2006-Fall-Midterm_s

# 2006-Fall-Midterm_s - More PastPaper...

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Unformatted text preview: More PastPaper: http://ihome.ust.hk/~cs_gxx Solution Mid06 1. 2.(a) The uncertainty before an experiment causes the variation of outcomes after the experiment. Probability measures the uncertainty before the experiments and thus it can explain the variation after the experiment. (b) P(higher for the year | higher first week)= 29/34=0.853. (c) P(Higher for the year)= 39/54=0.722 not equal 0.853, therefore dependent. (d) H L FH 28 2 P(H | FH)= 28/30 = 0.933 FL 11 13 3. (a) P(X>1.6) = P(Z> (1.6 – 1.2)/0.3)=P(Z>1.33)=0.0918 (b) 1.2 ± 0.67(0.3) =(1.0, 1.4) (c) m+0.44sd=1.8 m-0.44sd=1.2 ⇒ m=1.5, sd=0.68. (d) Q3 = 2.3, Q1 = 2.3 – 2*(2.3 – 1.6) = 0.9, inter-quartile range = 2.3 – 0.9 = 1.4 ⎛7⎞ 4. (a) ⎜ ⎟(0.3) 4 (0.7) 3 = 0.097 ⎜ 4⎟ ⎝⎠ ⎛ 6⎞ ⎛ 6⎞ ⎛6⎞ ⎛ 6⎞ (b) ⎜ ⎟(0.3) 3 (0.7) 3 + ⎜ ⎟(0.3) 4 (0.7) 2 + ⎜ ⎟(0.3) 5 0.7 + ⎜ ⎟(0.3) 6 = 0.2557 ⎜ 3⎟ ⎜ 4⎟ ⎜5⎟ ⎜ 6⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎟⎜ ⎟ ⎟ 5.(a) Let X be the performance rating and Y be your fee, then Y = 120000 ⋅ X E ( X ) = 1 ⋅ 0.1 + 2 ⋅ 0.3 + 3 ⋅ 0.3 + 4 ⋅ 0.2 + 5 ⋅ 0.1 = 2.9 , E (Y ) = 120000 × 2.9 = 348000 E ( X 2 ) = 1(0.1) + 4(0.3) + 9(0.3) + 16(0.2) + 25(0.1) = 9.7 var( X ) = 9.7 − (2.9) 2 = 1.29 , var(Y ) = (1.2) 2 × 1010 × 1.29 = 1.858 × 1010 (b) E (Y ) = 300000 + 100000 × 0.3 = 330,000 var(Y ) = 1010 × [3 2 ⋅ 0.7 + 4 2 ⋅ 0.3 − (3.3) 2 ] = 0.21 × 1010 ht tp :// ih om e. us t.h k/ ~c s_ var(Y ) = 4 × 1010 ⋅ var( X ) = 4 × 1.29 × 1010 = 5.16 × 1010 (d) Choose option 1. If he faces similar situations repeatedly, then on the average you get the expectation. Thus in the long run it is best to choose the option with highest expectation. gx x/ (c) E (Y ) = 750,000 − E (5 − X ) × 200,000 = 750,000 − 2.1 × 200,000 = 330,000 ...
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