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

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 1.284 in currency A (to currency B) and standard deviation 0.013 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 more than 1.284 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 1.323 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 1.271 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 1.253 units of currency A or more than 1.317 units of currency A? ( Less than 1.253 is more unusual O More than 1.317 is more unusual

Explanation / Answer

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 1.284
standard Deviation ( sd )= 0.013
a.
probabilty that a randomly selected day during this period , a unit of currency B worth was more than 1.284 units of currency
P(X > 1.284) = (1.284-1.284)/0.013
= 0/0.013 = 0
= P ( Z >0) From Standard Normal Table
= 0.5
= 50%
b.
probabilty that a randomly selected day during this period , a unit of currency B worth was more than 1.323 units of currency
P(X > 1.323) = (1.323-1.284)/0.013
= 0.039/0.013 = 3
= P ( Z >3) From Standard Normal Table
= 0.0013
= 0.13%
c.
probabilty that a randomly selected day during this period , a unit of currency B worth was less than 1.271 units of currency
P(X < 1.271) = (1.271-1.284)/0.013
= -0.013/0.013= -1
= P ( Z <-1) From Standard Normal Table
= 0.2
=20%
d.
probabilty that a randomly selected day during this period , a unit of currency B worth was less than 1.253 units of currency
P(X < 1.253) = (1.253-1.284)/0.013
= -0.031/0.013= -2.3846
= P ( Z <-2.3846) From Standard Normal Table
= 0.0085 = 0.85%
probabilty that a randomly selected day during this period , a unit of currency B worth was more than 1.317 units of currency
P(X > 1.317) = (1.317-1.284)/0.013
= 0.033/0.013 = 2.5385
= P ( Z >2.5385) From Standard Normal Table
= 0.0056 = 0.56%
a unit of currency B worth was less than 1.253 units of currency is more unusual

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