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Math Help - Probability question

  1. #1
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    Probability question

    The diameter of an orange is normally distributed with mean 7cm and standard deviation 1cm. Using a series of sieves, we wish to sort the oranges into three classes. Class 1 should contain large oranges, class 2 should contain medium oranges, and class 3 should contain small oranges in the proportion of 20%, 50%, and 30%.

    Please see the attached image.

    Q1) How should we choose the gaps d1 and d2?
    Q2) If the gaps are appropriately choses, what is the probability that out of 10 oranges, 6 or more oranges fall into class 2?
    Attached Thumbnails Attached Thumbnails Probability question-q3-image-chapter-2-review.jpg  
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by woody198403 View Post
    The diameter of an orange is normally distributed with mean 7cm and standard deviation 1cm. Using a series of sieves, we wish to sort the oranges into three classes. Class 1 should contain large oranges, class 2 should contain medium oranges, and class 3 should contain small oranges in the proportion of 20%, 50%, and 30%.

    Please see the attached image.

    Q1) How should we choose the gaps d1 and d2?
    Q2) If the gaps are appropriately choses, what is the probability that out of 10 oranges, 6 or more oranges fall into class 2?
    a) Let d_1, d_2 be the desired diameters for class 1 and class 2 oranges, respectively, and X the diameter of a randomly selected orange (random variable).
    For d_1 we must have

    \begin{array}{lcll}<br />
\mathrm{P}(d_1\leq X) &=&0.2\\<br />
1-\mathrm{P}(X\leq d_1) &=& 0.2\\<br />
\mathrm{P}(X\leq d_1) &=& 0.8\\<br />
\mathrm{P}\left(\frac{X-7}{1}\leq \frac{d_1-7}{1}\right) &=& 0.8 &\big| \Phi^{-1}\\<br />
d_1-7 &=& \Phi^{-1}(0.8)\\<br />
d_1 &=& 7+\Phi^{-1}(0.8)\approx 7.84\mathrm{cm}<br />
\end{array}

    Similarly, for d_2 we must have that
    \begin{array}{lcll}<br />
\mathrm{P}(d_2\leq X) &=&0.2+0.5\\<br />
1-\mathrm{P}(X\leq d_2) &=& 0.7\\<br />
\mathrm{P}(X\leq d_2) &=& 0.3\\<br />
\mathrm{P}\left(\frac{X-7}{1}\leq \frac{d_2-7}{1}\right) &=& 0.3 &\big| \Phi^{-1}\\<br />
d_2-7 &=& \Phi^{-1}(0.3)\\<br />
d_2 &=& 7+\Phi^{-1}(0.3)\approx 6.45\mathrm{cm}<br />
\end{array}

    b) Let Y be the number of Class 2 oranges in a selection of n=10 oranges. Since the probability of a single, randomly selected orange to be in Class 2 is p=0.5 we get

    \mathrm{P}(6\leq Y) = \sum_{y=6}^{10}\binom{10}{y} 0.5^y\cdot (1-0.5)^{10-y}=\sum_{y=6}^{10}\binom{10}{y} 0.5^{10}\approx 0.38=38\%
    Last edited by Failure; August 8th 2009 at 08:27 PM.
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  3. #3
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    Thanks Failure,
    but I am a little confused with what you have written for part (a) for d2.

    Should it be
     1 - P(X \leq d_2) = 0.7
     P(X \leq d_2) = 0.3

    or am I not looking at it the right way?

    Also, can you please explain why you have used  \Phi^{-1} ?
    Isn't it the symbol for the pdf of a normal distribution?
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by woody198403 View Post
    Thanks Failure,
    but I am a little confused with what you have written for part (a) for d2.

    Should it be
     1 - P(X \leq d_2) = 0.7
     P(X \leq d_2) = 0.3

    or am I not looking at it the right way?
    No, you are absolutely right: that was a terrible typo - I am going to modify that answer right away...

    Also, can you please explain why you have used  \Phi^{-1} ?
    I had to get rid of \mathrm{P}(\ldots) on the left side of the equation in order to be able to isolate d_1-7 and d_2-7 respectively.
    Isn't it the symbol for the pdf of a normal distribution?
    Not quite. I use the definition \Phi(x) := \mathrm{P}(X\leq x) for a normally distributed random variable X with mean 0 and variance 1. So \Phi is the pdf (more precisely, the cumulative distribution function) of a normally distributed random variable with mean 0 and variance 1 ("standard normal cdf"). And \Phi^{-1} is simply the inverse of that (monotonously increasing) function \Phi:\mathbb{R}\to (0,1) (also called the quantile function of a normally distributed random variable). You get it by simply reading a table for \Phi the other way around...
    Last edited by Failure; August 8th 2009 at 11:09 PM.
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