Comments on: What is the length of time to complete a door assembly on an automobile factory? http://www.cargearusa.com/blog/what-is-the-length-of-time-to-complete-a-door-assembly-on-an-automobile-factory/ Thu, 09 Feb 2012 19:24:33 +0000 http://wordpress.org/?v=2.7 hourly 1 By: Anthony R http://www.cargearusa.com/blog/what-is-the-length-of-time-to-complete-a-door-assembly-on-an-automobile-factory/comment-page-1/#comment-8996 Anthony R Wed, 10 Mar 2010 23:04:36 +0000 http://www.cargearusa.com/blog/what-is-the-length-of-time-to-complete-a-door-assembly-on-an-automobile-factory/#comment-8996 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<a href="http://www.safedebthelp.com/consolidate-debt-help.htm"> Anthony R</a> 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

]]> By: Josh S http://www.cargearusa.com/blog/what-is-the-length-of-time-to-complete-a-door-assembly-on-an-automobile-factory/comment-page-1/#comment-8995 Josh S Mon, 08 Mar 2010 12:42:27 +0000 http://www.cargearusa.com/blog/what-is-the-length-of-time-to-complete-a-door-assembly-on-an-automobile-factory/#comment-8995 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<a href="http://www.resumeminers.com/Resume/Submit-Resume.htm"> Josh S</a> 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

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