Question 204448: An analysis of public emergency response times to calls made in a small city indicates that these responses are normally distributed with a mean μ = 240 seconds and a standard deviation σ = 40 seconds.
a. what is the probability that a particular call lasted between 180 and 300 seconds?
b. what is the probability that a particular call lasted between 110 and 180 seconds?
c. what is the probability that a particular call lasted less than 180 seconds?
d. what is the probability that a particular call lasted more than 280 seconds?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! An analysis of public emergency response times to calls made in a small city indicates that these responses are normally distributed with a mean μ = 240 seconds and a standard deviation σ = 40 seconds.
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For each of these problems you need to find the z-values corresponding to
the numbers, then find the probability relating to those z-values.
For example on part a see below:
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a. what is the probability that a particular call lasted between 180 and 300 seconds?
P(180<=x<-300) = P(z(180)<= z <= z(300)) = normalcdf(180,300,240,40) = 0.8664
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b. what is the probability that a particular call lasted between 110 and 180 seconds?
Use the same procedure as on "a"
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c. what is the probability that a particular call lasted less than 180 seconds?
Use normalcdf(0,180,240,40) = 0.0668
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d. what is the probability that a particular call lasted more than 280 seconds?
Use normalcdf(280,1000,240,40) = 0.1587
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Comment: I used a TI calculator to get the probabilities.
Let me know if this feedback was of no help.
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Cheers,
Stan H
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