Question

Expert Q8A Done 5. Suppose that a researcher approximates the density of the CAD/USD exchange rate in January 2018 by z~N(0.75, 0.024) (so the mean is 0.75 and the standard deviation is 0.024, measured in US dollars). (a) Find the probability of observing an exchange rate less than 0.762. (b) Find the probability of observing an exchange rate between 0.75 and 0.78. (c) Find a value m such that 0.75 + m > z > 0.75-m 95% of the tine. (d) Find a value m such that 0.75 + m > z > 0.75-m 99% of the time.
I don’t understand what c and d are asking, please answer these.

Answer 1

basically you need to find confidence interval here and it will help in calculating m.

We have, mean = u = 0.75, $\sigma$= 0.024

c) 95% confidence interval z score = 1.96

Lower limit of confidence interval = u - z*$\sigma$

= 0.75 - 1.96*0.024

= 0.75 - 0.04704

Upper limit of confidence interval = u + z*$\sigma$

= 0.75 + 1.96*0.024

= 0.75 + 0.04704

Thus, (0.75 - 0.04704) < u < (0.75 + 0.04704) for 95% of times

Thus, m = 0.04704

d)

99% confidence interval z score = 2.58

Lower limit of confidence interval = u - z*$\sigma$

= 0.75 - 2.58*0.024

= 0.75 - 0.06192

Upper limit of confidence interval = u + z*$\sigma$

= 0.75 + 2.58*0.024

= 0.75 + 0.06192

Thus, (0.75 - 0.06192) < u < (0.75 + 0.06192) for 99% of times

Thus, m = 0.06192