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 1.466 in currency A​ (to currency​ B) and standard deviation 0.041 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 is the probability that on a randomly selected day during this​ period, a unit of currency B was worth less than 1.466 units of currency​ A? The probability is nothing​%. ​(Type an integer or a​ decimal.) ​b) What is the probability that on a randomly selected day during this​ period, a unit of currency B was worth more than 1.548 units of currency​ A? The probability is nothing​%. ​(Type an integer or a​ decimal.) ​c) What is the probability that on a randomly selected day during this​ period, a unit of currency B was worth less than 1.343 units of currency​ A? The probability is nothing​%. ​(Type an integer or a​ decimal.) ​d) Which would be more​ unusual, a day on which a unit of currency B was worth less than 1.388 units of currency A or more than 1.533 units of currency​ A? More than 1.533 is more unusual. Less than 1.388 is more unusual.

Answer 1

Edutan's Gruen data mean u = 1.466 0.041 Standard deuration coloul a) P(x 21.408) :- PCxcluso) =P ( e 1466 – louso ) =P (zco) = 0.5 b) p(x > 1.548) - p(x>1.sus) = 1-p (xel.su8 ) = 1-p(*l. S48 – 1.466) si-pif222) : 1-0.97725 = 0.0227 c) P(x: I. 343): P(x< 11343) = p (komen < 1,343 = 1.466 ) =? (22-3) 0.001 1.343 -1.6166 0-001 0.0013 d) less than 1. 388 p(* 21.388) - (Rolf 1.358 . 1.466 ) = p (72-10976) 0.041 more than 1.533 :- p(x>1.533 ) - | (x4.5 .4667 =P(231.6341) oooul

Since abs (-1.0976) > 1.6341 less than 1.388 is more unasual.