SOLUTION: consider that the historical return of an investment follows a normal distribution with a mean of 2000 dollars and a standard deviation of 300 dollars. what is the lower limit for

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Question 1100887: consider that the historical return of an investment follows a normal distribution with a mean of 2000 dollars and a standard deviation of 300 dollars. what is the lower limit for which 40% of all investment return values are located given that the upper limit is at 2700 dollars?
What are the limits for which 91% of all values will be within?
The time required to prepare a certain specialty coffee at a local ship is uniformly distributed between 25 and 45 seconds. Assuming a customer just ordered one of these specialty coffee's. What percentage of these specialty coffee's will be prepared within 23 seconds?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
consider that the historical return of an investment follows a normal distribution with a mean of 2000 dollars and a standard deviation of 300 dollars. what is the lower limit for which 40% of all investment return values are located given that the upper limit is at 2700 dollars?
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Find the z-value of 2700::
z(2700) = (2700-2000)/300 = 7/3
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Find the left tail of z = 7/3
normalcdf(-100,7/3) = 0.99
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40% below 0.99 = 0.59
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Find the corresponding investment value
x = 0.59*300+2000 = $2177
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consider that the historical return of an investment follows a normal distribution with a mean of 2000 dollars and a standard deviation of 300 dollars.
What are the limits for which 91% of all values will be within?
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Half of 91% = 45.5%
Add and subtract 45.5% from 50% to get::
Lower limit:: 0.045
Upper limit:: 0.955
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Find the corresponding z-values:
Lower z-value:: invNorm(0.045) = -1.6954
Upper z-value:: +1.6954
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Find the corresponding investment values::
Lower:: -1.6954*300 + 2000
Upper:: +1.6954*300 + 2000
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The time required to prepare a certain specialty coffee at a local shop is uniformly distributed between 25 and 45 seconds. Assuming a customer just ordered one of these specialty coffee's. What percentage of these specialty coffee's will be prepared within 23 seconds?
mean = 25+(45-25)/2 = 35
std = (1/6)(45-25) = 20/6 = 3 1/3
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Find the z-value of 23:: z(23) = (23-35)/(3 1/3) = -3.6
Find invNorm(-3.6) = 0.000159 = 0.0159%
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Cheers,
Stan H.
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