Solutions to Practice Exam 2 - Solutions to selected...

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Unformatted text preview: Solutions to selected problems in Practice Exam 2 (Modified) 1. (i) {O, B, W} (ii) Outcomes O B W Probabilities 3/15 7/15 5/15 k 0 1 2 3 P(X=k) .2917 .525 .1750 .00833 (iii) .324 2. (ii) (iii) E(X)=µ=.89999 (iv) σ=.7 3. (ii) k 0 1 2 3 (iii) E(X)=µ=.9 (iv) σ=.794 4. (i) .2 (ii) .3 (iii) .6, Not independent, .3, .375, .75, .4167 P(X=k) .343 .441 .189 .027 5. (ii) Outcomes (in dollars) 2 11 Probabilities 9/11 2/11 (iii) E(X)= $3.64 6. Sample Space: {28, 30, 31} Outcomes 28 30 31 7. (ii) E(X)=4.472 (iii) σ= 1.404 8. (i) 2/3 (ii) ¼ (iii) 35/36 (iv) 2/3 Probabilities 1/12 4/12 7/12 (v) 4/27 9. E(X) = $1.38 10. (ii) Sample Space: {{R,R}; {R,W}; {R,B}; {W,W}; {W,B}} Outcomes R,R R, W R, B W, W W, B (iii) 2/3 Probabilities 1/5 2/5 1/5 1/15 2/15 11. (ii) (a) z=‐.67 (b) .2499 or .2525 Depending on whether you use the table [Average of A(‐.70) and A(‐.65)] or use cdf(lower limit, upper limit, µ, σ) 12. Gerald is better. Z‐score for Eleanor is 1.75, for Gerald is 1.833. 13. (i) P(X<10)=A(‐1.714) ≈ A(‐1.7) = .0446 (ii) .0808 14. (i) .2417 (ii) 47 (iii) .0797 (iv) 4.99 ...
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