CAN SOMONE PLEASE EXPLAIN THIS AND BREAK IT DOWN FOR ME. I AM CONFUSED. QUESTION:

1. You observe that the value of Range Resources stock is 75, and that a 9 month call option on Range at 73 sells for 10.476. The interest rate is 3%. The initial (first) M-K estimate of the standard deviation of this stock is 36.34%.

A: (10 points) Using the Black-Sholes formula, determine if this first estimate is too high or too low.

B: (10 points) Using the M-K update formula, determine the 2^nd estimate of the standard deviation of this stock.

ANSWER:

UPDATE: Commenting Standard dev doesnt really help me . I need help solving sigma, pt(t), d1, d2, N(d), N(d2), call, difference in call values

B-S value S K r T 0.5996 3.1416 2.5066 0.0495 1.0508 64.9519 0.0243 sigma pt(t) Present value strike update 75.000 difference in call values 73.000 pi 0.030 sqrt(2*pi) 0.750 d^2/2 0.363 exp(d^2/2) 0.978 S*sqrt(T) 71.376 1st term*2nd term/3rd term subtract from previous estimate yields new 1.051 estimate of std 0.050 0.315 0.315 0.000 0.6235 0.500 46.764 35.688 11.076 S/pt(t)*X In (S/pt(t*X) sig*sqrt(time) d= d2= N(d)=(delta)= N(D2) S*N(d) P(t)*X*N(D2) Call= price is too high, sigma is too high difference in call values 33.91% 0.5996 B-S value S K r T 0.5996 3.1416 2.5066 0.0495 1.0508 64.9519 0.0243 sigma pt(t) Present value strike update 75.000 difference in call values 73.000 pi 0.030 sqrt(2*pi) 0.750 d^2/2 0.363 exp(d^2/2) 0.978 S*sqrt(T) 71.376 1st term*2nd term/3rd term subtract from previous estimate yields new 1.051 estimate of std 0.050 0.315 0.315 0.000 0.6235 0.500 46.764 35.688 11.076 S/pt(t)*X In (S/pt(t*X) sig*sqrt(time) d= d2= N(d)=(delta)= N(D2) S*N(d) P(t)*X*N(D2) Call= price is too high, sigma is too high difference in call values 33.91% 0.5996