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Math Help - normal distribution

  1. #1
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    normal distribution

    Hi all,

    I've solved the question but I'm not sure whether or not my answer is correct. I need your help.

    Thanks in advance.



    The length of time it takes to be seated at a local restaurant on Friday night is normally distributed with a mean of 15 minutes and a standard deviation of 4.75 minutes.


    a. What is the probability that you have to wait more than 20 minutes to be seated?
    b. What is the probability that you have to wait between 13 and 16 minutes to be seated?

    a)

    x= length of time

    μ= 15
    σ= 4.75
    P(x>20)

    For x=20, z=(20-15)/4.75 -> 1.052
    P(x>15)= P(z> 1.052) = [total area] - [area to the left of z = 1]
    1- 0.8536 = 0.1464

    b)
    x= length of time
    μ = 15
    σ= 4.75

    P(13<x<16) For x=13, z= (13-15)/4.75 -> -0.421
    For x=16, z= 16-15/4.75 -> 0.210
    P(13<x16) = P(-0.412<z< 0.210)
    0.416-0.340= 0.076


    Attached Thumbnails Attached Thumbnails normal distribution-2-3-4-5.png  
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  2. #2
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    Re: normal distribution

    Quote Originally Posted by Liliann View Post
    Hi all,

    I've solved the question but I'm not sure whether or not my answer is correct. I need your help.

    Thanks in advance.



    The length of time it takes to be seated at a local restaurant on Friday night is normally distributed with a mean of 15 minutes and a standard deviation of 4.75 minutes.


    a. What is the probability that you have to wait more than 20 minutes to be seated?
    b. What is the probability that you have to wait between 13 and 16 minutes to be seated?

    a)

    x= length of time

    μ= 15
    σ= 4.75
    P(x>20)

    For x=20, z=(20-15)/4.75 -> 1.052
    P(x>15)= P(z> 1.052) = [total area] - [area to the left of z = 1]
    1- 0.8536 = 0.1464


    This is correct


    b)
    x= length of time
    μ = 15
    σ= 4.75

    P(13<x<16) For x=13, z= (13-15)/4.75 -> -0.421
    For x=16, z= 16-15/4.75 -> 0.210
    P(13<x16) = P(-0.412<z< 0.210)
    0.416-0.340= 0.076


    This is incorrect. You correctly computed the z-scores but did not convert those into probabilities correctly.
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  3. #3
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    Re: normal distribution

    Quote Originally Posted by romsek View Post
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    This is correct



    This is incorrect. You correctly computed the z-scores but did not convert those into probabilities correctly.
    Thanks a lot.

    0.416+0.340 = 0.756 is correct?
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  4. #4
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    Re: normal distribution

    Quote Originally Posted by Liliann View Post
    Thanks a lot.

    0.416+0.340 = 0.756 is correct?
    why would you add?

    The numbers I get are

    $pr[16]\approx 0.583372$

    $pr[13]\approx 0.336858$ you seem to have gotten this value correct

    the probability you are after is the difference of these two numbers

    $pr[16]-pr[13]\approx 0.246513$
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  5. #5
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    Re: normal distribution

    Quote Originally Posted by romsek View Post
    why would you add?

    The numbers I get are

    $pr[16]\approx 0.583372$

    $pr[13]\approx 0.336858$ you seem to have gotten this value correct

    the probability you are after is the difference of these two numbers

    $pr[16]-pr[13]\approx 0.246513$
    Thanks again.
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  6. #6
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    Re: normal distribution

    Pr [X>20]=Pr[N(0,1)>20-15/4.75]=Pr[N(0,1)>1.0526]=1-.8537=.1463
    Pr[X<16]-Pr[X<13]=Pr[N(0,1)<16-15/4.75]-Pr[N(0,1)<13-15/4.75]=
    Pr[N(0,1)<.2105]-Pr[N(0,1)<-.4211]=.5834-.3368=.2466
    Last edited by RShnay; June 6th 2014 at 06:53 AM.
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  7. #7
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    Re: normal distribution

    Quote Originally Posted by RShnay View Post
    Pr [X>20]=Pr[N(0,1)>20-15/4.75]=Pr[N(0,1)>1.0526]=1-.8537=.1463
    Pr[X<16]-Pr[X<13]=Pr[N(0,1)<16-15/4.75]-Pr[N(0,1)<13-15/4.75]=
    Pr[N(0,1)<.2105]-Pr[N(0,1)<-.4211]=.5834-.3368=.2466
    Thank you very much.
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