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The daily exchange rates for the five-year period 2003 to 2008 between currency

ID: 3066412 • Letter: T

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 1.044 in currency A (to currency B) and standard deviation 0.043 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 1.044 units of currency A?

The probability is ----------%. (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 1.130 units of currency A? The probability is ---------%. (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 0.915 units of currency A? The probability is ---------%. (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 0.945 units of currency A or more than 1.114 units of currency A?

a. Less than 0.945 is more unusual.

b. More than 1.114 is more unusual.

Explanation / Answer

mean = 1.044,s = 0.043

a)
P(X <1.044)
z = (x -mean)/s
= (1.044 -1.044)/0.043
= 0
P(X <1.044) P(z <0) = 0.5 = 50% by using z standard normal table

b)

P(X >1.130)
z = (x -mean)/s
= (1.130 -1.044)/0.043
= 2
P(X >1.130) P(z > 2) = 0.0228= 2.28% by using z standard normal table

c)

P(X < 0.915)
z = (x -mean)/s
= (0.915 -1.044)/0.043
= -3
P(X <0.915) P(z <-3) = 0.0013= 0.13% by using z standard normal table

d)
P(X < 0.945)
z = (x -mean)/s
= (0.945 -1.044)/0.043
= -2.3023

P(X <0.945) P(z <-2.3023) = 0.0107= 1.07% by using z standard normal table

P(X >1.114)
z = (x -mean)/s
= (1.114 -1.044)/0.043
= 1.6279

P(X >1.114) P(z > 1.6279) = 0.0518= 5.18% by using z standard normal table

a. Less than 0.945 is more unusual.

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