SOLUTION: The time spent waiting in the line is approximately normally distributed. The mean waiting time is 4 minutes and the variance of the waiting time is 1. Find the probability that

Algebra ->  Probability-and-statistics -> SOLUTION: The time spent waiting in the line is approximately normally distributed. The mean waiting time is 4 minutes and the variance of the waiting time is 1. Find the probability that      Log On


   

Question 1204473: The time spent waiting in the line is approximately normally distributed. The mean waiting time is 4 minutes
and the variance of the waiting time is 1. Find the probability that a person will wait for more than 5
minutes. Round your answer to four decimal places.
Answer by ikleyn(52814) About Me  (Show Source):
You can put this solution on YOUR website!
The time spent waiting in the line is approximately normally distributed. The mean waiting time is 4
minutes and the variance of the waiting time is 1. Find the probability that a person will wait
for more than 5 minutes. Round your answer to four decimal places.
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This probability is the area under the specified normal curve on the right of the z-score z= 5 minutes.


To find the value, use the standard function normalcdf on your regular calculator TI-83/84.

The format is

    P = normalcdf(z1, z2, mean, SD)


In this problem, take z1 = 5 minute,  z2 = 9999 (as infinity); mean = 4 minutes (given);  SD = 1 = sqrt%28variance%29.


You will get  P = normalcdf(5, 9999, 4, 1) = 0.1587  (rounded).

Solved.

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Alternatively, you can use free of charge online calculator at this web-page

https://davidmlane.com/hyperstat/z_table.html

Use it in the mode "Area from a value".