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# fm9 3 - 9-5 d1 = ln(P/X[rRF 2 2]t t = ln(\$30/\$35[0.05(0.25...

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Answers and Solutions: 9 - 3 9-5 . 3319 . 0 33333 . 0 5 . 0 ) 333333 . 0 )]( 2 / 25 . 0 ( 05 . 0 [ ) 35 /\$ 30 (\$ ln t σ )]t 2 / 2 ( σ RF [r (P/X) ln 1 d = + + = + + = d 2 = d 1 σ (t) 0.5 = -0.3319 – 0.5(0.33333) 0.5 = -0.6206. N(d 1 ) = 0.3700 (from Excel NORMSDIST function). N(d 2 ) = 0.2674 (from Excel NORMSDIST function). V = P[N(d 1 )] - t r RF Xe [N(d 2 )] = \$30(0.3700) - \$35e (-0.05)(0.33333) (0.2674) = \$11.1000 - \$9.2043 = \$1.8957 \$1.90. 9-6 The stock’s range of payoffs in one year is \$26 - \$16 = \$10. At expiration, the option will be worth \$26 - \$21 = \$5 if the stock price is \$26, and zero if the stock price \$16. The range of payoffs for the stock option is \$5 – 0 = \$5.
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