Question

The daily exchange rates for the​ five-year period 2003 to 2008 between currency A and currency B are well modeled by a normal distribution with mean 1.303 in currency A​ (to currency​ B) and standard deviation 0.036 in currency A. Given this​ model, and using the​ 68-95-99.7 rule to approximate the probabilities rather than using technology to find the values more​ precisely, complete parts​ (a) through​ (d).

​a) What would the cutoff rate be that would separate the lowest 0.15% of currency​ A/currency B​ rates?

b) What would the cutoff rate be that would seperate the highest 50%

c) What would the cutoff rate be that would seperate the middle 95%

d What would the cutoff rate be for the upper cut off?

e0 What would the cutoff rate be for the lowest 2.5%

Answer 1

solution: el = s = 1.303 0.036 of observations are 3.sd below a) Bs lowest 0.15% meam = le-36 = 1.303 - 3x0.036 = 1.195 / above b) Highest sol. of observations of mean - 11-303 c) Middle 95%. a) observation are between 2.sd of mean = le-20 and et ar = 1.303 - 2x 0.036 and 1.303 + 2x 0.036 1.231 and 11:375 1.231 1. 375 d) Problem in complete. e lowest 2.5% of observation are below 2.5d of meon li-20 = 1.303 - 2x0-0361 - 11.231