Question 373749: The length of time to complete a door assembly on an automobile factory assembly line is normally distributed with mean 6.7 minutes and standard deviation 2.2 minutes. For a door selected at random, what is the probability the assembly line time will be between 5 and 10 minutes? Place your answer, rounded to 4 decimal places, in the blank.
***So I figure z1=5 minutes and z2=10 minutes. I need to figure out both separately first. So Z1=5-6.7/2.2 which = 0.7727. Then I figured out
z2=10-6.7/2.2 which = 1.5. So I know z1=0.7727 and z2=1.5... But I do not know how to proceed from here?**
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! You're basically finding the area under the curve of a standard normal distribution from -0.7727 to 1.5
So either use a table or a calculator to find that area.
Preferably, a calculator will do. If you have a TI83, then simply hit "2nd" then "Vars" to get into the "Distribution" menu (which is for stats). Then find "normalcdf" and select that function.
So on your home screen, you should have "normalcdf(" (without quotes)
From there, type in -0.7727,1.5 (your boundaries). You can then optionally include the mean 0 and the standard deviation 1, but it's not necessary. Finally, close off the expression.
So you should have: normalcdf(-0.7727,1.5)
You can also type in : normalcdf(-0.7727,1.5,0,1) , but it's not necessary.
This then gives you
normalcdf(-0.7727,1.5) = 0.71334
Note: For more info on the "normcdf" function, check out this page
So the area under the standard normal curve form z=-0.7727 to z=1.5 is approximately 0.71334
So the probability that the assembly line time will be between 5 and 10 minutes is about 0.71334 (which is a 71.334% chance)
Alternatively, you can use the normal tables to give you the area. However, keep in mind that you're going to need to subtract.
If you need more help, email me at jim_thompson5910@hotmail.com
Also, feel free to check out my tutoring website
Jim
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