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Math Help - Probability calculations using normal distribution.

  1. #1
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    Probability calculations using normal distribution.

    The life-time of a popular computer flatscreen is known to follow a normal distribution with a mean μ = 2000 hours and standard deviation σ = 200 hours. If a computer flatscreen is selected at random, find the probability that it will last for:
    Less than 2500 hours.
    More than 1900 hours.
    Between 2000 and 2400 hours.


    need a head start
    can someone guide me tru this problem
    Last edited by mr fantastic; April 6th 2011 at 12:35 AM. Reason: Re-titled.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    For example

    p(x\leq 2500)=p(z\leq (2500-2000)/200)=\ldots

    Now, use the table of the standardised normal distribution N(0,1) .
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  3. #3
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    Quote Originally Posted by math321 View Post
    The life-time of a popular computer flatscreen is known to follow a normal distribution with a mean μ = 2000 hours and standard deviation σ = 200 hours. If a computer flatscreen is selected at random, find the probability that it will last for:
    Less than 2500 hours.
    More than 1900 hours.
    Between 2000 and 2400 hours.


    need a head start
    can someone guide me tru this problem
    If you need further help, please show your working and say where you are stuck. And since this type of question is so routine, you would clearly benefit from reviewing examples from your class notes and textbook.
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    so after i get to


     p(z\leq 2.5
    wat do i do from there

    got 0.4938 on the table for 2.5
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    Quote Originally Posted by math321 View Post
    so after i get to


     p(z\leq 2.5
    wat do i do from there

    got 0.4938 on the table for 2.5
    This is clearly wrong. You should know that Pr(Z < 0) = 0.5, so obviously Pr(Z < 2.5) must be greater than 0.5. You need to get one-on-one help from your instructor on how to use the tables you are using. It is the same sort of question as the ones you posted here: http://www.mathhelpforum.com/math-he...on-176941.html.
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    can u just break down the question into steps so i can understand it better

    if that is wrong wat do i do from there
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    Quote Originally Posted by math321 View Post
    can u just break down the question into steps so i can understand it better

    if that is wrong wat do i do from there
    Remember one thing, the table values are for
    P(Z>z) where z is your specific Z value.
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    stuck here

    Z = (2500 - 2000) / 200

    z = 2.5
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  9. #9
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    is it this
    P( X < 2500)
    = P(z < 2.5)
    = .9953
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by math321 View Post
    is it this
    P( X < 2500)
    = P(z < 2.5)
    = .9953

    No, it is .9938 because p(z\leq a)=p(z<a) .
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  11. #11
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    and for the 2nd part of the question
    z = (-100 / 200)

    z = -0.5
    =P(Z > -.5)
    =P(Z < .5)

    = .6915

    1- .6915
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  12. #12
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by math321 View Post
    =P(Z > -.5)
    =P(Z < .5)
    = .6915
    1- .6915

    No, it is

    p(z>-0.5)=p(z<0.5)=0.6915
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  13. #13
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    Quote Originally Posted by math321 View Post
    can u just break down the question into steps so i can understand it better

    if that is wrong wat do i do from there
    You are asking the same sort of questions every time in all your threads and getting given the same help. It is time you went back and reviewed the examples in your textbook and class notes and got one-on-one help in person from someone (eg. your instructor).

    Thread closed.
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