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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 1.094 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 1.094 units of currency​ A? The probability is nothing​%. ​(Type an integer or a​ decimal.)

User AdrianBR
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1 Answer

3 votes

Answer:

50% probability that on a randomly selected day during this​ period, a unit of currency B was worth more than 1.094 units of currency​ A.

Explanation:

The Empirical Rule(68-95-99.7 rule) states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 1.094

Standard deviation = 0.013

a) What is the probability that on a randomly selected day during this​ period, a unit of currency B was worth more than 1.094 units of currency​ A?

The normal distribution is symmetric, which means that 50% of the units of currency B are more than 1.094 of currency A and 50% are below.

So

50% probability that on a randomly selected day during this​ period, a unit of currency B was worth more than 1.094 units of currency​ A.

User Mukul
by
5.8k points
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