What is the length of time to complete a door assembly on an automobile factory?
FeFe asked:
The length of time to complete a door assembly on an automobile factory assembly line is normally distributed with a mean MU=7.9 minutes and standard deviation δ = 1.9 minutes. For a door selected at random, what is the probability the assembly line time will be
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The length of time to complete a door assembly on an automobile factory assembly line is normally distributed with a mean MU=7.9 minutes and standard deviation δ = 1.9 minutes. For a door selected at random, what is the probability the assembly line time will be
a.) 7 minutes or less?
b.) 10 minutes or less?
c.) between 7 and 10 minutes?
Please explain and Thank you so much!
March 8th, 2010 at 5:42 am
A) P(T < 7) = P(Z < (7-7.9)/1.9)
P( Z < -.4736842105)
Area below Z(-.4736842105) = .317863 = probability.
B) P(T < 10) P(Z < (10-7.9)/1.9)
P( Z < 1.105263158)
Area below Z(1.105263158) = .865477 = probability.
C) Between 7 and 10 minutes = P(10) - P(7)
= .865477 - .317863 = 0.547614 Josh S
March 10th, 2010 at 4:04 pm
The probability density function for the normal distribution is
e^(-(x-m)^2/2δ^2)
———————
δ Sqrt(2Pi)
We are given m=7.9 and δ=1.9 so we can plug those in and get
0.209e^(-0.139(-7.9+ x)^2)
a) to find the probability the time will be 7 min or less we integrate our function from - Infinity to 7 which gives us 0.318
b) integrate our function from -Infinity to 10 which is 0.865
c) integrate from 7 to 10 which gives us 0.865 - 0.318 = 0.548 Anthony R