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Thread: [SOLVED] Normal Distribution

  1. #1
    Member looi76's Avatar
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    [SOLVED] Normal Distribution

    Question:
    The length of sweetpea flower stems are normally distrusted with mean $\displaystyle 18.2 cm$ and standard deviation $\displaystyle 2.3 cm$.
    (a) Find the probability that the length of a flower stem is between $\displaystyle 16 cm$ and $\displaystyle 20 cm$.
    (b) $\displaystyle 12\%$ of the flower stems are longer than h cm. $\displaystyle 20\%$ of the flower stems are shorter than k cm. Find $\displaystyle h$ and $\displaystyle k$.
    (c) Stem lengths less than $\displaystyle 14 cm$ are unacceptable at a florist's shop. In a batch of $\displaystyle 500$ sweetpeas estimate how many would be unacceptable.


    Attempt:

    $\displaystyle \mu = 18.2$ , $\displaystyle \sigma = 2.3$

    $\displaystyle (a)$$\displaystyle P( 16 < X < 20 )$

    $\displaystyle P \left( \frac{(16 -18.20}{2.3} < \frac{X - \mu}{\sigma} < \frac{(20-18.2)}{2.3} \right)$

    $\displaystyle P( -0.957 < Z < 0.783)$
    $\displaystyle = \phi(0.783) - ( 1 - \phi(0.8289 + 0.0018))$
    $\displaystyle = 0.6139$

    Answer in textbook is 0.614

    $\displaystyle (b)$ Don't know how to start, need help!
    Attached Thumbnails Attached Thumbnails [SOLVED] Normal Distribution-normal_distrubtion_table.gif  
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:
    The length of sweetpea flower stems are normally distrusted with mean $\displaystyle 18.2 cm$ and standard deviation $\displaystyle 2.3 cm$.
    (a) Find the probability that the length of a flower stem is between $\displaystyle 16 cm$ and $\displaystyle 20 cm$.
    (b) $\displaystyle 12\%$ of the flower stems are longer than h cm. $\displaystyle 20\%$ of the flower stems are shorter than k cm. Find $\displaystyle h$ and $\displaystyle k$.
    (c) Stem lengths less than $\displaystyle 14 cm$ are unacceptable at a florist's shop. In a batch of $\displaystyle 500$ sweetpeas estimate how many would be unacceptable.

    Attempt:

    $\displaystyle \mu = 18.2$ , $\displaystyle \sigma = 2.3$

    $\displaystyle (a)$$\displaystyle P( 16 < X < 20 )$

    $\displaystyle P \left( \frac{(16 -18.20}{2.3} < \frac{X - \mu}{\sigma} < \frac{(20-18.2)}{2.3} \right)$

    $\displaystyle P( -0.957 < Z < 0.783)$
    $\displaystyle = \phi(0.783) - ( 1 - \phi(0.8289 + 0.0018))$
    $\displaystyle = 0.6139$

    Answer in textbook is 0.614

    $\displaystyle (b)$ Don't know how to start, need help!
    Well if your answer was off by .0001 I think you can safely assume you are correct
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  3. #3
    Member looi76's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Well if your answer was off by .0001 I think you can safely assume you are correct
    I need help in the second step!... I don't know how to start
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Hi

    You want to find $\displaystyle h=\phi^{-1}(88\%)$ which is exactly what you had to do in another exercise involving the life time of an object. (substitute $\displaystyle u=\frac{x-\mu}{\sigma}$ in the integral and solve)
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  5. #5
    Member looi76's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    You want to find $\displaystyle h=\phi^{-1}(88\%)$ which is exactly what you had to do in another exercise involving the life time of an object. (substitute $\displaystyle u=\frac{x-\mu}{\sigma}$ in the integral and solve)
    So is $\displaystyle \phi^{-1}(88\%) = .8800$
    $\displaystyle \phi = 8106$

    Is this step right?
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  6. #6
    Super Member flyingsquirrel's Avatar
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    I don't understand why you picked up .8106 instead of .8810 ?

    Having done the substitution, we are looking for $\displaystyle h$ such that $\displaystyle \int_{-\infty}^{\frac{h-\mu}{\sigma}} \frac{1}{\sqrt{2\pi}}\exp{-u^2}\,\mathrm{d}u=0.88$ and the table gives $\displaystyle \frac{h-\mu}{\sigma}=1.18$ for 0.8810, doesn't it ?
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  7. #7
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    Quote Originally Posted by looi76 View Post
    So is $\displaystyle \phi^{-1}(88\%) = .8800$
    $\displaystyle \phi = 8106$

    Is this step right?
    $\displaystyle z = \frac{x - \mu}{\sigma}$.

