Question 1182236: Q.1 Global financial institution transfers a large data file every evening form offices around the world to its London headquarters. Once the file is received, it must be cleaned and partitioned before being stored in the company’s data warehouse. Each file is the same size and the time required to transfer, clean, and partition a file is normally distributed, with a mean of 1.5 hours and a standard deviation of 15 minutes.
a. if one file is selected at random, what is the probability that it will take longer than 1 hour and 55 minutes to transfer, clean, and partition the file?
b. If a manager must be present until 85% of the files are transferred, cleaned, and partitioned, how long will the manager need to be there?
c. What percentage of the data files will take between 63 minutes and 110 minutes to be transferred, cleaned, and partitioned?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 1.5 hr is 90 minutes mean
sd is 15 minutes
z=(x-mean)/sd
=(115-90)/15 since 1hr55min is 115 min.
25/15=1.67
prob z>1.67is 0.0478
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z(0.85)=1.036. Use 2ndVARS3invnorm(0.85,0,1) ENTER
1.036=(x-90)/15
15.54=x-90
x=105.54 min
or about 1h45 min
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z=(63-90)/5=-1.8
z=(110-90)/15=1.33
probability z is between those two is 0.8729
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also 2ndVARS2normalcdf(63,110,90,15)ENTER =0.8729 or 87.29%
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