0.94229
    4. HCDF(X=20)=0.9423
  6. Cargo pallets
    1. 1-PCDF(L=1/5,T=2,X=0)=0.3297
    2. 1-PCDF(L=1/5,T=2,X=1)=0.0616
  7. Redundancy
    1. 1-PCDF(L=1/4,T=3,X=0)=0.5276
    2. 1-BCDF(p=0.5276,N=10,X=2)=0.9622
  8. pdf
    1. mean=1.1666
    2. E(X^2)=1.6667, StdDev=0.5528
  9. Mail
    1. Mean=40+15+35, StdDev=7.071
    2. 32-48 mins
    3. Mean=55, StdDev=5.831, About 36-60 mins
    4. Virtually none
  10. Computer replacement
    1. 1-ZCDF(M=220,S=13.27,X=245.5)=0.0273
    2. Solve ZCDF(220,13.27,X)=0.95, -> X=242
    3. 1-ZCDF(M=220,S=13.27,X=200.5)=0.9291
  11. Clearing charges
    1. 24
    2. 4.85
    3. ZCDF(X=20.5)=0.2353
    4. 1-ZCDF(X=30.5)=0.0901
  12. Y=X^2
    1. 4/3
    2. E(Y^2)=16/5, StdDev=1.193
    3. 0+4/3