    Pr(X > h) = 0.12 => Pr(X < h) = 0.88.
    Use your tables to find a value $\displaystyle z_{0.88}$ of z corresponding to $\displaystyle \Pr(z < z_{0.88}) = 0.88$: $\displaystyle z_{0.88} = 1.175$. Therefore h = .......

    Pr(X < k) = 0.2 => Pr(X > -k) = 0.2 => Pr(X < -k) = 0.8.
    Use your tables to find a value $\displaystyle z_{0.8}$ of z corresponding to $\displaystyle \Pr(z < z_{0.8}) = 0.8$: $\displaystyle z_{0.8} = 0.842$. Therefore k = .......
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  8. #8
    Member looi76's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    I don't understand why you picked up .8106 instead of .8810 ?

    Having done the substitution, we are looking for $\displaystyle h$ such that $\displaystyle \int_{-\infty}^{\frac{h-\mu}{\sigma}} \frac{1}{\sqrt{2\pi}}\exp{-u^2}\,\mathrm{d}u=0.88$ and the table gives $\displaystyle \frac{h-\mu}{\sigma}=1.18$ for 0.8810, doesn't it ?
    Quote Originally Posted by mr fantastic View Post
    $\displaystyle z = \frac{x - \mu}{\sigma}$.

    Pr(X > h) = 0.12 => Pr(X < h) = 0.88.
    Use your tables to find a value $\displaystyle z_{0.88}$ of z corresponding to $\displaystyle \Pr(z < z_{0.88}) = 0.88$: $\displaystyle z_{0.88} = 1.175$. Therefore h = .......

    Pr(X < k) = 0.2 => Pr(X > -k) = 0.2 => Pr(X < -k) = 0.8.
    Use your tables to find a value $\displaystyle z_{0.8}$ of z corresponding to $\displaystyle \Pr(z < z_{0.8}) = 0.8$: $\displaystyle z_{0.8} = 0.842$. Therefore k = .......
    Thanks flyingsquirrel and mr fantastic

    $\displaystyle = P( h \geq 12\% )$
    $\displaystyle = P( h \geq -88\% )$████$\displaystyle 12\% - 100\% = -88\%$
    $\displaystyle = \phi^{-1}(88\%)$
    $\displaystyle = \phi^{-1}(0.8800)$
    $\displaystyle = 1.175$

    $\displaystyle = P( k \leq 20\% )$
    $\displaystyle = P( k \leq -80\% )$████$\displaystyle 20\% - 100\% = -80\%$
    $\displaystyle = \phi^{-1}(-80\%)$
    $\displaystyle = \phi^{-1}(1- 0.8000)$
    $\displaystyle = 1 - 0.842$
    $\displaystyle = 0.158$

    In textbook answer for h is 20.9 and k is 16.3 ?
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  9. #9
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by looi76 View Post
    $\displaystyle P( h \geq 12\% )$
    What does this mean ? The probability that the length $\displaystyle h$ is greater than 12% ? Writing $\displaystyle P(X\geq h)=12\%$ would be OK : "the probability that the length of a stem is greater than $\displaystyle h$ equals 12%"

    You'd better written what Mr Fantastic suggested :
    $\displaystyle P(X>h)=12\%$ hence $\displaystyle P(x<h)=100\%-12\%=88\%=0.88$
    $\displaystyle \frac{h-\mu}{\sigma}=\phi^{-1}(88\%)=1.18$
    Whence $\displaystyle h=1.18\times2.3+18.2=20.9\,\mathrm{cm}$

    I let you redo the second one accordingly.
    In textbook answer for h is 20.9 and k is 16.3 ?
    Yes...
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    $\displaystyle z = \frac{x - \mu}{\sigma}$.

    Pr(X > h) = 0.12 => Pr(X < h) = 0.88.
    Use your tables to find a value $\displaystyle z_{0.88}$ of z corresponding to $\displaystyle \Pr(z < z_{0.88}) = 0.88$: $\displaystyle z_{0.88} = 1.175$. Therefore h = .......

    Mr F continues: $\displaystyle {\color{red}z = \frac{x - \mu}{\sigma} \Rightarrow 1.175 = \frac{h - 18.2}{2.3} \Rightarrow h = 20.90}$ (correct to two decimal places).

    You should check that Pr(X > 20.9) = 0.12 ......

    Pr(X < k) = 0.2 => Pr(X > -k) = 0.2 => Pr(X < -k) = 0.8.
    Use your tables to find a value $\displaystyle z_{0.8}$ of z corresponding to $\displaystyle \Pr(z < z_{0.8}) = 0.8$: $\displaystyle z_{0.8} = 0.842$. Therefore k = .......
    The other is done in the same way.
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